２次元可解量子系のエンタングルメント特性

２次元可解量量子系のエンタングルメント特性田中宗 (京大基研) Valence-‐‑‒Bond-‐‑‒Solid (VBS) 状態や Quantum hard-‐‑‒square モデルの基底状態についてエンタングルメント特性を検討したその結果２次元系のエンタングルメント特性と１次元量量子系の間に対応関係があることを見見出した

VBS on symmetric graphs J Phys A 43 255303 (2010) ldquoVBSCFT correspondencerdquo Phys Rev B 84 245128 (2011) Quantum hard-square model Phys Rev A 86 032326 (2012) Nested entanglement entropy Interdisciplinary Information Sciences 19 101 (2013)

共同研究者の方々

審査希望分野

No

1

申請研究課題

物理学Ⅱ

K1412007

注入同期レーザー量子計算機の高効率化及び汎用性の向上を志向した理論研究

田村

亮

姓

論文での使用姓(該当者のみ)

名

ミドルネーム

氏名

性別 5 国籍男

生年月日年月日

独立行政法人物質材料研究機構

若手国際研究センター

ポスドク研究員

日月年博士号取得(見込み)年月日

大学院名学位名

センター名

研究室名

所属長

受入希望研究室

電話番号

審査結果通知先Email

創発物性科学研究センター

量子情報エレクトロニクス部門量子光学研究グループ

山本喜久

tamuraryonimsgojp

2

3

4

6

7

8

9

10

11

12

日本

現職機関名(大学院名)

所属部署(学科名)

職　　名(課程名年次)

1984 303

2012 3 22

平成27年度採用基礎科学特別研究員採用申請書

特記事項

登録番号

080-6801-9078

東京大学大学院理学系研究科博士（理学）

桂法称 (東大院理理)

Anatol N Kirillov (RIMS 京大)

Vladimir E Korepin (YITP Stony Brook USA)

川島直輝 (東大物性研)

Lou Jie (Fudan Univ 中国)

田村亮亮 (NIMS)

ダイジェスト

正方格子VBS状態ラダー上量量子格子気体模型

排除体積効果

注目する物理理系 VBS状態

エンタングルメント特性

正方格子 1D AF Heisenberg

蜂の巣格子 1D F Heisenberg

注目する物理理系量量子格子気体模型


正方形ラダー 2D Ising

三角形ラダー 2D 3-state Potts

1次元量量子系物理理特性 (臨臨界現象)

２次元量量子系エンタングルメント特性

蜂の巣格子VBS状態

JPA 43 255303 (2010) PRB 84 245128 (2011) PRA 86 032326 (2012) Interdisciplinary Information Sciences 19 101 (2013)

導入エンタングルメント

目的

数学的準備

5

FIG 5 (Color online) Tensor network structure with a singleER level and a top tensor for six coarse-grained sites Bigsolid circles denote positions of sites on the coarse-grainedlattice The top tensor is put on the parallelogram frameAll parallelogram frames are the same by a skew periodicarrangement

for the infinite lattice This type of MERA is called afinite-correlation MERA22 because reduced correlationsbecome exactly zero for large distances The distancelimit for finite reduced correlations is roughly the sizeof the unit cell The main dicrarrerence between the twoschemes lies in the causal cones In the PBC schemeall the causal cones are limited to one unit cell becauseof the PBC However in the infinite-size scheme somecausal cones extend into multiple unit cells Thus thecomputational time for the infinite-size scheme may belonger than that for the PBC scheme

We assume that all tensors are independent within theunit cell to have more variational freedom with less biasWe optimized them to minimize the total energy of thetensor network states The tensors are iteratively up-dated by the singular value decomposition method22 Al-though this minimization problem may have some localminimum states we obtained stable results starting fromrandom initial tensors

2 Energy

Our ER tensor network has the spatial structure shownin Fig 3 As we mentioned above if we use the fi-nite dimension of the tensor index the network-structurebias may cause a significant systematic error To checkwhether or not this is the case we try several sets ofdimensions of tensor indices Figure 6 shows local en-ergies on finite and infinite lattices for various tensorsizes The value of the local energy on a triangle pla-quette defined by nearest neighbor sites i j and kSi middot Sj + Sj middot Sk + Sk middot Si is shown on a color scaleIn the 120 state which we believe is the ground statein the present case the local energy is homogeneousTherefore we expect that the whole system should beuniformly colored if the error of the calculation is su-ciently small Figures 6(a) and 6(c) show the results of

FIG 6 (color online) Local energies on the triangular plaque-ttes The value of the local energy on a triangle plaquette de-fined by nearest neighbor sites i j and k SimiddotSj+Sj middotSk+SkmiddotSiis shown by the color scale The bottom bar shows the cor-respondence between the value of local energy and color Re-sults of the PBC scheme are shown in (a) and (b) Those ofthe infinite-size scheme are shown in (c) and (d) The size oftensors in (a) and (c) is (123) = (2 2 2) In (b) and(d) (123) = (2 8 8)

small tensor size (123) = (2 2 2) for PBC andinfinite-size schemes of N = 114 respectively There areclear spatial patterns resulting from the structure of ERin both cases Increasing the tensor size should improvethe quality of the tensor network states In fact as shownin Figs 6(b) and 6(d) the patterns are clearly more ho-mogeneous than for small tensors We notice that thepatterns in the infinite-size scheme are quite similar tothose of the PBC scheme Thus the assumption of directproduct states for infinite systems does not acrarrect the spa-tial pattern of local energies in the tensor network states

Figure 7 shows the energy per site E for the tensorsets (123) = (2 2 2) (2 4 4) (2 8 4) and (2 8 8)for both PBC and infinite-size schemes When the ten-sor size increases the energy of tensor networks is in-deed improved The lowest energy per site at (2 8 8)is EPBC(N = 114) = 054181 and Einf(N = 114) =054086 These values compare well with results ob-tained from other methods (see Table III in Ref 23)In particular the result of a Greenrsquos function quantum

Monte Carlo (GFQMC) calculation with stochastic re-

configuration (SR)24 is E(N = 144) = 05472(2)Direct comparison may be dicult because our skewboundary is not equal to the periodic one used by theauthors of Ref 24 and our lattice (N = 114) is smallerthan their lattice (N = 144) However our result isclose to their result (1 lower than ours) Our resultfor the infinite-size scheme also agrees well with previ-ous estimates for the thermodynamic limit In particu-lar it compares favorably to that of a series expansion23E = 05502(4) and to that from a GFQMC with SRcalculation24 05458(1) in the thermodynamic limit

Kenji Harada

Multiscale entanglement renormalization ansatz (MERA)

University of Stuttgart

Quantum spin systems

Yazdani lab

Topological insulator

Joel E Moore

D-Wave Systems

Quantum information processing

Karlsruhe Institute of Technology

Single molecule magnet

Computation physicsCondensed matter physics amp statistical physics

H Bombin et al

Toric code

EQUINOX GRAPHICS

Cluster state

Quantum Hall systemsStrongly correlated fermionicbosonic systems

Li-Haldane conjecture

CFT correspondence

Density matrix renormalization group (DMRG)Tensor networkProjected entangled pair states (PEPS)

NV center in diamond

S Benjamin and J Smith

Entanglement２スピン系ハイゼンベルク模型 H = J 1 middot 2 (J gt 0)

J

1 2

部分系 B部分系 A

Energy

縮約密度度行行列列 (片方の部分系の自由度度に関する部分和を実行行)A = TrB = 2 |ss| 2 + 2 |ss| 2 =

12

1 00 1

Entanglement entropy (e-entropy)

Entanglement entropy全系

Schmidt 分解

規格化された状態ベクトル

von Neumann entanglement entropy

| 部分系に分割部分系 A

部分系 B

各々の部分系における正規直交基底

| =

=1

|[A] |[B]

縮約密度度行行列列

A = TrB | | =

=1

2 |[A]

[A] |

非負実数

最大エンタングル状態

S = ln

12

(| | + | |)例例

S = TrA ln A =

=1

2 ln2

2 =

1

Symmetric case

部分系 A

部分系 B

| =

|[A] |[B]

Pre-‐‑‒Schmidt 分解


各々の部分系における線形独立立な基底 (一般に非直交)

部分系Aと部分系Bとを(鏡映)対称軸で分割した場合を考える

重なり行行列列 (overlap matrix)

(M [A]) = [A] |[A]

(M [B]) = [B] |[B]

部分系Aと部分系Bとが対称M [A] = M [B] = M

p =d2

d2

S =

p ln p

e-entropy in symmetric cases

重なり行行列列 M の固有値 d

e-entropy in 1D systemse-entropy の熱力力学的極限での振る舞いと物理理特性の関係

jgi Thus the aim in the following is to compute theentropy of entanglement Eq (1) for the state jgi ac-cording to bipartite partitions parametrized by L

SL trLlog2L$ (4)

where L tr BBLjgihgj is the reduced density matrix

for BL a block of L spins The motivation behind thepresent approach is straightforward by considering theentanglement SL of a spin block as a function of its size Land by characterizing it for large L one expects tocapture the large-scale behavior of quantum correlationsat a critical regime

We start off with a description of the calculations tothen move to the analysis and discussion of the results asummary of which is provided by Fig 1 The XXZ modelEq (2) can be analyzed by using the Bethe ansatz [15]We have numerically determined the ground state jgi ofHXXZ for a chain of up to N 20 spins from which SLcan be computed We recall that in the XXZ model anddue to level crossing the nonanalyticity of the ground-state energy characterizing a phase transition alreadyoccurs for a finite chain Correspondingly already for achain of N 20 spins it is possible to observe a distinctcharacteristic behavior of SL depending on whether thevalues $ in Eq (2) belong or not to a critical regime

The XY model Eq (3) is an exactly solvable model(see for instance [13]) and this allows us to consider aninfinite chain N 1 The calculation of SL as sketchednext also uses the fact that the ground state jgi of thechain and the density matrices L for blocks of spins areGaussian states that can be completely characterized bymeans of certain correlation matrix of second moments

For each site l of the N-spin chain we consider twoMajorana operators c2l and c2lamp1 defined as

c2l

Yl1

m0

zm

xl c2lamp1

Yl1

m0

zm

yl (5)

Operators cm are Hermitian and obey the anticommuta-tion relations fcm cng 2$mn Hamiltonian HXY can bediagonalized by first rewriting it in terms of these newvariables HXYf

l g$ HXYfcmg$ and by then canoni-cally transforming the operators cm The expectationvalue of cm when the system is in the ground state iehcmi hgjcmjgi vanishes for all m due to the Z2symmetry x

l yl

zl $ x

l yl

zl $8l of the ori-

ginal Hamiltonian HXY In turn the expectation valueshcmcni $mn amp i$mn completely characterize jgi forany other expectation value can be expressed throughWickrsquos theorem in terms of hcmcni Matrix $ reads [16]

$

2

6666664

0 1 N1

1 0

1N 0

3

7777775

l

0 glgl 0

(6)

with real coefficients gl given for an infinite chain(N 1) by

gl 1

2amp

Z 2amp

0deil a cos 1 ia( sin

ja cos 1 ia( sinj (7)

From Eqs (6) and (7) we can extract the entropy SLof Eq (4) as follows First from $ and by eliminatingthe rows and columns corresponding to those spins of thechain that do not belong to the block BL we compute thecorrelation matrix of the state L namely $mn amp i$L$mnwhere

$L

2

666664

0 1 L1

1 0

1L 0

3

777775

(8)

Now let V 2 SO2L$ denote an orthogonal matrix thatbrings $L into a block-diagonal form [19] that is

10 20 30 40NUMBER OF SITES minus L minus

1

15

2

25

ENTROPY

minusS

minus

FIG 1 Noncritical entanglement is characterized by a satu-ration of SL as a function of the block size L noncritical Isingchain (empty squares) HXYa 11( 1$ noncritical XXZchain (filled squares) HXXZ 25 0$ Instead the en-tanglement of a block with a chain in a critical model displaysa logarithmic divergence for large L SL ( log2L$=6 (stars) forthe critical Ising chain HXYa 1( 1$ SL ( log2L$=3(triangles) for the critical XX chain with no magnetic fieldHXYa 1( 0$ in a finite XXX chain of N 20 spinswithout magnetic field (diamonds) HXXZ 1 0$ SLcombines the critical logarithmic behavior for low L with afinite-chain saturation effect We have also added the lines)camp cc$=6)log2L$ amp amp [cf Eq (12)] both for free conformalbosons (critical XX model) and free conformal fermions (criti-cal Ising model) to highlight their remarkable agreement withthe numerical diagonalization

P H Y S I C A L R E V I E W L E T T E R S week ending6 JUNE 2003VOLUME 90 NUMBER 22

227902-2 227902-2

G Vidal et al PRL 90 227902 (2003)

XXZ( = 25 = 0)

XY(a = 11 = 1)

L A B

XY(a = 1 = 1)

XY(a = = 0)

XXZ( = 1 = 0)

HXXZ =

i

x

i xi+1 + y

i yi+1 + z

i zi+1 z

i

HXY =

i

a

2(1 + )x

i xi+1 + (1 )y

i yi+1

+ z

i

S(L) =c

3lnL + S0

中心電荷

臨界系の場合

ギャップ系の場合S(L) const

Entanglement spectrum全系

Schmidt 分解



部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数

=

=1

e |[A] [A]

|

A = eHE

熱平衡状態の密度度行行列列とみなす

Entanglement Hamiltonian (e-Hamiltonian)

Entanglement spectrum (e-spectrum)e-Hamiltonian の固有値

= ln2

H Li and M Haldane PRL 101 010504 (2008)

e-spectrum in topological systems

out Any quasihole excitation in region A necessarilypushes electrons into region B and vice versa Howeverthe electron density anywhere on the sphere must remainconstant which can be achieved if the quasihole excita-tions in A and B are correlated (entangled) This gives theempirical rules of counting the levels Take the spectrum inFig 1(a) as an example The partitioning P0j0 results inthe root configuration 110011001100110 on the northernhemisphere (region A) and it corresponds to the singlelsquolsquolevelrsquorsquo at the highest possible value of LAz LAzmax 64We measure the LAz by its deviation from LAzmax ie L LAzmax $ LAz which has the physical meaning of being thetotal z-angular momentum carried by the quasiholes AtL 1 the levels correspond to edge excitations upon theL 0 root configuration There is exactly one edgemode in this case represented by the MR root configura-tion 110011001100101

The number of L 2 levels can be counted in exactlythe same way There are three of them of which the rootconfigurations are

110011001100100111001100110001101100110010101010

while for L 3 the five root configurations are

1100110011001000111001100110001010110011001010011001100110010101001011001010101010100

The counting for the levels at small L for P0j1 andP1j1 can be obtained similarly

For an infinite system in the thermodynamic limit theabove idea gives an empirical counting rule of the numberof levels at any L ie it is the number of independentquasihole excitations upon the semi-infinite root configu-ration uniquely defined by the partitioning For a finitesystem this rule explains the counting only for smallL for large L the finite-size limits the maximalangular momentum that can be carried by an individualquasihole Therefore the number of levels at large L in afinite system will be smaller than the number expectedin an infinite system Not only is this empirical rule con-sistent with all our numerical calculation but it also ex-plains why P0j0 and P0j1 have essentially identicallow-lying structures This is because the (semi-infinite)configuration lsquolsquo 1100110rsquorsquo is essentially equivalentto lsquolsquo 11001100rsquorsquo (with an extra lsquolsquo0rsquorsquo attached to theright) We expect that P0j0 and P0j1 become exactlyidentical in the thermodynamic limit

For completeness we list the root configurations asso-ciated with the first few low-lying levels in Fig 1(c)

L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100

Figure 2 shows the spectra of the system of the same sizeas in Fig 1 ieNe 16 andNorb 30 but for the groundstate of the Coulomb interaction projected into the secondLandau level obtained by direct diagonalizationInterestingly the low-lying levels have the same countingstructure as the corresponding Moore-Read case We iden-tify these low-lying levels as the lsquolsquoCFTrsquorsquo part of the spec-

FIG 2 The low-lying entanglement spectra of the Ne 16and Norb 30 ground state of the Coulomb interaction projectedinto the second Landau level (there are levels beyond the regionsshown here but they are not of interest to us) The insets showthe low-lying parts of the spectra of the Moore-Read state forcomparison [see Fig 1] Note that the structure of the low-lyingspectrum is essentially identical to that of the ideal Moore-Readstate

PRL 101 010504 (2008) P H Y S I C A L R E V I E W L E T T E R S week ending4 JULY 2008

010504-3

分数量量子ホール効果

H Li and M Haldane PRL 101 010504 (2008) F Pollmann et al PRB 81 064439 (2010)

T Ohta S Tanaka I Danshita and K Totsuka in preparation

we arrive at Uz2=1 The combined operation RxRz however

may give rise to a nontrivial phase factor By repeating thissymmetry twice the associated unitary matrix UxUz can beshown in the same way as above to satisfy UxUz=eixzUzUx Since the phases of Ux and Uz have been de-fined the phase factor eixz is not arbitrary and can have aphysical meaning Clearly eixz = 1 If eixz =minus1 then thespectrum of is doubly degenerate since commutes withthe two unitary matrices Ux and Uz which anticommuteamong themselves For the AKLT state Ux=$x and Uz=$ztherefore UxUz=minusUzUx and the Haldane phase is protectedif the system remains symmetric under both Rx and Rz

V EXAMPLES

We now demonstrate how the symmetries discussedabove stabilize the Haldane phase We use the iTEBDmethod to numerically calculate the ground state of themodel given by Eq 3 augmented by various symmetry-breaking perturbations We used MPSrsquos with a dimension of

=80 for the simulations The double degeneracy of the en-tanglement spectrum is used to identify the Haldane phase

Example 1 We begin with the original Hamiltonian H0 inEq 3 This model is translation invariant invariant underspatial inversion under eminusiampSx

and under eminusiampSyTR Using

the above argument we know that inversion symmetry aloneis sufficient to protect a Haldane phase The phase diagram isshown in Fig 1a and agrees with the results of Ref 3 Adiverging entanglement entropy indicates a phase transitionsee for example Ref 34 and we use observables such asSy and Sz to reveal the nature of the phases In our approachthe Haldane phase can be easily identified by looking at thedegeneracy of the entanglement spectrum as shown in Fig 2the even degeneracy of the entanglement spectrum occurs inthe entire phase

In the entanglement spectrum of the trivial insulatingTRI phase there are both singly and multiply degeneratelevels We have checked that the multiple degeneracies in theTRI spectrum can be lifted by adding symmetry-breakingperturbations to the Hamiltonian while preserving inversionsymmetry In the Haldane phase on the other hand thedouble degeneracy of the entire spectrum is robust to addingsuch perturbations

The colormap in Fig 3a shows the difference of the twolargest Schmidt values for the whole Bx-Uzz phase diagramIn the Haldane phase the whole spectrum is at least twofolddegenerate and thus the difference is zero

Example 2 The Hamiltonian H0 has in fact more sym-metries than are needed to stabilize the Haldane phase Todemonstrate this we add a perturbation H1 of the form

H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17

H1 is translation invariant and symmetric under spatial inver-sion but breaks the eminusiampSx

and the eiampSyTR symmetry The

phase diagram for fixed Bz=01J and Uxy =01J as a functionof Bx and Uzz is shown in Fig 1b As predicted by thesymmetry arguments above we find a finite region of stabil-ity for the Haldane phase This region is characterized asbefore by a twofold degeneracy in the entanglement spec-trum as shown in Fig 3b

FIG 1 Color online The colormaps show the entanglemententropy S for different spin-1 models Panel a shows the data forHamiltonian H0 in Eq 3 panel b for H0 plus a term whichbreaks the eminusiampSy

TR symmetry $Eq 17 and panel c for H0plus a term which breaks inversion symmetry $Eq 18 The bluelines indicate a diverging entanglement entropy as a signature of acontinuous phase transition The phase diagrams contain four dif-ferent phases a TRI phase for large Uzz two symmetry breakingantiferromagnetic phases Z2

z and Z2y and a Haldane phase which is

absent in the last panel

minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI

FIG 2 Color online Entanglement spectrum of HamiltonianH0 in Eq 3 for Bx=0 only the lower part of the spectrum isshown The dots show the multiplicity of the Schmidt valueswhich is even in the entire Haldane phase

ENTANGLEMENT SPECTRUM OF A TOPOLOGICAL PHASEhellip PHYSICAL REVIEW B 81 064439 2010

064439-5

J Phys Soc Jpn DRAFT

0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)

Subsystem A Subsystem BSubsystem B

Lsub

Fig 2 (a) Division of the system We calculated ES along (b) the horizon-tal dotted line and (c) vertical dotted line in Fig 1 (d) with N = 503 andLsub = 249 (b) (c) ES We show the number of the degeneracy of ES bydots on the levels (b) The degeneracy in the lowest levels is four one (no de-generacy) and two in C N and AF phases (c) The degeneracy in the lowestlevels is four in both C and C phases At the critical points the degeneratestructure disappears

These exotic phases result from the competition among sev-eral couplings Similar behavior was reported in the cluster-XY model47) and in the transverse Ising model in frustratedlattices48)

The Hamiltonian in the momentum space has the form ofBogoliubov de-Gennes (BdG) Hamiltonian with time reversalsymmetry We can define topological invariant called wind-ing number that is an integer value49) for the mapping fromone-dimensional Brillouin Zone (circle) to Hilbert space (cir-cle) The winding number changes only if the phase transitionoccurs Therefore each phase has a definite value of windingnumber We confirm that the winding numbers of C C AFF and N phases are two two one one and zero respectivelyThese numbers correspond to the number of fermion appear-ing at an edge of the system50)

Let us study the phase diagram in terms of entanglementWe divide the total system into two subsystems A and B sym-metrically around the center with the length of A being Lsub(Fig 2 (a)) We calculate the eigenvalues λi of the reduceddensity matrix ρA of A defined by tracing out the subsystemB ρA = TrBρ where ρ is the density matrix of the groundstate of total system14) ES is defined by ξi = minus ln λi Firstwe focus on the horizontal dotted thin line depicted in Fig 1As shown in Fig 2 (b) the degree of degeneracy in the lowestvalue of ES stays constant in each phase We see that the num-ber is four one and two for C N and AF phases respectivelySimilarly the degree of degeneracy is four in both C and dualC phases shown in Fig 2 (c) From all of these results thedegree of degeneracy is equal to the number of the Majoranazero-modes existing at the end of the system We have thusconfirmed that ES reflect the fictitious edge modes appearingat the position where we cut the system

We study the dynamics during a interaction sweep acrossthe critical point depicted in Fig 1 with open boundarycondition Starting from the initial interaction parametersJYYJXZX = JYZYJXZX = 0 we calculate the time evolu-tion of the initial Bogoliubov vacuum according to the time-

dependent Hamiltonian

H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)

where the interaction parameter changes linearly with finitespeed as

J(t)JXZX = 2tτ 0 le t le τ (16)

We call τ as the sweep time The system go through the crit-ical point at t = 05τ Near the critical point the divergenceof the relaxation time prevents the system to change adiabati-cally which implies the final state after the sweep has a finitenumber of defects We study how the defects affect the lengthdependence of the string correlation function and the entan-glement entropy using time-dependent Bogoliubov transfor-mation51) We take L adjacent sites as the subsystem and thecorrelation between the end sites herein the length L is mea-sured from the center of the system

We show the length dependence of the dual string correla-tion function and entanglement entropy with τ = 25 100 200in Fig 3 (a) (b) As τ becomes larger a characteristic struc-ture of period four appears in the length dependence of bothof them To examine how many Bogoliubov quasiparticlesthere are in the final state we calculate the expectation val-ues of the Bogoliubov number of instantaneous Hamiltonianat t = τ We can see from Fig 3 (c) that the low energy Bo-goliubov excitations becomes dominant as τ becomes largerThus we see that the periodic structure stems from the Bo-goliubov quasiparticle excitations generated near the criticalpoint This suggests that there is a close relationship betweenthe string correlation function and entanglement entropy ofthe excited states

The periodic structure is actually seen in the length depen-dence of the string correlation function of one Bogoliubovquasiparticle excitation It appears for one Bogoliubov quasi-particle with finite energy at JYYJXZX = 0 0 lt JYZYJXZX lt1 in C phase (not shown here) The Bogoliubov excitationswith zero-energy cause the ground-state degeneracy whichwe confirm donrsquot change the behavior of the correlation func-tion for the Bogoliubov vacuum

In this paper we have studied a one-dimensional clustermodel with several interactions to understand the dynamicsin the topological phase First we determine the ground-statephase diagram and characterize each phase by the degener-acy in the lowest levels of ES which corresponds to the edgemodes existing at the edge of the system Second we haveinvestigated the dynamics during the interaction sweep withfinite speed across the critical point to observe the periodic-ity in the length dependence of both correlation functions andentanglement entropy We found this stems from the Bogoli-ubov quasiparticles excited when the system is close to thecritical point It is interesting that a similar structure appearsin the string correlation function and entanglement entropyfor excited states

We have studied dynamic properties in the top0logical sys-tem and suggested the relationship between the string corre-lation functions and entanglement entropy for excited statesSince it not clear whether this relation holds for other modelswhich show topological phases further investigations shouldbe done in the future

3

量量子スピン系

トポロジカル相非Landau型相 emsp　局所秩序変数による emsp　検出不不可 emsp　非局所秩序変数による記述 emsp　ストリング秩序変数等 emsp　 e-spectrum の基底状態の emsp　縮退数

e-spectrum の準位構造と端に生じる低エネルギー励起構造の類似性

目的２次元量量子系における新しいエンタングルメント特性の発見見基底状態が厳密に構成できる統計力力学模型に着目

２次元 Valence-Bond-Solid (VBS) 状態

量量子 hard-‐‑‒square modelの基底状態

e-spectrum に続く新しい概念念の検討e-Hamiltonian の基底状態に注目しエンタングルメントを見見る

ダイジェスト


排除体積効果













VBS状態VBS状態の性質

エンタングルメントの性質 (e-entropy e-spectrum nested e-entropy)

１次元量量子系との対応

VBS on symmetric graphs J Phys A 43 255303 (2010) ldquoVBSCFT correspondencerdquo Phys Rev B 84 245128 (2011)

Nested entanglement entropy Interdisciplinary Information Sciences 19 101 (2013)

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排除体積効果













Valence-Bond-Solid (VBS) 状態Valence bond = singlet pair

12

(| |)

VBS 状態 Valence bond で敷き詰められた状態

Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)

１次元VBS状態を基底状態とするHamiltonian (AKLT model)

I Affleck T Kennedy E Lieb and H Tasaki PRL 59 799 (1987)

Valence-Bond-Solid (VBS) 状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)


I Affleck T Kennedy E Lieb and H Tasaki PRL 59 799 (1987) M Yamashita et al Coordination Chemistry Reviews 198 347 (2000)

S=1 VBS状態が厳密な基底状態である

S=1 VBS状態が唯一の基底状態である

Haldane gapの厳密な証明を与える

反強磁性相関関数が指数関数的に減衰する

MnCl3bpy NINO


F Verstraete and J I Cirac PRA 70 060302 (2004)D Gross and J Eisert PRL 98 220503 (2007) T-C Wei et al PRL 106 070501 (2011) A Miyake Ann Phys 326 1656 (2011)

正方格子蜂の巣格子

VBS状態を用いた測定型量量子計算(MBQC)

VBS state = singlet-covering state


Schwinger boson representation (スピンと２種類のボゾンの関係)

| = adagger|vac | = bdagger|vac


n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2

adaggerkak + bdaggerkbk = 2Sk

a ボゾンの個数

b ボゾンの個数

ab ボゾンの個数の和に関する拘束条件

|VBS =

kl

adaggerkbdaggerl bdaggerkadaggerl

|vac

VBS state




部分系 A

部分系 B

鏡映対称軸で分割



Valence-Bond-Solid (VBS) 状態|VBS =

kl


|vac

=

|[A] |[B]

局所ゲージ変換

鏡映対称性


= 1 middot middot middot |A|

補助空間のスピンi = plusmn12端ボンドの本数

重なり行行列列行行列列M 2|A| 2|A|

重なり行行列列の各要素モンテカルロ法で求める SU(N)の場合にも容易易に適用可能な方法を構築

J Lou S Tanaka H Katsura and N Kawashima PRB 84 245128 (2011)H Katsura J Stat Mech P01006 (2015)

エンタングルメントe-entropy e-spectrum nested e-entropy



２次元 VBS 状態





Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2

１次元 VBS 状態の e-entropy

H Katsura T Hirano and Y Hatsugai PRB 76 012401 (2007)


|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac

S = ln ( edge states)

熱力力学的極限

e-entropyS = TrA ln A =

=1

2 ln2

２次元 VBS 状態の e-entropy S

|A| = ln 2

端ボンドの本数

0 1D = 0 square gt hexagonal

square gt hexagonalH Katsura T Hirano and Y Hatsugai PRB 76 012401 (2007)



部分系 B部分系 ALx

Ly

開放端

周期境界条件

H Katsura N Kawashima ANKirillov V EKorepin S Tanaka JPA 43 255303 (2010) J Lou S Tanaka H Katsura and N Kawashima PRB 84 245128 (2011)

２次元 VBS 状態の e-entropy ENTANGLEMENT SPECTRA OF THE TWO-DIMENSIONAL PHYSICAL REVIEW B 84 245128 (2011)

058

06

062

064

066

SL

y

(a) (b)Lx=1Lx=2Lx=3Lx=4Lx=5

0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3

FIG 2 (Color online) Entanglement entropy S per valence bondacross the boundary as a function of Ly for (a) square lattices withrectangular geometry (OBC) (b) square lattices with cylindricalgeometry (PBC) (c) hexagonal lattices with OBC and (d) hexagonallattices with PBC A further increase of Lx does not affect the resultsmuch The extrapolations to infinite size are indicated by the brokenlines

spectrum of the antiferromagnetic (ferromagnetic) HeisenbergHamiltonian in one dimension We also introduce a newquantity nested entanglement entropy and further elucidatethe relationship between the holographic spin chain and theHamiltonian for the Heisenberg chain

A Entanglement entropy

Let us first study the entanglement entropy (EE) which canbe obtained from

S = minusTr[ρA ln ρA] = minus

α

pα ln pα (10)

where pα (α = 12 2Ly ) are the eigenvalues of ρA Theobtained EE divided by the boundary length for the squareand hexagonal lattices is shown in Fig 2 In both cases weinvestigate lattices with both PBC and OBC results of whichare expected to be equivalent in the thermodynamic limit(Ly rarr infin) The EE per unit boundary length SLy in the 2DAKLT model is found to be strictly less than ln 2 = 0693 147which is in contrast to the result for the 1D AKLT model181923

Note that the area law is satisfied as SLy approaches to aconstant in the thermodynamic limit

The difference between 1D and 2D EE results can beunderstood in terms of ldquovalence bond loopsrdquo (VBL) whichare formed by sequences of singlets connected head-to-tail (Aprecise definition of VBL in the context of Schwinger bosoncan be found in Appendix B) The existence of such loops isa unique feature of 2D VBS states which provides additionalcorrelations between two separated singlets As a result weexpect extra correlations between boundary bosons in the 2DVBS states which lead to modification of the EE from the 1Dresult The possibility of forming a valence bond loop decaysexponentially fast with respect to its length (The reason isdiscussed in Appendix B) This can be related to multipleobserved phenomena of EE from our numerical calculation

TABLE I Obtained fitting parameters for entanglement entropyby the method of least squares

Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)

First of all in VBS states on the square lattice themagnitude of deviation from the 1D result (SLy = ln 2)is significantly larger than that of the hexagonal lattice Tounderstand this we observe that elemental loops (shortestVBLs that have largest possibility) connect two boundary sitesdirectly in the square lattice while an extra site is in betweentwo boundary spins in the hexagonal lattice As a result theamplitude of elemental loops is higher in the square latticeIn the end we obtain considerably more VBLs which theninduce a larger amount of extra correlations on the boundaryin the square lattice than in the hexagonal lattice

Meanwhile increasing the number of sites Lx in thebulk direction provides additional valence bond loops whichinvolve sites not only on the boundary but also inside thetwo-dimensional lattice Hence the EE is further reduced whenexpanding bulk width Lx The effect converges exponentiallyfast however The 2D limit has been almost reached at Lx = 5and 3 for square and hexagonal VBS states respectively It isagain related to the fact that the amplitude of a VBL decreaseswith its length

We also observe that the thermodynamic limit (Ly rarr infin)of EE is achieved with totally different behaviors for latticeswith rectangular (OBC) and cylindrical (PBC) geometries asshown in Fig 2 For VBS states on a lattice with OBC SLy

is found to scale with the boundary length (Ly) as

SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)

where C1C2 (gt0) are the fitting parameters and σ denotesthe EE per boundary length in the thermodynamic limit Themagnitude of C2 is significantly less than C1 as shown inTable I In contrast for a lattice wrapped on a cylinder (PBC)the thermodynamic limit is approached exponentially fast inLy ie σ minus SLy prop exp(minusLyξ ) where ξ (also shown in thetable) is of the order of the lattice spacing

Such distinctively different behaviors of EE for PBCand OBC lattices are closely related to VBL formationsFor example for the square lattice the leading correction

245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)

ENTANGLEMENT SPECTRA OF THE TWO-DIMENSIONAL PHYSICAL REVIEW B 84 245128 (2011)

058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

(ln 2 = 0693147 middot middot middot )EE is less than ln 2

S = TrA ln A

正方格子 (PBC)正方格子 (OBC)

２次元 VBS 状態の e-entropy

Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)

(ln 2 = 0693147 middot middot middot )

S = TrA ln A

EE is less than ln 2

蜂の巣格子 (PBC)蜂の巣格子 (OBC)

２次元 VBS 状態の e-spectrum e-spectrum

e-Hamiltonian の固有値 = ln2

LOU TANAKA KATSURA AND KAWASHIMA PHYSICAL REVIEW B 84 245128 (2011)

FIG 3 (Color online) Entanglement spectra of the (left) squareand (right) hexagonal VBS states with cylindrical geometry (PBC)In both cases Ly = 16 and Lx = 5 in which case results haveconverged to the two-dimensional limit The ground-state energy ofHE is denoted by λ0 The total spins S of the states are marked bydifferent symbols

to boundary spin correlations and EE is associated with agroup of elemental valence bond loops mentioned aboveeach connecting two nearest-neighbor boundary spins Ina lattice with rectangular geometry (OBC) the number ofsuch elemental loops is always 1 less than the number ofboundary sites Ly As a result the leading correction toentropy per boundary spin is proportional to (Ly minus 1)Ly which approaches 1 in the thermodynamic limit Accordinglywe expect that the entanglement entropy converges slowlyfollowing the function 1Ly as observed in Fig 2

For a lattice wrapped on a cylinder (PBC) winding loopsassociated with periodic boundary provide extra spin-spincorrelations to the system The EE in a finite system is lowerthan the result in the thermodynamic limit due to the factthat such winding valence bond loops are more prominentin a smaller lattice compared with those in a larger oneFurthermore amplitudes associated with winding loops decayexponentially fast according to the number of boundary spinsLy which is of the same order of winding loopsrsquo length As aresult we observe a exponentially converging behavior of theentanglement entropy in a PBC lattice

B Entanglement spectrum

Let us now consider the ES defined as a set of eigenvaluesof the entanglement Hamiltonian HE = minus ln ρA In a systemwith PBC we are able to study the ES as a function of mo-mentum in the y direction (k in Fig 3) As shown in Fig 3 theES obtained from the square (hexagonal) VBS state resemblesthe spectrum of the spin-12 antiferromagnetic (ferromagnetic)Heisenberg chain The reason for the dependence of the ES onthe lattice structure is as follows Although both the square andhexagonal lattices are bipartite neighboring boundary spinsbelong to the different sublattices in the case of the square VBSstate As a result the entanglement Hamiltonian is reminiscentof the AFM Heisenberg chain In contrast in the hexagonallattice all boundary spins belong to the same sublattice hence

the FM spectrum appears The lowest-lying modes in theleft panel can be identified as the des Cloizeaux-Pearsonspectrum in the AFM Heisenberg chain24 In the right panelthe excitations with the total spin S = 7 look like the ordinaryspin-wave spectrum in the FM Heisenberg chain In fact dueto the translational symmetry in the y direction the Blochtheorem applies and the single-magnon states in the FM chainare exact eigenstates of the RDM for the hexagonal VBS state

C Nested entanglement entropy

In order to further establish the correspondence betweenthe holographic spin chain and the Hamiltonian for theHeisenberg chain we introduce a measure which we callnested entanglement entropy (NEE) and study its scalingproperties One might think that the finite-size scaling analysisof the ground-state energy λ0 is sufficient to reveal the CFTstructure in the holographic spin chain However there isa subtle point here The spectrum of the RDM can onlytell us that the entanglement Hamiltonian can be expressedas HE = βeffHhol with the Hamiltonian for the holographicchain Hhol There is no unique way to disentangle a fictitioustemperature βeff from Hhol Assuming that Hhol is gapless andits low-energy dispersion is given by vk we can estimate βeffvfrom the slope of the modes at k = 0 in the left panel ofFig 3 (see Table II) The obtained slope shows saturation atabout Lx = 45 which suggests that the fictitious temperatureis presumably nonzero (βeff lt infin) in the infinite 2D systemAs we will see however the NEE provides a more clear-cutapproach to investigate the critical behavior of Hhol withoutany assumption

Let us now give a precise definition of the NEE Basedon the ground state of the entanglement Hamiltonian HE weconstruct the nested RDM for the subchain of length ℓ in theholographic spin chain as

ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)

where |ψ0⟩ denotes the normalized ground state of HE and thetrace is taken over the remaining sites excluding the subchainThe NEE is then obtained from the nested RDM as

S(ℓLy) = minusTr1ℓ[ρ(ℓ) ln ρ(ℓ)] (13)

where the trace is over the sites on the subchain Since thelow-energy physics of the AFM Heisenberg chain is describedby c = 1 CFT the NEE is expected to behave as25

SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)

for a lattice with PBC where c is the central charge and s1 isa nonuniversal constant In Fig 4(a) we show NEE obtainedfrom the square VBS state with Lx = 5 and Ly = 16 The

TABLE II Slopes (βeffv) of modes at k = 0 for Ly = 16

Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)

H Li and M Haldane PRL 101 010504 (2008) J I Cirac et al PRB 83 245134 (2011)

J Lou S Tanaka H Katsura and N Kawashima PRB 84 245128 (2011)













ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)



des Cloizeaux-Pearson mode1D AF Heisenberg model













ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)




Spin wave1D F Heisenberg model

Nested e-entropyEntanglement ground state = e-Hamiltonianの基底状態

HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE



() = Tr+1middotmiddotmiddot L [|0 0|]Nested reduced density matrix

Nested e-entropyS( L) = Tr1middotmiddotmiddot [() ln ()]

１次元量量子臨臨界系(周期境界条件)

P Calabrese and J Cardy J Stat Mech (2004) P06002

Nested e-entropyENTANGLEMENT SPECTRA OF THE TWO-DIMENSIONAL PHYSICAL REVIEW B 84 245128 (2011)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting

FIG 4 (Color online) Nested entanglement entropy S(ℓLy)as a function of the subchain length ℓ for Lx = 5 and Ly = 16(a) and (b) show results obtained for square VBS states with PBC andOBC respectively Fits to the CFT predictions Eqs (14) and (16)are indicated by open circles The lines are guides to the eye

CFT prediction Eq (14) well explains the spatial profile ofthe NEE The fit yields c = 101(7) which is reasonably closeto c = 1

The NEE for ℓ = Ly2 (half-NEE) is simplified to beSPBC(Ly2Ly) = (c3) ln(Ly) + const From the data withthe finite-size scaling form of the NEE we can also extractthe central charge c as summarized in Table III For the caseLx = 1 c is remarkably close to unity This further supportsour conjecture a correspondence between the entanglementHamiltonian of the 2D AKLT model and the physical Hamil-tonian of the 1D Heisenberg chain A slight modification of cis observed as the bulk width increases from Lx = 1 to 5

In the case of the rectangular geometry (OBC) a staggeredpattern of NEE is expected from studies of the standard EE inopen spin chains2627

SOBC(ℓLy) = c

6ln[g(2ℓ + 1)] + a minus πc1

2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)

In Fig 4(b) we show the NEE for the square VBS state onthe rectangle with Lx = 5 and Ly = 16 We take the centralcharge to be c = 1 and the Tomonaga-Luttinger parameterK = 1 and tune a and c1v as fitting parameters The fitis reasonably good except for the boundaries This deviationmight be due to logarithmic corrections which appear in theSU(2) symmetric AFM spin chains26 but are not taken intoaccount in Eq (16)

To provide further evidence for the holographic spin chainwe now consider the hybrid system comprised of squaresand hexagons shown in Fig 5(a) It is expected that theentanglement Hamiltonian is described by the FM-AFMalternating Heisenberg model Figure 5(b) shows the ES forthis system It is clearly seen that there is a gap between the

TABLE III The obtained central charge from half-NEE

Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

FIG 5 (Color online) (a) (Left) VBS state on a hybrid lattice Thered (dotted) line indicates the reflection axis (Right) Correspondingholographic chain The thick and dashed lines indicate AFM and FMbonds respectively (b) Entanglement spectrum of the VBS state in(a) with Ly = 16 The ground state energy is denoted by λ0 Thedashed curve indicates the spectrum of triplet-pair bound states(c) Nested entanglement entropy S(ℓLy) as a function of ℓ forLy = 16

ground state and the continuum The gap rapidly saturates withincreasing Ly The behavior of the ES is totally consistent withthe energy spectrum of the alternating Heisenberg chain theinteractions of which are denoted J and J prime28 By comparing thebound-state spectrum in Fig 5(b) with that of the triplet-pairexcitation in the alternating Heisenberg chain28 we estimatethe ratio of the FM exchange divided by the AFM one asJ primeJ sim 05 Note that one can further manipulate J primeJ of theholographic spin chain by tuning the lattice structure and thenumber of valence bonds on each edge the lattice The NEEassociated with the ground state of HE is shown in Fig 5(c)The NEE is 2 log 2 when two AFM bonds are cut whereas itbecomes about log 2 when both FM and AFM bonds are cutThe obtained result is in good agreement with the standard EEin the alternating Heisenberg chain studied in Ref 29

IV CONCLUSION

We have studied both the entanglement entropy andspectrum associated with the VBS ground state of the AKLTmodel on various 2D lattices It was shown that the reduceddensity matrix of a subsystem can be interpreted as a thermaldensity matrix of the holographic spin chain the spectrumof which resembles that of the spin-12 Heisenberg chainTo elucidate this relationship we have introduced the conceptnested entanglement entropy (NEE) which allows us to clarifythis correspondence in a quantitative way without informationon the fictitious temperatures The finite-size scaling analysisof the NEE revealed that the low-energy physics of theholographic chain associated with the square lattice VBS iswell described by c = 1 conformal field theory The NEEwas also applied to the hybrid VBS state the entanglement

245128-5

正方格子 (PBC) Square ladder (OBC)

A B

中心電荷 c=1 1D AF Heisenberg model

VBSCFT correspondence

ダイジェスト


排除体積効果













量量子格子気体模型基底状態


２次元古典系との対応

Quantum hard-square model Phys Rev A 86 032326 (2012) Nested entanglement entropy Interdisciplinary Information Sciences 19 101 (2013)

ダイジェスト


排除体積効果













Rydberg Atom

Max Planck Institute

Ground state

Rydberg atom (excited state)

Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |

ハードコア格子気体模型ハードコア１つのサイトに１粒粒子まで入ることが可能

排除体積効果

排除体積効果のあるハードコア格子気体模型

コントロールパラメータ化学ポテンシャル

排除体積効果の距離離

D S Gaunt and M E Fisher J Chem Phys 43 2840 (1965) R J Baxter et al J Stat Phys 22 465 (1980)

S Todo and M Suzuki Int J Mod Phys C7 811 (1996) R J Baxter J Phys A 13 L61 (1980)

R J Baxter Exactly Solved Models in Statistical Mechanics (1982)


排除体積効果








排除体積効果







正方形の敷き詰め問題 Hard-square model

量量子ハードコア格子気体模型基底状態が厳密に得られる模型を構築排除体積効果が強い場合に妥当な理理論論模型


Pi =

jGi

(1 nj)冷冷却原子系

ground state 0 darrexcited state (Rydberg atom) 1 uarr

ni i

S Ji et al PRL 107 060406 (2011) I Lesanovsky PRL 108 105301 (2012)

Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i

hdaggeri (z)hi(z) hi(z) = [i

z(1 ni)]Pi

Rydberg原子の生成消滅 Rydberg原子間相互作用(排除体積効果) 化学ポテンシャル

量量子ハードコア格子気体模型


H =L

i=1

P

zxi + (1 3z)ni + zni1ni+1 + z

P1次元鎖の場合

拘束条件のある横磁場イジング模型

ゼロエネルギー状態 Positive semi-definite Hamiltonian 固有値は非負

真空状態Perron-Frobeniusの定理より非縮退

|z =1(z)

i

exp

z+i Pi

| middot middot middot

基底状態が厳密に得られる模型を構築排除体積効果が強い場合に妥当な理理論論模型

Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi


規格化された基底状態

|(z) =

(z) |z =

CSznC2 |C

拘束条件(排除体積効果)のもとでの許される古典的配置の集合

古典的配置Cにおける粒粒子数

古典的配置C|C = CC

基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




規格化因子=古典ハードコア格子気体模型の分配関数

(z) = (z)|(z) =

CSznC

化学ポテンシャル

基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B

正方形はしご格子三角形はしご格子

|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)

局所 Boltzmann 重み

1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14

重なり行行列列 M =1

(z)[T (z)]T T (z)

S Tanaka R Tamura and H Katsura PRA 86 032326 (2012)

基底状態の e-entropy e-entropy

5

(a)

(b)

FIG 3 Two-row transfer matrix for the square ladder (a)and that for the triangular ladder (b)

square and triangular ladders with periodic boundaryconditions (see Fig 1) In order to avoid boundary con-ditions incompatible with the modulation of the dens-est packings we assume that L = 2m and L = 3m(m isin N) for the square and triangular ladders respec-tively The dependence of EE on the activity z forboth models is shown in Figs 4 (a) and 4 (b) Inthe limit z rarr infin the EEs for square and triangularladders become ln 2 and ln 3 respectively irrespectiveof the system size L This can be understood as fol-lows for large z the ground state is approximately givenby a superposition of the ordered states with the max-imum density of particles These ordered states are re-lated to each other by translations Therefore we haveρA sim 1radic

2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the

square ladder and a similar one with period 3 for thetriangular ladder As a result we obtain the observedsaturation values of EE In the opposite limit z rarr 0 theEEs become zero since the ground state is the vacuumstate (see Eq (6)) In the intermediate region betweenthese two limits the EE shows a non-monotonic depen-dence on z and has a peak in both square and triangularladders as shown in Figs 4 (a) and 4 (b) In both casesthe peak position is at about z = zc ie the critical ac-tivity in the corresponding classical model and remainsalmost unchanged with increasing L

Let us focus on the entanglement properties of themodel at z = zc Figures 4 (c) and 4 (d) show the scalingof the EE S(L) for both the square and triangular lad-ders From these plots it is clear that the EEs at z = zc

scale linearly with the system size L (corresponding tothe length of the boundary between A and B) and thusobey the area law

S(L) = αL + S0 (17)

where α and S0 are the fitting parameters inde-pendent of L For square and triangular lad-ders (α S0) = (02272(3)minus0036(6)) and (α S0) =(04001(3) 0020(5)) respectively In both cases S0 isnearly zero suggesting that the topological EE intro-duced in Refs [48 49] is zero in our system This is

0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)

FIG 4 (color online) (a) EE (S) of the state Eq (8) on thesquare ladder as a function of activity z and (b) the same forthe triangular ladder As indicated by the arrows S increaseslinearly with increasing system size L in both cases The dot-ted horizontal lines indicate respectively ln 2 and ln 3 whichare the EEs at z = infin (c) Size-dependence of S at the crit-ical activity z = zc for the square ladder with different sizesL = 4 minus 22 and (d) the same for the triangular ladder withL = 3 minus 21 The dotted lines are least squares fits to the lastfour data points using Eq (17) The fitting parameters areshown in the figures Correlation length divided by systemsize ξ(z)L versus z for the square ladder and (f) that for thetriangular ladder

consistent with the fact that CFTs describing the entan-glement Hamiltonians of our system are non-chiral as wewill see later

As discussed in the previous subsection the Gram ma-trix M can be interpreted as a transfer matrix in the cor-responding 2D classical model Therefore we expect thatthe anomalous behavior of EE at z = zc is attributed tothe phase transition in the classical model To study thenature of the phase transitions in both square and tri-angular cases we perform a finite-size scaling analysis ofthe correlation length The correlation length is definedin terms of the entanglement gap as

ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)

where p(1)(z) and p(2)(z) are the largest and the second-largest eigenvalues of M at z respectively Figures 4(e) and 4 (f) show the correlation length divided by thesystem size L as a function of z for both square andtriangular cases Clearly the curves for different systemsizes cross at the same point z = zc which implies thatthe dynamical critical exponent is given by 1 Near the

e-entropyは極大値をもつ

z=0でゼロになる(基底状態は真空) z が大きい極限で一定値に収束(基底状態の縮退数)

S = TrA ln A =

=1

2 ln2

臨臨界点 zc の決定(z) =

1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)

重なり行行列列emsp　の最大固有値M

5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)


重なり行行列列emsp　の第二固有値M


相関長

転移点

転移点

zc 3796 zc 11090

有限サイズスケーリング 6

0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21

FIG 5 (color online) Finite-size scaling plot of ξ(z)L forthe square ladder (a) and that for the triangular ladder (b)The correlation length exponents of the square and triangularladders are ν = 1 and ν = 56 respectively

critical point we expect that ξ(z)L obeys the scalingrelation

ξ(z)L = f((z minus zc)L1ν) (19)

where ν is the correlation length exponent and f(middot) is ascaling function Figures 5 (a) and 5 (b) show plots ofξ(z)L versus the scaling variable (z minus zc)L1ν For thesquare ladder we find a good data collapse with ν = 1which agrees with the correlation length exponent of the2D Ising model (see Fig 5 (a)) On the other hand forthe triangular ladder the exact value ν = 56 can be ob-tained by noting that the two-row transfer matrix shownin Fig 3 (b) is exactly equivalent to that of the hard-hexagon model [50] We obtain an excellent data collapseas shown in Fig 5 (b) Note that the exponent ν = 56coincides with that of the three-state Potts model [51]

C Entanglement spectrum (ES)

The anomalous behavior of the correlation length atz = zc suggests that the entanglement gap at this pointvanishes linearly with 1L To further elucidate thegapless nature of the entanglement Hamiltonian HE =minus lnM we calculate the excitation spectrum of HE atthe critical point The ES λαα=1NL can be obtainedfrom the relation λα = minus ln pα Each eigenstate is la-beled by the total momentum k due to the translationalsymmetry in the leg direction Fig 6 shows the ES ofboth square and triangular ladders at z = zc In bothcases there is the minimum eigenvalue λ0(= minα λα)at k = 0 and the ES is symmetric about k = 0 and k = π(mod 2π) For the square ladder the gapless modes atmomenta k = 0 and k = π are clearly visible On theother hand for the triangular ladder they are at k = 02π3 and 4π3

These towers of energy levels can be identified as thoseof CFTs describing the low-energy spectra of HE As op-posed to topologically ordered systems such as fractional

quantum Hall states the underlying CFTs are non-chiralin our case ie there are both left- and right-movingmodes In general the excitation energies from whichtowers are generated have the form

λα minus λ0 =2πv

L(hLα + hRα) (20)

where v is the velocity and hLα and hRα are (holomor-phic and anti-holomorphic) conformal weights The sumhLα + hRα is called scaling dimension hLα minus hRα isrelated to the total momentum k Comparing the en-ergy spectrum obtained numerically with the above re-lation we can identify conformal weights of low-lyingstates Scaling dimensions of several low-lying states areindicated in Fig 6 From the obtained scaling dimen-sions we conclude that low-energy spectrum of HE forthe square ladder model at z = zc is described by theCFT with central charge c = 12 [52] whereas that forthe triangular ladder model is described by the c = 45CFT The former CFT is identical to that of the 2D crit-ical Ising model while the latter describes the universal-ity class of the critical three-state Potts model Theseare all consistent with the analysis of correlation lengthexponent in the previous subsection

We note that the exact scaling dimensions for the tri-angular ladder model at the critical point can be ob-tained analytically by exploiting the integrability of thehard-hexagon model [53] Here integrability means theexistence of a family of commuting transfer matrices Infact for the model on the triangular ladder we can showthat there is a one-parameter family of commuting ma-trices including the Gram matrix M in Eq (15) as aspecial case (see Appendix B for the proof) This prop-erty holds even away from the critical point Thereforethe entanglement Hamiltonian of this system at arbitraryz is integrable in the same sense as in the original hard-hexagon model

D Nested entanglement entropy (NEE)

To provide further evidence that the entanglementHamiltonians at z = zc are described by minimal CFTswith central charge c lt 1 we next consider NEE whichwas first introduced in Ref [22] As demonstrated in theprevious work one can obtain the central charge directlyfrom the scaling properties of the NEE without evaluat-ing the velocity v in Eq (20) Let us first define the NEEfrom the ground state of HE We divide the system onwhich the holographic model lives into two subsystemsa block of ℓ consecutive sites and the rest of the chainThe nested reduced density matrix is then defined as

ρ(ℓ) = Trℓ+1middotmiddotmiddot L[|ψ0⟩⟨ψ0|] (21)

where |ψ0⟩ denotes the normalized ground state of HE

and the trace is taken over the degrees of freedom outsidethe block Using ρ(ℓ) the NEE is expressed as

s(ℓ L) = minusTr1middotmiddotmiddot ℓ[ρ(ℓ) ln ρ(ℓ)] (22)

2D Ising = 1 = 56

2D 3-state Potts


e-spectrum zce-spectrum



7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π

FIG 6 (color online) Low-energy spectra of the entangle-ment Hamiltonian HE at z = zc for the square ladder (toppanel) and for the triangular ladder (bottom panel) In bothcases the system size is L = 18 The ground state energyof HE is denoted by λ0 The red open circles indicate posi-tions of the primary fields of the corresponding CFTs whilethe green open squares imply the positions of the descendantfields Discrepancies between the numerical results and theCFT predictions are due to finite-size effects

where the trace is taken over the states in the blockLet us now analyze the NEE obtained numerically in

detail Since the low-energy spectra of HE at z = zc showgood agreement with those of CFT predictions the NEEis expected to behave as

s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)

where s1 is a non-universal constant [54] Figure 7 showsthe NEE for both square and triangular ladders as a func-tion of ln[g(ℓ)] The data for L = 24 are obtained by thepower method The slopes of the dotted lines in Fig 7 arec3 where c = 12 for the square ladder and c = 45 forthe triangular ladder Clearly the results are in excellentagreement with the formula Eq (23) and provide furtherevidence for the holographic minimal models ie the en-tanglement Hamiltonians associated with the square andtriangular ladders at z = zc are described by unitaryminimal CFTs with c lt 1

IV CONCLUSION

We have presented a detailed analysis of the entan-glement entropy and entanglement spectrum (ES) in theground state of the quantum lattice-gas model on two-leg ladders The exact ground state of the model can be

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45

FIG 7 (color online) NEE for the square ladder (a) and thatfor the triangular ladder (b) Dotted lines are a guide to theeye and indicate the slope of c3 where c = 12 and c = 45are used for the square and triangular ladders respectivelyThe solid circle indicates NEE for L = 24 The other symbolsare the same as in Fig 5

obtained as a superposition of states each of which islabeled by a classical configuration with nearest neigh-bor exclusion We have shown that the reduced densitymatrix of one of the legs can be written in terms of thetransfer matrix in the classical lattice-gas model in twodimensions We then numerically studied the entangle-ment properties of the models on both square and tri-angular ladders Both models exhibit critical ES whenthe parameter (activity) z is chosen so that the corre-sponding classical model is critical From the finite-sizescaling analysis we found that the critical theory describ-ing the gapless ES of the square ladder is the CFT withc = 12 while that of the triangular ladder is the CFTwith c = 45 This was further confirmed by the analysisof the nested entanglement entropy We also showed thatthe model on the triangular ladder is integrable at arbi-trary z in the sense that there is a one-parameter familyof matrices commuting with the reduced density matrix

It would be interesting to calculate other quantitiesby exploiting the close connection between the entan-glement Hamiltonian and the two-dimensional classicalmodel It is likely that the virtual-space transfer matrixmethod [55] can be applied to the calculation of Renyientropies Finally it would also be interesting to ex-tend our analysis to the case of truly two-dimensionallattices We note however that the model on N -leg lad-ders (N gt 2) cannot be treated on the same footing sincethe Gram matrix cannot be written by the transfer ma-trix alone Thus new methods need to be developed toanalyze two-dimensional systems

Acknowledgment

The authors thank Shunsuke Furukawa Tohru Komaand Tsubasa Ichikawa for valuable discussions This workwas supported by Grant-in-Aid for JSPS Fellows (23-7601) and for Young Scientists (B) (23740298) RTis partly supported financially by National Institute for

7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0

M Henkel ldquoConformal invariance and critical phenomenardquo (Springer)

正方形はしご格子

三角形はしご格子Primary field Descendant field

0 =2v

L(hL + hR)

v

hL

hR

Velocity Holomorphic conformal weight Antiholomorphic conformal weight

hL + hR

Scaling dimension

2D Ising c=12

2D 3-state Potts c=45

Nested e-entropy zcEntanglement ground state = e-Hamiltonianの基底状態

HE|0 = Egs|0

(e-ground state)





SHU TANAKA RYO TAMURA AND HOSHO KATSURA PHYSICAL REVIEW A 86 032326 (2012)

states the underlying CFTs are nonchiral in our case (ie thereare both left- and right-moving modes) In general the excita-tion energies from which towers are generated have the form

λα minus λ0 = 2πv

L(hLα + hRα) (20)

where v is the velocity and hLα and hRα are (holomorphic andantiholomorphic) conformal weights The sum hLα + hRα

is called scaling dimension hLα minus hRα is related to thetotal momentum k Comparing the energy spectrum obtainednumerically with the above relation we can identify conformalweights of low-lying states Scaling dimensions of several low-lying states are indicated in Fig 6 From the obtained scalingdimensions we conclude that low-energy spectrum of HE forthe square ladder model at z = zc is described by the CFT withcentral charge c = 12 [52] whereas that for the triangularladder model is described by the c = 45 CFT The formerCFT is identical to that of the 2D critical Ising model while thelatter describes the universality class of the critical three-statePotts model These are all consistent with the analysis ofcorrelation length exponent in the previous subsection

We note that the exact scaling dimensions for the triangularladder model at the critical point can be obtained analyticallyby exploiting the integrability of the hard-hexagon model [53]Here integrability means the existence of a family of commut-ing transfer matrices In fact for the model on the triangularladder we can show that there is a one-parameter family ofcommuting matrices including the Gram matrix M in Eq (15)as a special case (see Appendix B for the proof) This propertyholds even away from the critical point Therefore the entan-glement Hamiltonian of this system at arbitrary z is integrablein the same sense as in the original hard-hexagon model


To provide further evidence that the entanglement Hamil-tonians at z = zc are described by minimal CFTs with centralcharge c lt 1 we next consider NEE which was first introducedin Ref [22] The NEE is the entanglement (von Neumann)entropy associated with the ground state of HE One mightthink that the scaling analysis presented in the previoussubsection suffices to find the central charges However theNEE provides a simpler and more clear-cut approach Asdiscussed in Sec III A the Gram matrix M = exp(minusHE) canbe interpreted as a thermal density matrix of the holographicmodel in one dimension This implies that HE can bewritten as HE = βeffHhol with Hhol being the Hamiltonianfor the holographic model Here βeff is the fictitious inversetemperature and is not universal We are interested in Hholrather than HE However there is no unique way to disentangleβeff from Hhol To overcome this we first note that the groundstate of HE is the same as that of Hhol corresponding to theeigenvector of M associated with the largest eigenvalue Wethen recall that for 1D critical systems one can read offthe underlying CFT solely from the entanglement entropyobtained from the ground state of Hhol [54] Therefore one canobtain the central charge directly from the scaling propertiesof the NEE without evaluating βeff or the velocity v in Eq (20)as demonstrated in the previous work [22]

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45

FIG 7 (Color online) NEE for the square ladder (a) and thatfor the triangular ladder (b) Dotted lines are a guide to the eye andindicate the slope of c3 where c = 12 and c = 45 are used for thesquare and triangular ladders respectively The solid circles indicateNEE for L = 24 The other symbols are the same as in Fig 5

Let us now give the definition of the NEE We dividethe system on which the holographic model lives into twosubsystems a block of ℓ consecutive sites and the rest of thechain The nested reduced density matrix is then defined as

ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)

where |ψ0⟩ denotes the normalized ground state of HE and thetrace is taken over the degrees of freedom outside the blockUsing ρ(ℓ) the NEE is expressed as

s(ℓL) = minusTr1ℓ[ρ(ℓ) ln ρ(ℓ)] (22)

where the trace is taken over the states in the blockLet us now analyze the numerically obtained NEE in detail

Since the low-energy spectra of HE at z = zc show goodagreement with those of CFT predictions the NEE is expectedto behave as

s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)

where s1 is a nonuniversal constant [54] Figure 7 shows theNEE for both square and triangular ladders as a functionof ln[g(ℓ)] The data for L = 24 are obtained by the powermethod The slopes of the dotted lines in Fig 7 are c3where c = 12 for the square ladder and c = 45 for thetriangular ladder Clearly the results are in excellent agreementwith the formula Eq (23) and provide further evidencefor the holographic minimal models (ie the entanglementHamiltonians associated with the square and triangular laddersat z = zc are described by unitary minimal CFTs with c lt 1)

IV CONCLUSION

We have presented a detailed analysis of the entanglemententropy and entanglement spectrum (ES) in the ground state ofthe quantum lattice-gas model on two-leg ladders The exactground state of the model can be obtained as a superposition ofstates each of which is labeled by a classical configuration withnearest-neighbor exclusion We have shown that the reduceddensity matrix of one of the legs can be written in terms ofthe transfer matrix in the classical lattice-gas model in twodimensions We then numerically studied the entanglementproperties of the models on both square and triangular ladders

032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6

正方形はしご格子 L=6-‐‑‒24

三角形はしご格子 L=6-‐‑‒24

2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果













共同研究者の方々

審査希望分野

No

1

申請研究課題

物理学Ⅱ

K1412007

注入同期レーザー量子計算機の高効率化及び汎用性の向上を志向した理論研究

田村

亮

姓

論文での使用姓(該当者のみ)

名

ミドルネーム

氏名

性別 5 国籍男

生年月日年月日

独立行政法人物質材料研究機構

若手国際研究センター

ポスドク研究員

日月年博士号取得(見込み)年月日

大学院名学位名

センター名

研究室名

所属長

受入希望研究室

電話番号

審査結果通知先Email

創発物性科学研究センター

量子情報エレクトロニクス部門量子光学研究グループ

山本喜久

tamuraryonimsgojp

2

3

4

6

7

8

9

10

11

12

日本

現職機関名(大学院名)

所属部署(学科名)

職　　名(課程名年次)

1984 303

2012 3 22

平成27年度採用基礎科学特別研究員採用申請書

特記事項

登録番号

080-6801-9078

東京大学大学院理学系研究科博士（理学）

桂法称 (東大院理理)

Anatol N Kirillov (RIMS 京大)

Vladimir E Korepin (YITP Stony Brook USA)

川島直輝 (東大物性研)

Lou Jie (Fudan Univ 中国)

田村亮亮 (NIMS)

ダイジェスト


排除体積効果














目的

数学的準備

5




2 Energy







Kenji Harada




Yazdani lab


Joel E Moore

D-Wave Systems





H Bombin et al

Toric code

EQUINOX GRAPHICS

Cluster state



CFT correspondence





J

1 2


Energy


12

1 00 1



Schmidt 分解




部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数


S = ln

12

(| | + | |)例例

S = TrA ln A =

=1

2 ln2

2 =

1

Symmetric case

部分系 A

部分系 B

| =

|[A] |[B]






(M [A]) = [A] |[A]

(M [B]) = [B] |[B]


p =d2

d2

S =

p ln p





SL trLlog2L$ (4)






c2l

Yl1

m0

zm

xl c2lamp1

Yl1

m0

zm

yl (5)



l yl

zl $ x

l yl

zl $8l of the ori-


$

2

6666664

0 1 N1

1 0

1N 0

3

7777775

l

0 glgl 0

(6)


gl 1

2amp

Z 2amp




$L

2

666664

0 1 L1

1 0

1L 0

3

777775

(8)



1

15

2

25

ENTROPY

minusS

minus



227902-2 227902-2

G Vidal et al PRL 90 227902 (2003)

XXZ( = 25 = 0)

XY(a = 11 = 1)

L A B

XY(a = 1 = 1)

XY(a = = 0)

XXZ( = 1 = 0)

HXXZ =

i

x

i xi+1 + y

i yi+1 + z

i zi+1 z

i

HXY =

i

a

2(1 + )x

i xi+1 + (1 )y

i yi+1

+ z

i

S(L) =c

3lnL + S0

中心電荷

臨界系の場合



Schmidt 分解



部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数

=

=1

e |[A] [A]

|

A = eHE




= ln2





110011001100100111001100110001101100110010101010


1100110011001000111001100110001010110011001010011001100110010101001011001010101010100




L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100




010504-3






V EXAMPLES









H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17








minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI



064439-5


0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)


Lsub







H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)








3








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12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果













ダイジェスト


排除体積効果














目的

数学的準備

5




2 Energy







Kenji Harada




Yazdani lab


Joel E Moore

D-Wave Systems





H Bombin et al

Toric code

EQUINOX GRAPHICS

Cluster state



CFT correspondence





J

1 2


Energy


12

1 00 1



Schmidt 分解




部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数


S = ln

12

(| | + | |)例例

S = TrA ln A =

=1

2 ln2

2 =

1

Symmetric case

部分系 A

部分系 B

| =

|[A] |[B]






(M [A]) = [A] |[A]

(M [B]) = [B] |[B]


p =d2

d2

S =

p ln p





SL trLlog2L$ (4)






c2l

Yl1

m0

zm

xl c2lamp1

Yl1

m0

zm

yl (5)



l yl

zl $ x

l yl

zl $8l of the ori-


$

2

6666664

0 1 N1

1 0

1N 0

3

7777775

l

0 glgl 0

(6)


gl 1

2amp

Z 2amp




$L

2

666664

0 1 L1

1 0

1L 0

3

777775

(8)



1

15

2

25

ENTROPY

minusS

minus



227902-2 227902-2

G Vidal et al PRL 90 227902 (2003)

XXZ( = 25 = 0)

XY(a = 11 = 1)

L A B

XY(a = 1 = 1)

XY(a = = 0)

XXZ( = 1 = 0)

HXXZ =

i

x

i xi+1 + y

i yi+1 + z

i zi+1 z

i

HXY =

i

a

2(1 + )x

i xi+1 + (1 )y

i yi+1

+ z

i

S(L) =c

3lnL + S0

中心電荷

臨界系の場合



Schmidt 分解



部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数

=

=1

e |[A] [A]

|

A = eHE




= ln2





110011001100100111001100110001101100110010101010


1100110011001000111001100110001010110011001010011001100110010101001011001010101010100




L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100




010504-3






V EXAMPLES









H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17








minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI



064439-5


0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)


Lsub







H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)








3








ダイジェスト


排除体積効果


















ダイジェスト


排除体積効果














12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



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排除体積効果

















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排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

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結論論


排除体積効果














目的

数学的準備

5




2 Energy







Kenji Harada




Yazdani lab


Joel E Moore

D-Wave Systems





H Bombin et al

Toric code

EQUINOX GRAPHICS

Cluster state



CFT correspondence





J

1 2


Energy


12

1 00 1



Schmidt 分解




部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数


S = ln

12

(| | + | |)例例

S = TrA ln A =

=1

2 ln2

2 =

1

Symmetric case

部分系 A

部分系 B

| =

|[A] |[B]






(M [A]) = [A] |[A]

(M [B]) = [B] |[B]


p =d2

d2

S =

p ln p





SL trLlog2L$ (4)






c2l

Yl1

m0

zm

xl c2lamp1

Yl1

m0

zm

yl (5)



l yl

zl $ x

l yl

zl $8l of the ori-


$

2

6666664

0 1 N1

1 0

1N 0

3

7777775

l

0 glgl 0

(6)


gl 1

2amp

Z 2amp




$L

2

666664

0 1 L1

1 0

1L 0

3

777775

(8)



1

15

2

25

ENTROPY

minusS

minus



227902-2 227902-2

G Vidal et al PRL 90 227902 (2003)

XXZ( = 25 = 0)

XY(a = 11 = 1)

L A B

XY(a = 1 = 1)

XY(a = = 0)

XXZ( = 1 = 0)

HXXZ =

i

x

i xi+1 + y

i yi+1 + z

i zi+1 z

i

HXY =

i

a

2(1 + )x

i xi+1 + (1 )y

i yi+1

+ z

i

S(L) =c

3lnL + S0

中心電荷

臨界系の場合



Schmidt 分解



部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数

=

=1

e |[A] [A]

|

A = eHE




= ln2





110011001100100111001100110001101100110010101010


1100110011001000111001100110001010110011001010011001100110010101001011001010101010100




L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100




010504-3






V EXAMPLES









H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17








minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI



064439-5


0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)


Lsub







H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)








3








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12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



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排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

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排除体積効果













結論論


排除体積効果













5




2 Energy







Kenji Harada




Yazdani lab


Joel E Moore

D-Wave Systems





H Bombin et al

Toric code

EQUINOX GRAPHICS

Cluster state



CFT correspondence





J

1 2


Energy


12

1 00 1



Schmidt 分解




部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数


S = ln

12

(| | + | |)例例

S = TrA ln A =

=1

2 ln2

2 =

1

Symmetric case

部分系 A

部分系 B

| =

|[A] |[B]






(M [A]) = [A] |[A]

(M [B]) = [B] |[B]


p =d2

d2

S =

p ln p





SL trLlog2L$ (4)






c2l

Yl1

m0

zm

xl c2lamp1

Yl1

m0

zm

yl (5)



l yl

zl $ x

l yl

zl $8l of the ori-


$

2

6666664

0 1 N1

1 0

1N 0

3

7777775

l

0 glgl 0

(6)


gl 1

2amp

Z 2amp




$L

2

666664

0 1 L1

1 0

1L 0

3

777775

(8)



1

15

2

25

ENTROPY

minusS

minus



227902-2 227902-2

G Vidal et al PRL 90 227902 (2003)

XXZ( = 25 = 0)

XY(a = 11 = 1)

L A B

XY(a = 1 = 1)

XY(a = = 0)

XXZ( = 1 = 0)

HXXZ =

i

x

i xi+1 + y

i yi+1 + z

i zi+1 z

i

HXY =

i

a

2(1 + )x

i xi+1 + (1 )y

i yi+1

+ z

i

S(L) =c

3lnL + S0

中心電荷

臨界系の場合



Schmidt 分解



部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数

=

=1

e |[A] [A]

|

A = eHE




= ln2





110011001100100111001100110001101100110010101010


1100110011001000111001100110001010110011001010011001100110010101001011001010101010100




L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100




010504-3






V EXAMPLES









H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17








minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI



064439-5


0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)


Lsub







H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)








3








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12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














J

1 2


Energy


12

1 00 1



Schmidt 分解




部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数


S = ln

12

(| | + | |)例例

S = TrA ln A =

=1

2 ln2

2 =

1

Symmetric case

部分系 A

部分系 B

| =

|[A] |[B]






(M [A]) = [A] |[A]

(M [B]) = [B] |[B]


p =d2

d2

S =

p ln p





SL trLlog2L$ (4)






c2l

Yl1

m0

zm

xl c2lamp1

Yl1

m0

zm

yl (5)



l yl

zl $ x

l yl

zl $8l of the ori-


$

2

6666664

0 1 N1

1 0

1N 0

3

7777775

l

0 glgl 0

(6)


gl 1

2amp

Z 2amp




$L

2

666664

0 1 L1

1 0

1L 0

3

777775

(8)



1

15

2

25

ENTROPY

minusS

minus



227902-2 227902-2

G Vidal et al PRL 90 227902 (2003)

XXZ( = 25 = 0)

XY(a = 11 = 1)

L A B

XY(a = 1 = 1)

XY(a = = 0)

XXZ( = 1 = 0)

HXXZ =

i

x

i xi+1 + y

i yi+1 + z

i zi+1 z

i

HXY =

i

a

2(1 + )x

i xi+1 + (1 )y

i yi+1

+ z

i

S(L) =c

3lnL + S0

中心電荷

臨界系の場合



Schmidt 分解



部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数

=

=1

e |[A] [A]

|

A = eHE




= ln2





110011001100100111001100110001101100110010101010


1100110011001000111001100110001010110011001010011001100110010101001011001010101010100




L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100




010504-3






V EXAMPLES









H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17








minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI



064439-5


0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)


Lsub







H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)








3








ダイジェスト


排除体積効果


















ダイジェスト


排除体積効果














12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














Schmidt 分解




部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数


S = ln

12

(| | + | |)例例

S = TrA ln A =

=1

2 ln2

2 =

1

Symmetric case

部分系 A

部分系 B

| =

|[A] |[B]






(M [A]) = [A] |[A]

(M [B]) = [B] |[B]


p =d2

d2

S =

p ln p





SL trLlog2L$ (4)






c2l

Yl1

m0

zm

xl c2lamp1

Yl1

m0

zm

yl (5)



l yl

zl $ x

l yl

zl $8l of the ori-


$

2

6666664

0 1 N1

1 0

1N 0

3

7777775

l

0 glgl 0

(6)


gl 1

2amp

Z 2amp




$L

2

666664

0 1 L1

1 0

1L 0

3

777775

(8)



1

15

2

25

ENTROPY

minusS

minus



227902-2 227902-2

G Vidal et al PRL 90 227902 (2003)

XXZ( = 25 = 0)

XY(a = 11 = 1)

L A B

XY(a = 1 = 1)

XY(a = = 0)

XXZ( = 1 = 0)

HXXZ =

i

x

i xi+1 + y

i yi+1 + z

i zi+1 z

i

HXY =

i

a

2(1 + )x

i xi+1 + (1 )y

i yi+1

+ z

i

S(L) =c

3lnL + S0

中心電荷

臨界系の場合



Schmidt 分解



部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数

=

=1

e |[A] [A]

|

A = eHE




= ln2





110011001100100111001100110001101100110010101010


1100110011001000111001100110001010110011001010011001100110010101001011001010101010100




L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100




010504-3






V EXAMPLES









H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17








minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI



064439-5


0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)


Lsub







H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)








3








ダイジェスト


排除体積効果


















ダイジェスト


排除体積効果














12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果













Symmetric case

部分系 A

部分系 B

| =

|[A] |[B]






(M [A]) = [A] |[A]

(M [B]) = [B] |[B]


p =d2

d2

S =

p ln p





SL trLlog2L$ (4)






c2l

Yl1

m0

zm

xl c2lamp1

Yl1

m0

zm

yl (5)



l yl

zl $ x

l yl

zl $8l of the ori-


$

2

6666664

0 1 N1

1 0

1N 0

3

7777775

l

0 glgl 0

(6)


gl 1

2amp

Z 2amp




$L

2

666664

0 1 L1

1 0

1L 0

3

777775

(8)



1

15

2

25

ENTROPY

minusS

minus



227902-2 227902-2

G Vidal et al PRL 90 227902 (2003)

XXZ( = 25 = 0)

XY(a = 11 = 1)

L A B

XY(a = 1 = 1)

XY(a = = 0)

XXZ( = 1 = 0)

HXXZ =

i

x

i xi+1 + y

i yi+1 + z

i zi+1 z

i

HXY =

i

a

2(1 + )x

i xi+1 + (1 )y

i yi+1

+ z

i

S(L) =c

3lnL + S0

中心電荷

臨界系の場合



Schmidt 分解



部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数

=

=1

e |[A] [A]

|

A = eHE




= ln2





110011001100100111001100110001101100110010101010


1100110011001000111001100110001010110011001010011001100110010101001011001010101010100




L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100




010504-3






V EXAMPLES









H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17








minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI



064439-5


0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)


Lsub







H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)








3








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12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



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排除体積効果

















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排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果















SL trLlog2L$ (4)






c2l

Yl1

m0

zm

xl c2lamp1

Yl1

m0

zm

yl (5)



l yl

zl $ x

l yl

zl $8l of the ori-


$

2

6666664

0 1 N1

1 0

1N 0

3

7777775

l

0 glgl 0

(6)


gl 1

2amp

Z 2amp




$L

2

666664

0 1 L1

1 0

1L 0

3

777775

(8)



1

15

2

25

ENTROPY

minusS

minus



227902-2 227902-2

G Vidal et al PRL 90 227902 (2003)

XXZ( = 25 = 0)

XY(a = 11 = 1)

L A B

XY(a = 1 = 1)

XY(a = = 0)

XXZ( = 1 = 0)

HXXZ =

i

x

i xi+1 + y

i yi+1 + z

i zi+1 z

i

HXY =

i

a

2(1 + )x

i xi+1 + (1 )y

i yi+1

+ z

i

S(L) =c

3lnL + S0

中心電荷

臨界系の場合



Schmidt 分解



部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数

=

=1

e |[A] [A]

|

A = eHE




= ln2





110011001100100111001100110001101100110010101010


1100110011001000111001100110001010110011001010011001100110010101001011001010101010100




L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100




010504-3






V EXAMPLES









H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17








minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI



064439-5


0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)


Lsub







H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)








3








ダイジェスト


排除体積効果


















ダイジェスト


排除体積効果














12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














Schmidt 分解



部分系 B


| =

=1

|[A] |[B]


A = TrB | | =

=1

2 |[A]

[A] |

非負実数

=

=1

e |[A] [A]

|

A = eHE




= ln2





110011001100100111001100110001101100110010101010


1100110011001000111001100110001010110011001010011001100110010101001011001010101010100




L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100




010504-3






V EXAMPLES









H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17








minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI



064439-5


0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)


Lsub







H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)








3








ダイジェスト


排除体積効果


















ダイジェスト


排除体積効果














12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

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結論論


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110011001100100111001100110001101100110010101010


1100110011001000111001100110001010110011001010011001100110010101001011001010101010100




L 0 11001100110011001L 1 110011001100110001

110011001100101010L 2 1100110011001100001

110011001100101001011001100110010011001100110010101010100




010504-3






V EXAMPLES









H1 = Bzj

Sjz + Uxy

jSj

xSjy + Sj

ySjx 17








minus1 minus05 0 05 1 15 20

2

4

6

8

10

12

UzzJ

minus2log()

Z2

zHaldane TRI



064439-5


0

3

6

9

0 05 1 15 2

ξi

JYY

JXZX

JYZY

JXZX

0

2

4

6

05 1 15

ξi

(b)

(a)

(c)


Lsub







H(t) = minusJXZXN

i=1

σxi σ

zi+1σ

xi+2 + J(t)

N

i=1

σyiσ

zi+1σ

yi+2 (15)








3








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12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果

















ダイジェスト


排除体積効果


















ダイジェスト


排除体積効果














12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



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排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

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結論論


排除体積効果


















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12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



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ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














12

(| |)


Valence bond

S = 1への射影

１次元VBS状態

H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)




H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














H =

i

Si middot Si+1 +

13

Si middot Si+1

2

(S = 1)







MnCl3bpy NINO










n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果






















n(b)k = bdaggerkbk

n(a)k = adaggerkak

0 1 2 3 40

1

2

3

4

S=0 12 1 32 2





|VBS =

kl


|vac

VBS state




部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


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ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果















部分系 A

部分系 B





kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














kl


|vac

=

|[A] |[B]


鏡映対称性















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果





















Lx

Ly

開放端

周期境界条件

S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果













S=1

S=3

S=4

S=6

S=8

S=2




|VBS =N

i=0

adaggeri b

daggeri+1 bdaggeria

daggeri+1

S|vac




=1

2 ln2


|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














|A| = ln 2







Ly

開放端

周期境界条件



058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

2 4 6 8 10 12 14 16 2 4 6 8 10 12 14 16Bonds Bonds

S

Bon

ds

SLy

= +C1

Ly+

C2

Ly lnLy S

Ly exp(Ly)


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


S = TrA ln A



Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














Bonds Bonds

S

Bon

ds


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3


058

06

062

064

066

SL

y


0684

0685

0686

0687

2 4 6 8 10 12 14 16

SL

y

Ly

(c)

2 4 6 8 10 12 14 16Ly

(d)Lx=1Lx=2Lx=3






α

pα ln pα (10)





Square

OBC PBC

σ C1 C2 σ ξ

Lx = 1 061277(3) 00865(3) 00009(2) 06129(1) 154(5)Lx = 2 059601(4) 01127(4) 00034(3) 05966(2) 24(1)Lx = 3 059322(5) 01196(5) 00049(3) 05942(3) 29(2)Lx = 4 059259(5) 01223(5) 0006(3) 05936(4) 31(2)Lx = 5 059246(4) 01227(5) 0006(3) 05934(4) 31(2)

Hexagonal

OBC PBC

σ C1 C2 σ ξ

Lx = 1 068502(2) 00092(6) 00011(7) 068508(6) 04(1)Lx = 2 068469(3) 00098(6) 00012(8) 0684757(4) 06(1)Lx = 3 068468(3) 00098(6) 00013(7) 0684754(5) 07(1)





SLy

= σ + C1

Ly

minus C2

Ly log Ly

(11)



245128-3

SLy

= +C1

Ly+

C2

Ly lnLy

SLy

exp(Ly)


S = TrA ln A















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果

























ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果

























ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)
















ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果

























ρ(ℓ) = Trℓ+1Ly[|ψ0⟩⟨ψ0|] (12)




SPBC(ℓLy) = c

3ln[f (ℓ)] + s1 (14)

f (ℓ) = Ly

πsin

πℓ

Ly

(15)



Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

169000 202221 213535 217006 218755

245128-4

0

(Lx=5 Ly=16)(Lx=5 Ly=32)






HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














HE|0 = Egs|0

(e-ground state)


A = TrB | | =

=1

2 |[A]

[A] |

=

=1

e |[A] [A]

|

A = eHE








06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(a)

s1=077(4)c=101(7) Lx=5 Ly=16

fitting 05

06

07

08

0 2 4 6 8 10 12 14 16

S(l1

6)

l

(b)

a=0393(1)c1v=0093(3)

Lx=5 Ly=16fitting





SOBC(ℓLy) = c


2v

(minus1)ℓ

g(2ℓ + 1)12K

(16)

g(ℓ) = 2(Ly + 1)π

sin

πℓ

2(Ly + 1)




Lx = 1 Lx = 2 Lx = 3 Lx = 4 Lx = 5

1007(4) 1042(4) 1055(4) 1056(2) 1059(2)

-1 1

triplet excitation

x

y

2ln2

ln2

(a) (b)

(c)

06

08

1

12

14

0 2 4 6 8 10 12 14 16

S(l1

6)

l



IV CONCLUSION


245128-5


A B



ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果













ダイジェスト


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果

















ダイジェスト


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果













Rydberg Atom


Ground state


Interaction

H =

i

xi +

i

ni + V

ij

ninj

|ri rj |


排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














排除体積効果








排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














排除体積効果








排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














排除体積効果










Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果















Pi =

jGi



ni i


Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi




H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果















H =L

i=1

P






|z =1(z)

i

exp

z+i Pi



Hsol =

z

i

(+i + i )Pi +

i

[(1 z)ni + z]Pi

Hsol =

i


z(1 ni)]Pi


基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果













基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C




基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果













基底状態|z =

1(z)

i

exp

z+i Pi



|(z) =

(z) |z =

CSznC2 |C





(z) = (z)|(z) =

CSznC


基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果













基底状態1 2 L

1 2 L

1 2 L

1 2 L

部分系 A

部分系 B

部分系 A

部分系 B


|(z) =

[T (z)] | |

[T (z)] =L

i=1

w (ii+1 i+1 i)

a b

cd

= w(a b c d)


1 z14 z12z14 z14 z14 z12 1 z14 z12z14 z14 z14


(z)[T (z)]T T (z)



5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




S = TrA ln A =

=1

2 ln2


1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














1ln[p(1)(z)p(2)(z)]

p(1)(z)

p(2)(z)


5

(a)

(b)



2(|0101⟩⟨0101| + |1010⟩⟨1010|) for the




S(L) = αL + S0 (17)


0

1

2

3

4

5

0 10 20 30

S

z

(a)

0

1

2

3

4

5

0 5 10 15 20 25

S

L

α=02272(3)S0=-0036(6)

(c)

0

1

2

0 2 4 6

ξ(z

)L

z

(e)

0

2

4

6

8

10

0 50 100 150

S

z

(b)

0

2

4

6

8

10

0 5 10 15 20 25

S

L

α=04001(3)S0=0020(5)

(d)

0

1

2

0 5 10 15

ξ(z

)L

z

(f)




ξ(z) =1

ln[p(1)(z)p(2)(z)] (18)




相関長

転移点

転移点

zc 3796 zc 11090


0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














0

05

1

15

2

25

-10 0 10

ξ(z)

L

(z-zc)L1ν

(a)

L= 6L= 8L=10L=12L=14L=16L=18L=20L=22

0

05

1

15

2

-50 0 50

ξ(z)

L

(z-zc)L1ν

(b)

L= 6L= 9L=12L=15L=18L=21










L(hLα + hRα) (20)









2D Ising = 1 = 56

2D 3-state Potts





7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果
















7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


7

0

2

4

6

0 025 05 075 1

λ-λ

0

k2π

0

1

2

0 025 05 075 1

λ-λ

0

k2π




s(ℓ L) =c

3ln[g(ℓ)] + s1 (23)

g(ℓ) =L

πsin

πℓ

L

(24)


IV CONCLUSION


06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)c=45




Acknowledgment


0

0




0 =2v

L(hL + hR)

v

hL

hR


hL + hR

Scaling dimension

2D Ising c=12



HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果














HE|0 = Egs|0

(e-ground state)








L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6




L(hLα + hRα) (20)






06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(a)

c=12

06

08

1

12

14

0 05 1 15 2

s(lL

)

ln[g(l)]

(b)

c=45



ρ(ℓ) = Trℓ+1L[|ψ0⟩⟨ψ0|] (21)





s(ℓL) = c

3ln[g(ℓ)] + s1 (23)

g(ℓ) = L

πsin

πℓ

L

(24)


IV CONCLUSION


032326-6



2D Ising

2D 3-state Potts

ダイジェスト


排除体積効果













結論論


排除体積効果













ダイジェスト


排除体積効果













結論論


排除体積効果













結論論


排除体積効果











