excitations of the gapped xxz heisenberg spin-1/2 chaincaux) that are designed to e ectively...

Master Thesis

Excitations of the Gapped

XXZ Heisenberg Spin-1/2 Chain

In Partial Fulfillmentof the Requirements for the Degree

MSc Physics and AstronomyTheoretical Physics

At the Institute for Theoretical Physics, UvA

September 2017 - August 2018

60 ECTs

Author: Supervisor/Examiner:Rebekka Koch Prof. Dr. Jean-Sebastien Caux

Student number: Second Examiner:11410248 Dr. Vladimir Gritsev

Table of Contents

Acknowledgements iii

Abstract v

1 Introduction 11.1 The Power of the Bethe Ansatz . . . . . . . . . . . . . . . . . . . . . . . . . 1

Notion of Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . 1Role of Integrable Systems in Many-Body Physics . . . . . . . . . . 2Heisenberg Spin Chains . . . . . . . . . . . . . . . . . . . . . . . . . 2Correlation Functions and Form Factors . . . . . . . . . . . . . . . . 3Numerical Methods - ABACUS . . . . . . . . . . . . . . . . . . . . . 4Classification of States . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Goal and Outline of this Thesis . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 The XXZ Heisenberg Spin Chain 72.1 The One-dimensional Heisenberg Model . . . . . . . . . . . . . . . . . . . . 72.2 Symmetries of the XXZ Spin Chain and the

Dimensionality of its Hilbert Space . . . . . . . . . . . . . . . . . . . . . . . 92.2.1 Ground State and Excitations . . . . . . . . . . . . . . . . . . . . . . 11

3 The Coordinate Bethe Ansatz 133.1 Coordinate Bethe Ansatz for XXZ spin chains . . . . . . . . . . . . . . . . 13

A Single Flipped Spin . . . . . . . . . . . . . . . . . . . . . . . . . . 13Two Flipped Spins . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14Arbitrary Number of Flipped Spin . . . . . . . . . . . . . . . . . . . 15

3.1.1 Bethe Equations in Terms of Rapidities . . . . . . . . . . . . . . . . 17Bethe Equations of the gapped XXZ Heisenberg Spin Chain . . . . 17Logarithmic Bethe Equations . . . . . . . . . . . . . . . . . . . . . . 18Energy and Total Momentum Parametrized with Rapidities . . . . . 19

3.2 Complex Rapidities and String States . . . . . . . . . . . . . . . . . . . . . 203.3 Bethe-Takahashi Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

String Hypothesis and Bethe Equations . . . . . . . . . . . . . . . . 21String Hypothesis and Logarithmic Bethe Equations . . . . . . . . . 23Energy and Total Momentum for the String Hypothesis . . . . . . . 24

4 Classification of States 254.1 Possible Quantum Number Sets . . . . . . . . . . . . . . . . . . . . . . . . . 254.2 Possible Symmetry Transformations . . . . . . . . . . . . . . . . . . . . . . 27

Definition of the Symmetry Transformation Sj,α . . . . . . . . . . . 29Properties of the Symmetry Transformation Sj,α . . . . . . . . . . . 30

4.3 Restrictions on the Quantum Number Sets . . . . . . . . . . . . . . . . . . 31Restrictions for Case (a) . . . . . . . . . . . . . . . . . . . . . . . . . 32Restrictions for Case (b) . . . . . . . . . . . . . . . . . . . . . . . . . 33Restrictions for Case (c) . . . . . . . . . . . . . . . . . . . . . . . . . 34

i

TABLE OF CONTENTS

4.4 Uniqueness and Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . 36Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.5 Counting Sets of Quantum Numbers . . . . . . . . . . . . . . . . . . . . . . 40Rectangular Young Diagrams . . . . . . . . . . . . . . . . . . . . . . 40Young Diagrams with only Distinct Rows . . . . . . . . . . . . . . . 41A. Two Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41B. Three Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43C. Arbitrary Number of Particles . . . . . . . . . . . . . . . . . . . . 47Arbitrarily Shaped Young Diagram . . . . . . . . . . . . . . . . . . . 49

4.6 Cardinality and Completeness of Sets . . . . . . . . . . . . . . . . . . . . . 514.7 The Phase Space of the Anti-ferromagnet . . . . . . . . . . . . . . . . . . . 55

5 String Deviations 595.1 Completeness of Heisenberg spin chains . . . . . . . . . . . . . . . . . . . . 59

String Deviations in the Gapped XXZ Case . . . . . . . . . . . . . 60Narrow and Wide Pairs . . . . . . . . . . . . . . . . . . . . . . . . . 61String Deviations for 2-Strings . . . . . . . . . . . . . . . . . . . . . 62

6 Conclusion 67Summary and Conclusion . . . . . . . . . . . . . . . . . . . . . . . . 67Outlook . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

A Bethe Equations for the XXX and the Gapless XXZ Case 69The Isotropic Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69The Gapped Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

B Supplemental Equations to Determine String Deviations 73

Bibliography 77

ii

Acknowledgements

First of all, I want to thank my supervisor Jean-Sebastien Caux. With his energy andcontagious enthusiasm he reminded me once more, that doing physics is actually fun. Ienjoyed this year of collaboration and I am especially grateful for his patience when an-swering my questions - even the trivial ones -, his support in finding a PhD position, andabove all, making my attendance at the summer school in Les Houches possible.

I want to thank the whole group - or better say gang - of J.-S. for supporting me andhelping me out whenever they could. For instance, not a single member missed my testtalk for the interview in Dresden and them giving me extremely useful feedback helpedto make it a success. Moreover, it was a great atmosphere within the gang and I reallyenjoyed working alongside so many great young physicists.

Furthermore, I want to thank all my family! Specifically, my grandparents Helga, Ludwigand Hans, my parents Franz and Brigitte and my wonderful siblings Teresa, Sophia andJohannes. Everyone of you supported my stay in Amsterdam in one way or other andenabled me a joyful and successful two years. It is unfortunately not given to have a lov-ing and supporting family, but I know that I have one and I am extremely grateful for that.

I want to say thank you to all my friends, all these inspiring and fun people, especiallythe ones that I got to know during the last two years. You made my time in Amsterdamsuch a great experience and you are the reason, that my farewell from Amsterdam will beso immensely difficult.

Last but not least, I want to thank my boyfriend Julian for all his professional and mentalsupport. Without you, I would have never graduated. Thanks for all the days and nightsyou spent helping me out whenever I almost drowned in my self-made chaos or whenevermy confidence as a physicist threatened to leave me for good. You never let me down andI am so happy about all the time that we have spent and will spend together.

iii

Abstract

The magnetic properties of many materials are accurately captured by the Heisenbergmodel. If the lattice structure possesses a much stronger interaction in one spatial di-mension it behaves as an effective one-dimensional system described by a Heisenberg spinchain. Also in experiments with cold atoms Heisenberg spin chains have proven useful asthey provide the theoretical backbone for the spin-spin interactions of the cold atoms.

The exact eigenstates of Heisenberg spin chains are provided by the famous Bethe ansatz.The eigenstate basis is one of the key ingredients to calculate form factors, correlationfunctions and other physical observables that are then measured in experiments. In orderto obtain all eigenstates in the framework of the Bethe ansatz, a complete classification ofstates is needed.In the specific case of the gapped XXZ Heisenberg spin-1/2 chain the still missing classi-fication of states in terms of quantum numbers is presented in this thesis. Only by usingthe periodicity of the momenta describing the wave function of an eigenstate, the fullclassification is obtained. Counting the eigenstates then gives rise to the completeness ofthe eigenstate basis. However, deviations of bound states have to be taken into account,that hinder a rigorous proof of completeness by solely counting states. These deviationsare partially discussed as well.

Such a classification of states is especially useful for algorithms like ABACUS (by J.-S.Caux) that are designed to effectively calculate e. g. spin-spin correlation functions.For the gapped XXZ Heisenberg spin-1/2 chain, ABACUS is missing such a classificationwhich leads to an overcounting of states when calculating certain types of correlation func-tions. The main motivation of this thesis is to fix the overcounting problem of ABACUS.

v

Chapter 1

Introduction

”You can recognize truth by its beauty and simplicity.

When you get it right, it is obvious that it is right – at least if you have anyexperience – because usually what happens is that more comes out than goes in.“

Richard Feynman [1]

1.1 The Power of the Bethe Ansatz

It was Enrico Fermi’s simplicity in approaching problems [2, pp. 193-195] that inspiredHans Bethe to formulate his famous ansatz [3], nowadays known as coordinate Betheansatz. It was published almost a century ago, in 1931, during Bethe’s research visit inRome, which he primarily undertook to meet with Fermi. The Bethe ansatz reveals theexact eigenstates and eigenfunctions of a certain class of one-dimensional models. In [3],Bethe explicitly found, thanks to his great intuition, the solutions to one particular system- the isotropic Heisenberg spin-1/2 chain. Heisenberg spin chains are the one-dimensionalversions of the Heisenberg model, which is an effective model describing spin interactionsin atomic lattices, introduced around three years earlier by Heisenberg [4] (and almostsimultaneously by Dirac) in order to explain magnetism.Bethe wanted to generalize this idea to two or three dimensional lattices (at least he statesso in his paper [3]) but he either never tried or, more likely, tried and failed. Fact is, hismethod is not transferable to arbitrary models especially not higher dimensional modelsand the reason why is integrability.

Notion of Integrability

The notion of integrability comes from classical mechanics. A well-known example, thatis often used to illustrate the difference between integrability and non-integrability, is theKepler problem and its extension to a three body problem. For the former, a generalsolution to the differential equations for arbitrary initial conditions is found by integrationand is therefore coined an integrable model. For the latter, there is no general closed-formsolution [5]: The three body problem is in general non-integrable. The reason lies in thenumber of conservation laws, if one follows the Liouville notion of integrability [6]. Aclassical system with Hamiltonian H is integrable if it has a 2n-dimensional phase spaceand n first integrals in involution, e.g. n linearly independent functions f that are Poisson-commuting among each other and with the Hamiltonian H.The power of the Bethe ansatz - coordinate and algebraic - is that it solves most quantumintegrable models. In contrast to the classical case however, it is less clear what a goodnotion of integrability is [7]. It is clearly not enough to change the Poisson brackets intocommutation relations, since then every finite dimensional quantum systems would beintegrable. In fact, quantum integrable systems have infinitely many conservation lawsin contrast to non-integrable ones. A more accurate statement is thus, that systems areintegrable if they satisfy the Yang-Baxter equation (after the independent work of Yang [8]

1

Chapter 1 - Introduction

and Baxter [9]). In one dimensional integrable systems, the eigenstates are characterizedby a set of quantum numbers leading to quasi particles that represent the fundamentalexcitations of the system. Two particles scatter with each other by either preserving(identity) or exchanging (permutation) their momenta and thereby neither get creatednor destroyed in a scattering event. Additionally, many-body scattering factorizes intotwo-body scattering events in which the order does not matter. This is in words roughlywhat the Yang-Baxter equation states.The Young-Baxter equation is one of the key ingredients for the algebraic Bethe ansatzdeveloped in the 1980s by the Leningrad (St. Petersburg) school under the leadership ofFaddeev. Faddeev also developed the inverse scattering method for quantum integrablesystems [10, 11] which together with the algebraic Bethe ansatz allows for calculating formfactors and correlation functions that can eventually be measured in experiments.

Role of Integrable Systems in Many-Body Physics

Due to their exact eigenstates and eigenenergies, integrable models are ideal systems toplay with. Nevertheless, there are just a small number of integrable systems and one couldwonder why they might be of any interest?The fact that there are exactly solvable models in the past have led to developing betternumerical and analytical methods. Integrable models serve as reference models, able tocross-check algorithms and approximate methods, since they provide exact analytical an-swers as well. They not only play this role in quantum many body physics but they havealso proven very useful in high energy physics for similar reasons.Additionally, due to many breakthroughs in experiments with cold atoms in the last twodecades integrable models have had a new drive, as it is now possible to realize almost anyinteraction strength and geometry of quantum many-body systems in the lab. This haslead to new physical questions being raised. Especially in the strong interacting regime,for which methods like perturbation theory fail, tools like the Bethe ansatz are cruciallyneeded to find theoretical answers. For instance, the exact time evolution, thermodynam-ics or correlation functions can be derived for integrable Hamiltonians and have shown togive precise predictions for many experiments [12].One of the most famous examples is the Quantum Newton Cradle [13] and just recently,by using advanced integrable techniques, it has been shed some light on its dynamics [14].Out-of-equilibrium physics has gained more and more attention in the field of integrabil-ity, not least because of good experimental realizations: By switching lasers or magneticfields on and off the state of cold atoms during an experiment suddenly changes (quantumquench). Alternatively, a system can also be driven by applying periodic signals (Floquetdynamics). Both lead to interesting effects in integrable systems for which appropriate the-oretical tools have been developed (e.g. for quantum quenches [15] and Floquet dynamics[16]) and research is still ongoing.

Heisenberg Spin Chains

One of the most fruitful integrable quantum systems solved by the Bethe ansatz is theHeisenberg spin chain, in which only nearest neighbor spin couplings are taken into ac-count. Special cases are for instance the isotropic spin chain (XXX model) which isexactly the spin chain examined in Bethe’s original paper [3], or the gapless XXZ spinchain for which the coupling between spins in one spatial direction is reduced. Introducing

2


an anisotropy favoring the (anti)-alignment of spins in one spatial direction gives anotherexample - the so-called gapped XXZ model, which has shown to effectively describe manymaterials such as Sr2CuO3, SrCuO2 [17], CsCoCl3 [18] and CsCoBr3 [19], [20].The fundamental excitations of spin chains are delocalized spin waves that can be imag-ined as flipped spins over a ground state. These flipped spins can also form bound states,so-called string states. The ground state, ‘unbound’ states and string states then span theHilbert space of the Hamiltonian.String states are complicated to verify in experiments and are often only very indirectlymeasured, for instance by using quantum quenches [21]. Just recently string states forthe gapped XXZ spin chain have been directly measured for the first time in SrCo2V2O8

using high-resolution tetrahertz spectroscopy [22].

Correlation Functions and Form Factors

It was around the beginning of this millennium that great progress was made in extractingquantitative results from integrable models, especially from Heisenberg spin chains. Thatis, it was found by Kitanine et al. that form factors [23] and correlation functions [24]for finite size spin chains at zero temperature can be derived by using the algebraic Betheansatz.Of course, pioneering work beforehand has paved the way: We have already mentioned thealgebraic Bethe ansatz and the quantum inverse scattering method [25]. Moreover, theimportant norm of Bethe wave functions was already known since the late sixties [26], [27],and another crucial step was taken by Slavnov in 1989, when he calculated the overlap(inner product) of Bethe states [28].We will give some more explanations in the following, highlighting the importance ofstructure factors and correlation functions. The Fourier transformation of the two-pointcorrelation function 〈Saj (t)Saj′(0)〉 at zero temperature is the dynamical structure factor(DSF) Saa(k, ω)

Saa(k, ω) =1

N

N∑j,j′=1

exp(−ik(j − j′))∫ ∞−∞

dt exp(iωt)〈Saj (t)Saj′(0)〉 (1.1)

where Saj are the spin-1/2 operators of the jth site, N is the system size and α = z,+,−.A more detailed description of the spin operators is given later on. The notation Saj stands

for the hermitian conjugate (Saj )†. The dynamical structure factor is experimental acces-sible by measuring the cross section in neutron scattering experiments and is therefore ofgreat interest.Reshaping the dynamical structure factor makes it more suitable for numerical calcula-tions. Hence, another Fourier transform of the operators Sak =

∑j exp(−ikj)Saj is used

and after summation over the lattice sites and integration over time we get the so-calledLehmann series representation

Saa(k, ω) =2π

N

∑µ

|〈λ0|Sak |µ〉|2δ(ω − Eµ + E0). (1.2)

Here, |λ0〉 is the ground state with ground state energy E0, the states |µ〉 with energy Eµare called intermediate eigenstates and the matrix element is the form factor 〈λ0|Sak |µ〉. Itis summed over the whole Hilbert space, which for the spin-1/2 chain is of order 2N , hence

3


exponentially growing in system size. The convenient property however is, that not a lotof states are significantly contributing to the DSF. Therefore, applying certain sum rulesthat determine the important summands give the DSF up to 99, 9% even though only afew states were considered.

Numerical Methods - ABACUS

The calculation of the DSF including form factors or correlation functions remains a verydifficult task and advanced numerical methods have to be applied. Designed for thesecalculations is for instance ABACUS (Algebraic Bethe Ansatz-based Computation of Uni-versal Structure factors) [29], a state-of-the-art algorithm developed by J.-S. Caux tocalculate correlation functions for the prototypical Bethe ansatz solvable models includingspin-1/2 Heisenberg spin chains. A significant advantage of ABACUS is that it can dealwith a large number of sites N thanks to the efficient implementation of the sum rules.Finding the eigenenergies by exactly diagonalizing the Hamiltonian is in principle possible,but becomes in most cases unsuitable when larger systems are considered: As the Hilbertspace increases exponentially in system size one runs into the curse of dimensionality.Nevertheless, this method can serve as comparison for very small system sizes.ABACUS has shown to give excellent results for various systems: The first calculationsdone involved the dynamical structure factor for the gapless XXZ spin chain in a mag-netic field [30, 31]. Various calculations of correlation functions followed for XXX andXXZ spin chains [32], [33] as well as for other integrable models like the Lieb-Linigermodel [34]. Among others, ABACUS has also been applied to study quench dynamics inthe Lieb-Liniger model [35] and it was additionally involved for analyzing the dynamicsof the Quantum Newton Cradle [14].Furthermore, the spectrum of low energy excitations in the anti-ferromagnetic XXZ spinchain, so called spinons, and the corresponding correlators have been analyzed with thesupport of ABACUS [36], [37]. They had been measured in experiments [19], [20] before-hand. In the gapped regime however, only ’unbound’ states have been involved in thisanalysis.To yield quantifiable theoretical results for experiments involving string states (for instance[22]) the corresponding form factors have to be included in the DSF. Unfortunately, ABA-CUS runs into a problem with the representation of eigenstates in terms of quantumnumbers in the case of the gapped XXZ-Heisenberg spin chain. This is to be specified inthe following section.

Classification of States

One of the key ingredients needed to calculate the DSF, and in fact also almost any otherphysical quantity, is the eigenstate basis spanned by the mentioned excitations of the spinchain. Therefore, we have to find a unique description of quantum numbers that help toconstruct all the eigenstates. Broadly speaking, for a spin chain with N sites there areN possible quantum numbers that combined in all possible ways (there might be somerestrictions) describe the different states. They only have to obey the generalized Pauliprinciple, i.e. no two quantum numbers belonging to the same (the state defining) set cancoincide. In the case of the gapped spin chain however, there is too much of a choice forquantum number sets. It turns out that two different combinations still may describe thesame eigenstate and hence the description in terms of quantum numbers is not unique.

4


ABACUS is also still missing a correct classification of all states for the gapped XXZspin chain and as a consequence, the same eigenstates are either counted multiple timesor too few eigenstates are considered distorting the DSF. We therefore seek for stricterconditions on the set of quantum numbers. By doing so, we will classify the eigenstatesin terms of quantum numbers. Such a classification of states has already been done forthe unbounded excitations in the XXZ-Heisenberg spin chain [36]. We will complete theclassification of states in this thesis by extending it to the string states.

1.2 Goal and Outline of this Thesis

This thesis aims at completing ABACUS by providing a classification for all quantumnumber sets and hence, for all eigenstates of the gapped XXZ-Heisenberg spin chain.The full classification of states is of course not solely important for ABACUS but for everyapplication the complete eigenstate basis is needed for. However, ABACUS is the promptprofiteer and will be ready - once the correct choice of quantum numbers is implemented- to compute the DSF straight away.During the classification the question of completeness of states is inevitable. Thus, itis shown by counting admissible quantum number sets that their total number matchesthe expected number of states and hence gives rise to the completeness of the Betheequations for the XXZ-model. However, completeness of states has to be discussed handin hand with string deviations - deformations of string states - as has been pointed outby Hagemans et al [38]. Besides, in small system sizes (small number of sites) stringdeviations become more and more important and they have to be considered in order tocalculate the exact eigenstate basis.The structure of the thesis is as follows. In chapter 2, the model in our focus - the gappedXXZ Heisenberg spin-1/2 chain - is introduced and in chapter 3 the coordinate Betheansatz is presented, giving the analytical tools to calculate eigenstates. The classificationof quantum number sets and a proof of completeness of the sets is done in chapter 4.Chapter 5 deals with some aspects of string deviations before the thesis ends with aconclusion and an outlook in chapter 6.

5

Chapter 2

The XXZ Heisenberg Spin Chain

2.1 The One-dimensional Heisenberg Model

This chapter deals with some properties of the Heisenberg model in one dimension (1d)and the next chapter with the Coordinate Bethe Ansatz in order to find its exact Eigen-states. All of it is well known physics and covered by several papers, books, and lecturenotes. These two parts of the thesis were written based on mainly two lecture notes [39]and [40] and the book [41] that are to be read for more details and - more importantly- are not repeatedly cited. All other sources used are referenced at the appropriate textpassages.

In one dimension, the Heisenberg model describes spins with nearest neighbor interactionon a chain or on a ring, if periodic boundary conditions (PBC) are imposed. Its mostgeneral Hamiltonian (with PBC) reads

HXY Z =N∑j=1

(JxS

xj S

xj+1 + JyS

yj S

yj+1 + JzS

zjS

zj+1

)− hz

N∑j=1

Szj , (2.1)

where N is the number of sites and Sαj with α ∈ x, y, z are the spin projection operatorsin α direction. The spin operators act on the spin sitting on site j and due to the PBC,SN+j = Sj . Jx, Jy and Jz are the coupling constants of the different spacial dimensionsx, y and z, respectively. Finally, the last sum encodes the effect of an external magneticfield along the z-axis on the spin chain with field strength hz.If the three coupling constants Jx, Jy and Jz take distinct values, we call this model XY ZHeisenberg model, which was introduced by Sutherland [42] in 1970 by relating it to theeight-vertex model and exactly solved by Baxter [9] one year later. If only Jz differs fromJx = Jy, it is called the XXZ Heisenberg model and likewise, if all coupling constants areequal Jx = Jy = Jz, it is called the XXX Heisenberg model which represents the isotropicversion of the Heisenberg spin chain. The latter equals the model Bethe initially solved [3].If Jz = 0, one arrives at either the XY model, solved by Lieb, Schultz and Mattis in 1961[43], or the XX model, depending on whether Jx 6= Jy or Jx = Jy, respectively. The lattercorresponds to non interacting spinless fermions on a lattice which can by obtained by aJordan-Wigner transformation [44].In this thesis we consider only the XXZ Heisenberg model in zero external field in whichcase the Hamiltonian reads

HXXZ = J

N∑j=1

(Sxj S

xj+1 + Syj S

yj+1 + ∆

(SzjS

zj+1 −

1

4

)), (2.2)

where ∆ · J = Jz and we denote ∆ the anisotropy. Also, a non physical constant of−1/4 · Jz is added that will come in handy later on.We have not specified the multiplicity of the spin system yet. In general, Heisenberg spinchains can be solved for arbitrary spin [45, 46], but throughout this thesis, only spin-1/2

7

Chapter 2 - The XXZ Heisenberg Spin Chain

systems are considered. In this case, the spins can either have the magnetic quantumnumber 1/2 or −1/2 referred to as

”spin up“ or

”spin down“, respectively. The spin-1/2

operators follow the SU(2) symmetry relations

[Sαj , Sβk ] = i~δjkεαβγSγj with α = x, y, z, (2.3)

where δjk is the Kronecker delta and εαβγ is the totally antisymmetric Levi-Civita symbol.They furthermore can be rewritten in terms of annihilation and creation operators

S±j = Sxj ± iSyj (2.4)

with commutation relations

[Szj , S±j ] = ±~S±j , [S+

j , S−j ] = 2~Szj , (2.5)

that act on a state as

Szj |±〉j = ±~2|±〉j , S±j |∓〉j = ~|±〉j , S±j |±〉j = 0 (2.6)

where |+〉j denotes the spin up and |−〉j the spin down state of a spin on the jth site.They can be represented by the well known Pauli matrices σαj as

Sαj =~2σαj (2.7)

with the Pauli matrices defined as

σxj =

(0 1

1 0

), σyj =

(0 −i

i 0

), σzj =

(1 0

0 −1

). (2.8)

From now on we will set ~ = 1.

In terms of the raising and lowering operators, the Hamiltonian reads

HXXZ = JN∑j=1

(1

2

(S+j S−j+1 + S−j S

+j+1

)+ ∆

(SzjS

zj+1 −

1

4

)). (2.9)

Again, one can distinguish several cases in which the parameters J and ∆ take differentregions in parameter space. The sign of the coupling J determines whether the spin chainis ferromagnetic or anti-ferromagnetic. In the former case, J is negative and the spins haveto align to minimize the system’s energy. The added constant −1/4 sets the ferromagneticground state energy to zero. In the latter case, J is positive and the spins tend to anti-align. Here, the −1/4 ensures that only anti-aligned spins contribute to the overall sum.The Eigenstates of the ferromagnet and the anti-ferromagnet are the same but changingthe sign of J flips the spectrum and thus, their ground state energy and the low-energyexcitations differ greatly from one each other.Obviously, if the anisotropy |∆| = 1, one recovers the isotropic XXX Heisenberg spinchain. The sign of the anisotropy ∆ also determines the magnetic properties. If J > 0 but∆ ≤ −1, the system changes again from a ferromagnet to an anti-ferromagnet and viceversa (up to a rotation of 180◦ around the z-axis). It effectively boils down to changing

8


the sign in front of the planar term Sxj Sxj+1 +Syj S

yj+1 and Yang and Yang [47] have shown

that this does not change the spectrum of the Hamiltonian.If we take J > 0 for now, one can distinguish three phases that are crucially dependedon the value of the anisotropy ∆: the ferromagnetic (∆ ≤ −1), the anti-ferromagneticquantum critical (−1 < ∆ < 1) and the anti-ferromagnetic phase (∆ ≥ 1). In the lastphase, the spectrum of the anti-ferromagnet is gapped between the ground state and aquasi degenerate ground state [48]. If ∆ → ∞, we arrive at the classical Ising model in1d and the system is no longer gapped but the ground state energy is exactly degenerate(see next section). If |∆| < 1, the system has a continuous spectrum [48] and is referredto as ‘gapless’. It is then in its quantum-critical phase as the competing planar term(Sxj S

xj+1 + Syj S

yj+1) becomes more important than the term in z-direction. However, the

ground state is still anti-ferromagnetic.In this thesis, the main focus lies on the XXZ Heisenberg model without external fieldwith its Hamiltonian displayed in eq. (2.9) that is

• anti-ferromagnetic (J > 0)

• gapped (∆ > 1)

• a spin-1/2 system.

2.2 Symmetries of the XXZ Spin Chain and the

Dimensionality of its Hilbert Space

Lets define the α projection of the total spin operator as

Sαtot =∑j

Sαj with α ∈ x, y, z. (2.10)

The XXX Hamiltonian (∆ = 1) commutes with all of these three projection operators

[HXXX , Sαtot] = 0 ∀ α ∈ x, y, z, (2.11)

which is easily obtained by using the commutation relations (2.3) and the XXX Hamil-tonian (2.2) in terms of the spin projection operators Sαj . In particular, one uses

[Sαj Sαj+1 + Sβj S

βj+1 + Sγj S

γj+1, S

γj ] = [Sαj , S

γj ]Sαj+1 + [Sβj , S

γj ]Sβj+1

= i~εαβγ(Sαj Sβj+1 − S

βj S

αj+1) (2.12)

with α 6= β 6= γ and α, β, γ ∈ x, y, z. This vanishes because it is summed over all sites jand PBC were assumed.Therefore, the XXX Heisenberg spin chain has a global SU(2) symmetry. However, theXXZ Heisenberg spin chain’s symmetry is broken by the anisotropy ∆ and thus, theXXZ Hamiltonian only commutes with the z-projection of the total spin operator

[HXXZ , Sztot] = 0. (2.13)

It therefore possesses a global U(1) symmetry.In addition, the XXX and XXZ spin chains also possess a discrete translational sym-metry, i. e. the spins can be shifted by multiples of the lattice spacing from one site to

9


another.It is worthwhile mentioning that for certain values of the anisotropy ∆, the spin chain hasfurthermore the sl2 loop algebra as symmetry [49]. This only can happen in the quantumcritical phase, when the anisotropy is parametrized as

∆ = cos(η) =1

2(q + q−1) with 0 ≤ η ≤ π, (2.14)

at roots of unity

q2n = 1 with n ∈ N+, and n 6= 1. (2.15)

This will be briefly revisited in subsequent discussions about the completeness of the Betheansatz.

Another important property of the spin chain is its magnetization, determined by thez-projection of the total spin operator Sztot. Its eigenvalue gives the magnetization that isalso defined as

Sztot =N

2−M (2.16)

with M denoting the number of flipped spins. Because of (2.13), the Hilbert space H ofthe whole spin chain separates into subspaces HM of fixed magnetization, e. g. with afixed number of flipped spins.To analyze the dimensionality of the total Hilbert space H lets write down an immediatebasis - the basis spanned by the basis states |b〉 that are obtained by taking tensor productsof the individual states (spin up or spin down) of all the N sites

|b〉 =N⊗j=1

|±〉j . (2.17)

Here, |±〉j means either |+〉j or |−〉j depending on the state |b〉 and every |±〉j is inde-pendent of the state |±〉k of another site k such that there are 2N different combinationsto build the product in (2.17). Hence, there are 2N linearly independent basis states |b〉,indicating the dimensionality of the Hilbert space dim(H) = 2N .Each subspace HM with fixed M has then the dimensionality dim(HM ) =

(NM

), since

the binomial coefficient calculates the number of all the different possibilities to place Mflipped spins on N sites. Summing the dimensionality of the subspaces and using thebinomial theorem, one recovers the dimensionality of the whole Hilbert space

N∑M=0

dim(HM ) =N∑

M=0

(N

M

)= 2N = dim(H). (2.18)

It is important to note that the Hamiltonian HXXZ is not diagonal in this basis. Diago-nalizing the Hamiltonian HXXZ is exactly what is achieved by the Bethe Ansatz which isexplained and performed in the next chapter.

10


2.2.1 Ground State and Excitations

It is very intuitive, that the ferromagnet is in its ground state if all the spins are alignedand indeed, such a state is an eigenstate of the Hamiltonian. The ground state in zeroexternal field is degenerate and is given by the two states

|+,+,+, . . . ,+,+,+〉 and |−,−,−, . . . ,−,−,−〉,

where, again, ‘+’ stands for spin-up and ‘−’ for spin-down.The disturbances or excitations of the ferromagnetic order are spin waves, a concept whichwas firstly introduced by F. Bloch [50] in 1930 and almost 30 years later, in 1957, detectedfor the first time by Brockhouse [51] in a neutron scattering experiment. These low energyexcitations are quantized and coined magnons. They are delocalized all over the spin chainand, as they change the magnetization Sztot of the spin chain by one, are spin-1 particles(bosons) [52, p. 98]. More remarkably, the magnon states are the exact eigenstates ofthe Hamiltonian and the magnon-magnon-scattering generalizes in such a way to higherorder magnon-scattering that the whole eigenbasis of the Hamiltonian is found by studyingthe magnon-magnon scattering within the Bethe ansatz framework, discussed in the nextchapter. Instead of scattering off, magnons also can form a bound state or so called stringstate that travels as one entity around the spin chain.Naively one would think, that the ground state of the anti-ferromagnet is given whenall spins are anti-aligned. A state in such order is called Neel-ordered state, going backto Louis Neel who was the first to describe this type of ordering [53]. However, Neel-ordered states are not even eigenstates of the Hamiltonian. Only in the classical Isinglimit (∆→∞), the degenerate ground state is given by the two Neel-ordered states:

|+,−,+, . . . ,−,+,−〉 and |−,+,−, . . . ,+,−,+〉

The true ground state of the anti-ferromagnet cannot be visualized in a classical picturebut it has still magnetization Sztot = 0. One rather has to picture the anti-ferromagneticground state as the N/2-magnon state, when N is even. If N is odd, the situation is morecomplicated as there are two degenerate ground states with Sztot = ±1/2. We will stickto the easy case and assume from now on, that N is always even. The magnons in theanti-ferromagnetic ground state are all single excitations (no bound states).The exact ground state and its energy are rigorously obtained by the Bethe ansatz. Thishas been proven by Yang and Yang [54] and they furthermore showed that for the groundstate the solution to the Bethe equations is also unique. As mentioned before, the groundstate is non degenerate, but possesses the energy gap between the ground state and thequasi degenerate ground state which scales with system size as exp(−const(∆)N) [55,p. 7].In the anti-ferromagnetic case, the low-energy excitations are best understood when goingback to the Ising limit. Flipping a spin, which is essentially removing a magnon, followedby subsequent spin flips creates Neel-ordered domains on which boundaries two spins arepointed in the same direction:

| . . . ,−,+,−,+,−,+,+,+,−,+, . . .〉 → | . . . ,−,+,+,−,+,−,+,−,+,+, . . .〉

We see, that removing one quasi particle has led to two domain walls. We can assign quasiparticles - so-called spinons - to these domain walls, also occurring for finite ∆. They are

11


fractional emergent particles with spin 1/2 (fermions), first described by Faaddev andTakhtadzhyan [56] in 1981.The spinon spectrum is as well obtained by the Bethe ansatz. Therefore, let us finallyturn to unfold this very powerful ansatz.

12

Chapter 3

The Coordinate Bethe Ansatz

Hans Bethe presented in his famous paper”Zur Theorie der Metalle“ (English:

”About

the Theory of Metals“) [3] in 1931 a method that is nowadays known as”Coordinate Bethe

Ansatz“. With this method he was able to find the exact eigenvalues and -functions of theisotropic XXX Heisenberg spin chain. With Orbachs parametrization of the anisotropy∆ this exact ansatz was successfully generalized to the XXZ spin chain in 1958 [57]. Thecoordinate Bethe ansatz was later extended to a variety of other models, for instance toδ−interacting bosons in 1d (Lieb-Liniger model) by Lieb and Liniger [58] in 1963 (see [39]and [41]).In the following, we only present the coordinate Bethe ansatz for spin chains with mainfocus on the gapped XXZ case.

3.1 Coordinate Bethe Ansatz for XXZ spin chains

We seek the exact Eigenstates of the XXZ Hamiltonian (2.9), i.e. we want to diagonalizeit. In order to do so, lets start with a reference state: the state |0〉 with only spin up stateswith magnetization N/2 (number of flipped spins M = 0)

|0〉 =N⊗j=1

|+〉j . (3.1)

This state is trivially an eigenstate of the Hamiltonian and the only state of the subspaceHM=0 as

(N0

)= 1.

A Single Flipped Spin

Lets continue with the next subspace HM=1 that has exactly one flipped spin. To accountfor the translational invariance of the spin chain, we make an ansatz for the eigenstate|ψ1〉 and write it as a superposition of all N states with one flipped spin

|ψ1〉 =

N∑j=1

Ψ1(j) · S−j |0〉 =

N∑j=1

Ψ1(j)|j〉. (3.2)

Here, we have labeled the state |j〉 = S−j |0〉 as the state with one flipped spin on site jand Ψ1(j) is the corresponding amplitude.Projecting the Schrodinger equation HXXZ |ψ1〉 = E1|ψ1〉 onto the bra state 〈j| = 〈0|S+

j

reveals

E1〈j|ψ1〉 = 〈j|HXXZ |ψ1〉

E1Ψ1(j) = 〈j|JN∑k=1

(1

2

(S+k S−k+1 + S−k S

+k+1

)+ ∆

(SzkS

zk+1 −

1

4

))|ψ1〉

E1Ψ1(j) =J

2(Ψ1(j − 1) + Ψ1(j + 1))− J∆Ψ1(j), for 1 < j < N, (3.3)

13

Chapter 3 - The Coordinate Bethe Ansatz

Thus, |ψ1〉 is an Eigenstate if its amplitude Ψ1 satisfies equation (3.3) as well as theperiodic boundary condition Ψ(1) = Ψ(N + 1). The latter is revealed by doing the samederivation in (3.3) explicitly for site 1 and site N .A plane wave ansatz Ψ1(j) = exp(ikj) solves (3.3) and leads to the following expressionfor the energy E1

E1 exp(ikj) + J∆ exp(ikj) =J

2exp(ik(j − 1)) +

J

2exp(ik(j + 1))

⇔ E1 = J(cos(k)−∆). (3.4)

Furthermore, by imposing the boundary condition we get

exp(ikN) = 1

⇒ kI =2πI

N, with I = 0, 1, . . . , N − 1. (3.5)

From that requirement it is seen, that only quantized momenta are allowed. Additionally,there are N distinct momenta kI , each of which leads to a linearly independent Eigenstate

|Ψ1〉 =∑N

j=1 exp(ikIj)|j〉. This is in agreement with the dimensionality of the one magnon

subspace dim(HM=1) =(N1

)= N .

Two Flipped Spins

Lets move on to the next sector with two flipped spins - the two magnon sector. Eventhough the same steps are followed as in the last sector, this is more complicated butluckily, with the two magnon sector in hand, we are able to generalize the wave functionsto the arbitrary M ≤ N/2 sector.We start again with a superposition of all states with two flipped spins

|ψ2〉 =∑j1<j2

Ψ2(j1, j2)|j1, j2〉 =∑j1<j2

Ψ2(j1, j2) · S−j1S−j2|0〉. (3.6)

To avoid double counting states in the superposition, j1 is chosen to be always smallerthan j2. When projecting the Schrodinger equation HXXZ |ψ2〉 = E2|ψ2〉 onto the bra〈j1, j2| = 〈0|S+

j1S+j2

, one has to distinguish two cases: Either the flipped spins occupy sitesnext to each other or not. The former case gives

(E2 + J∆)Ψ2(j1, j2) =J

2(Ψ2(j1 − 1, j2) + Ψ2(j1, j2 + 1)) , for 2 < j1 + 1 = j2 < N,

(3.7)

and the latter

J

2(Ψ2(j1 − 1, j2) + Ψ2(j1, j2 − 1) + Ψ2(j1 + 1, j2) + Ψ2(j1, j2 + 1))

= (E2 + 2J∆)Ψ2(j1, j2), for 2 < j1 + 1 < j2 < N. (3.8)

Following Bethe’s logic [3], this is solved by a wave function generalized from the formersector that is not only a simple superposition of two plane waves but has also an interactingterm (momentum exchanging term)

Ψ2(j1, j2) = A12 exp(ik1j1 + ik2j2) +A21 exp(ik2j1 + ik1j2),

with 1 ≤ j1 < j2 ≤ N. (3.9)

14


Substituting this wave function into (3.8) gives the expression for the energy

(E2 + 2J∆)Ψ2(j1, j2) =JΨ2(j1, j2)

2(exp(ik1) + exp(−ik1) + exp(ik2) + exp(−ik2))

⇔ E2 = J(cos(k1) + cos(k2)− 2∆). (3.10)

Inserting it into (3.7), using this energy expression and recalling that j1 + 1 = j2 gives arelation for its amplitudes A12 and A21

A12

A21= −1 + exp(ik1 + ik2)− 2∆ exp(ik1)

1 + exp(ik1 + ik2)− 2∆ exp(ik2)≡ − exp(−iΦ(k1, k2)), (3.11)

where Φ(k1, k2) is the scattering phase shift function defined as

Φ(k1, k2) = 2 arctan

(∆ sin k1−k2

2

cos k1+k22 −∆ cos k1−k22

). (3.12)

By setting A12 = exp(− i2Φ) and A21 = − exp( i2Φ) the wave function (3.11) is rewritten

as

Ψ2(j1, j2) = exp(ik1j1 + ik2j2 −i

2Φ(k1, k2))− exp(ik2j1 + ik1j2 +

i

2Φ(k1, k2)). (3.13)

We have not yet imposed the periodic boundary conditions, given by the four equationsthat one obtains when deriving equation (3.7) and (3.8) for site 1 and N at a time. Theseequations boil down to the conditions

Ψ2(j2, j1 +N) = Ψ2(j1, j2), Ψ2(j2 −N, j1) = Ψ2(j1, j2). (3.14)

In analogy to the M = 1 sector, the PBC restrict the momenta, since

exp(ik1N) = − exp(−iΦ(k1, k2)), exp(ik2N) = − exp(iΦ(k1, k2)) (3.15)

and in logarithmic form

Nk1 + Φ(k1, k2) = 2πI1, Nk2 + Φ(k1, k2) = 2πI2, (3.16)

with I1, I2 =1

2, . . . , N − 1

2and I1 6= I2.

Note, that this time the quantum numbers are given by half odd integers.

Arbitrary Number of Flipped Spin

At this point, one can already generalize the ansatz to arbitrary M . However, we willrestrict ourselves to M ≤ N/2. The reason lies in the choice of the reference state: Wecould have started with the state of only downwards pointed spins as reference state and aswell recover all the magnon states by flipping spins up. Because of this mirror symmetryit is enough to only consider M ≤ N/2. The wave function for arbitrary M then reads

ΨM (j1, . . . , jM ) =∏

M≥a>b≥1

sgn(ja − jb)× (3.17)

∑PM

(−1)[P ] exp

i M∑a=1

kPaja +i

2

∑M≥a>b≥1

sgn(ja − jb)Φ(kPa , kPb)

(3.18)

15


with the eigenstate

|ΨM 〉 =∑

j1<j2···<jM

ΨM (j1, j2, . . . , jM )|j1, j2, . . . , jM 〉. (3.19)

and the PBC

ΨM (j1, j2, . . . , jM ) = ΨM (j2, j3 . . . , jM , j1 +N) (3.20)

in every index.From the PBC it follows that

exp(ikaN) = (−1)M−1 exp(−i∑b6=a

Φ(ka, kb)), with a = 1, . . . ,M. (3.21)

That are the so called Bethe equations. They are M such equations guaranteeing consis-tency. Every set of M momenta ka, that satisfy these equations, gives an eigenstate of thesystem. The momenta have fermionic properties and thus, the momenta in a set have tobe distinct.These Bethe equations are visualized as follows:

k1 →

(a) 1-Magnon Sector

k1 → k2

k3kM−1

kM

. . .

(b) M-Magnon Sector

The one magnon sector (depicted on the left) consist of one wave package with momentumk1. Moving it exactly once around the closed chain (depicted as circle) does not changethe wave. In the M -magnon sector however (shown on the right), the wave package withmomentum k1 will scatter with the remaining M − 1 waves while moving once around thechain. This explains the M−1 factors on the right hand side of equation (3.21) decryptingthe phase shifts the moved wave picks up from each scattering event.In logarithmic form the Bethe equations are

Nka +∑b6=a

Φ(ka, kb) = 2πIa, with a = 1, . . . ,M, (3.22)

connecting M quantum numbers to the momenta.The eigenenergy is given by

EM = JM∑a=1

(cos(ka)−∆) (3.23)

and we can define the total momentum as sum of the individual momenta

P =∑a

ka. (3.24)

16


Before we take the next step and write the Bethe equations in terms of rapidities (quasimomenta) - the probably better known formulation - a few things should be noticed.First, the momenta ka have a 2π periodicity, implying that the wave function does notchange when a momentum ka is shifted by ±π. Since we are describing quasi particles ona lattice and the Fourier transformation of a δ-function δ(x) is given by exp(ikx), this isnot surprising. Fixing the quantum numbers I to the range 0, . . . , N − 1 implies that weonly consider the first Brillouin zone.Second, the many-body-wave function is only dependent on a scattering shift function oftwo particles. That means in particular, the many-body-interaction factorizes into two-body interactions, a remarkable observation and the very manifestation of integrability.This is no longer true for the Heisenberg model on a 2d and 3d lattice, that are as a matterof fact non-integrable.

Third, up to now the anisotropy ∆ and the interaction strength J have not been specifiedyet, though they crucially determine the physics of the spin chain. Therefore, the Betheequation introduced here are valid for all XXX and XXZ models. A distinction of thedifferent models will happen in the next step, when a parametrization in terms of rapiditiesis introduced. In the next section, the Coordinate Bethe ansatz is further derived for thegapped XXZ spin chain to illustrate the method, but for the sake of completeness thesolutions to the XXX case and the gapless XXZ case are indicated as well.

3.1.1 Bethe Equations in Terms of Rapidities

It is handy to find a parametrization of the momenta in such a way, that the scatteringshift function is no longer dependent on two momenta (two variables) but only dependenton their difference (one variable). This has also the advantage of bringing the scattering-phase-shift function in a translational invariant form.The parametrization is dependent on the anisotropy ∆, therefore, the three cases (a)|∆| = 1, (b) |∆| > 1, and (c) |∆| < 1 are distinguished. We treat the first case in thefollowing, and the others are displayed in the appendix A.The different parametrization of these three cases goes back to Orbach [57] and theanisotropy ∆ for the anti-ferromagnet reads

(a) ∆ = 1, (b) ∆ = cosh(η), (c) ∆ = cos(η) (3.25)

It has the following important advantage: We don’t have to care about the phase transitionin ∆ of the spin chain as it is hidden inside the parametrization for real η.

Bethe Equations of the gapped XXZ Heisenberg Spin Chain

We start by setting

exp(ik) =sin(λ+ iη2 )

sin(λ− iη2 ). (3.26)

in which we denote λ as rapidity and we have set η = arcosh ∆. This choice becomes moreobvious when substituting the above parametrization (3.26) into equation (3.11), which

17


had indirectly defined the scattering phase shift function

exp(−iΦ(k1, k2)) = −1 +

sin(λ1+i η2

) sin(λ2+i η2

)

sin(λ1−i η2 ) sin(λ2−i η2 )− 2∆

sin(λ1+i η2

)

sin(λ1−i η2 )

1 +sin(λ1+i η

2) sin(λ2+i η

2)

sin(λ1−i η2 ) sin(λ2−i η2 )− 2∆

sin(λ2+i η2

)

sin(λ2−i η2 )

= −2 cos(λ1 − λ2)− 2∆ cos(λ1 − λ2 + iη) + 2(∆− cosh(η)) cos(λ1 + λ2)

2 cos(λ1 − λ2) + 2∆ cos(λ1 − λ2 − iη) + 2(∆− cosh(η)) cos(λ1 + λ2)

= −sin(λ1 − λ2 + iη)

sin(λ1 − λ2 − iη)≡ exp(−iΦ(λ1 − λ2)). (3.27)

After inserting (3.27) and (3.26) into the Bethe equations (3.21), they become

Bethe Equations(sin(λj + iη/2)

sin(λj − iη/2)

)N=

M∏k 6=j

(sin(λj − λk + iη)

sin(λj − λk − iη)

)with j = 1, . . . ,M. (3.28)

Logarithmic Bethe Equations

We can work on equation (3.26) to gather an explicit expression for the parametrizationof the momenta k(λ)

k(λ) = −i ln

(sin(λ+ iη2 )

sin(λ− iη2 )

)= −i ln

(sin(λ) cosh(η2 ) + i cos(λ) sinh(η2 )

sin(λ) cosh(η2 )− i cos(λ) sinh(η2 )

)= π − 2 arctan

(tan(λ)

tanh(η/2)

)− 2π

⌊λ

π+

1

2

⌋mod (2π)

≡ π −Θ1(λ) mod (2π), (3.29)

with Θj(λ) the below defined kernel. The floor-function b. . . c is added to take care of thebranch cuts of the complex logarithm. It furthermore ensures that the information aboutthe different branches of the tangent is not lost after taking the arctangent.A very similar calculation to (3.29) can be performed for the scattering phase shift function(see last line of (3.27))

Φ(λ1 − λ2) = 2 arctan

(tan(λ1 − λ2)

tanh(η)

)+ 2π

⌊λ1 − λ2

π+

1

2

⌋≡ Θ2(λ). (3.30)

Inserting both into the logarithmic Bethe equations (3.22) we arrive at the logarithmicBethe functions in terms of rapidities

18


Bethe Equations in Logarithmic Form

2π

NIj = Θ1(λj)−

1

N

M∑k=1

Θ2(λj − λk), with j = 1, . . . ,M (3.31)

and Θj(λ) = 2 arctan

(tan(λ)

tanh( j·η2 )

)+ 2π

⌊λ

π+

1

2

⌋

and with Ij ∈

− N

2+

1

2, . . . ,

N

2− 1

2if M even

− N

2+ 1, . . . ,

N

2if M odd.

Note, that the Bethe equations in their”normal“ form (3.28) as well as in their logarithmic

form (3.31) are transcendental equations that need to be solved numerically. This is inparticular hard in the first case. In the latter case however, we can generate solutions bytaking an arbitrary set of M quantum numbers out of the N possible ones, inserting theminto the M Bethe equations and solve for it numerically. The kernel Θj(λ) is a monotonicincreasing function which ensures to find a unique solution.Recall that the momenta have to differ to render the eigenstates independent and so dothe rapidities and thus the quantum numbers. One can build

(NM

)such sets, which is

exactly the expected dimensionality of the Hilbert subspace with a given magnetizationdim(HM ) =

(NM

). In turns out, that some of the eigenstates are equal due to the 2π

symmetry of the momenta reflected in the quantum number sets. How to avoid theredundant sets is intensively discussed in chapter 4.Other independent solutions to the Bethe equations, that endow us with the missingeigenstates, are provided by complex rapidities bound together to so called

”string states“.

Energy and Total Momentum Parametrized with Rapidities

Before turning to string states, lets quickly express the eigenenergy and the total mo-mentum in terms rapidities. Using the parametrization for the momentum k(λ) (3.29) torewrite the energy EM (3.23) we find

EM = J

M∑a=1

(cos(ka)−∆) = −JM∑a=1

(cos

(2 arctan

(tan(λ)

tanh(η/2)

))+ ∆)

= −JM∑a=1

(tanh2(η/2)− tan2(λ)

tanh2(η/2) + tan2(λ)+ ∆

)= −J

M∑a=1

(cosh(η) cos(2λ)− 1

cosh(η)− cos(2λ)+ ∆

)

= −JM∑a=1

(cosh(η) cos(2λ)− 1 + cosh2(η)− cosh(η) cos(2λ)

cosh(η)− cos(2λ)

)

= −JM∑a=1

(sinh2(η)

cosh(η)− cos(2λ)

). (3.32)

19


Similarly, we find for the total momentum P (3.24)

P =

M∑a=1

ka =

M∑a=1

π −Θ1(λa) mod (2π)

= Mπ − 2π

N

M∑a=1

Ia mod (2π) (3.33)

in which we have used the logarithmic Bethe equations and the fact that the skew sym-metric kernel Θ2 was summed over symmetrically.

3.2 Complex Rapidities and String States

Complex rapidities can occur, too, which was already realized by Bethe himself [3]. Itwas shown by Vladimirov [59], that the solutions to the Bethe equations in the XXXas well as in the XXZ case are invariant under complex conjugation. Consequently, allcomplex rapidities come in complex conjugate pairs and the imaginary parts cancel whenfor instance the eigenenergy is calculated. Several pairs of complex conjugates may forma so called string state, a bound state, that is considered as one effective quasi particle.The so called string hypothesis assumes that the strings occur in very regular patterns.This is qualitatively understood by going back to the Bethe equations (3.28). In (3.28),the left hand side might either become exponentially large or small for N � 1, dependingon weather its norm is larger or smaller than 1, respectively. Therefore, the right handside has also to increase or decrease exponentially. For fixed M , this is possible if forevery complex rapidity, there exists another rapidity with imaginary part differing byapproximately ηi.Hence, in case of the gapped XXZ Heisenberg spin chain, a string complex consisting ofj bound particles takes the form

λnj,α = λj,α +iη

2(j + 1− 2n) + dnj,α (3.34)

in which n = 1, . . . , j counts through the j particles in the complex. We refer to j asthe length of a string. The number of j-strings (strings with length j) is denoted as Mj ,and α = 1, . . . ,Mj labels the different j-strings with the same length j. We call λj,αthe string center and it gives the real part of the complex if one neglects the deviationdnj,α = εnj,α + iδnj,α. For a pair of complex conjugate rapidities, the deviations must becomplex conjugates as well because of Vladimirovs statement.Usually, the string deviations are exponentially small in system size N but there are someexceptions for large but finite sizes [38], [60]. For the purpose of calculating physicalquantities like correlation functions, the excessive deformed strings might be neglected asthere are only a small fraction of those [38], [60]. Nevertheless, it is of great importance tobe aware of the string deviations especially for small system sizes. String deviations arefurther discussed in chapter 5.For now, we use the string hypothesis and assume all eigenstates of the spin chain aredescribed by (undeformed) strings. We thus neglect all deviations for now. Because of theabove mentioned exceptions, it is doubted whether the whole Hilbert space is exhaustedonly with the string hypothesis in hand. However, in the thermodynamic limit N → ∞,

20


it has provided accurate results: The exact thermodynamics for the XXX chain wheregiven by Takahashi [61], for the gapped XXZ chain by Gaudin [62] and for the gaplessXXZ chain by Takahashi and Suzuki [63].As a start, we will use the string hypothesis for the gapped XXZ spin chain to derive theBethe-Takahashi equations in the next section. In the subsequent chapter, we reason bya simple counting argument that the number of Bethe-Takahashi solutions coincides withthe dimensionality of the Hilbert space.

3.3 Bethe-Takahashi Equations

Using the string hypothesis, we for now assume only string rapidities λnj,α parametrized as

λnj,α = λj,α +iη

2(j + 1− 2n). (3.35)

The bounded rapidities in a string form perfect patterns in which their imaginary partsdiffer by exactly iη.For a given number of flipped spins M , the strings can form different patterns defined bya set of {Mj} where each Mj indicates the number of j-strings. If e. g. Mk = 0, there areno strings with length k present. We will refer to the set of {Mj} as base and to a j-stringas string of level j. The base {Mj} has to satisfy

∞∑j=1

jMj = M. (3.36)

The number of string complexes Ns in a base is simply∑Mj = Ns and determines the

number of independent parameters.

String Hypothesis and Bethe Equations

To calculate the complex rapidities, the perfect strings (3.35) are inserted in the BetheEquations (3.28). The big advantage of using string hypothesis is that we can get ridof the complex part in the Bethe equations by multiplying together the j equations forrapidities in the same string. We thus end up with a system of Ns equations

j∏n=1

(sin(λnj,α + iη/2)

sin(λnj,α − iη/2)

)N=

j∏n=1

∏(k,β,m)6=(j,α,n)

(sin(λnj,α − λmk,β + iη)

sin(λnj,α − λmk,β − iη)

). (3.37)

We will treat both hand sides of this equation separately starting with the left hand side(LHS)

sinN(λj,α + iη

2 (j − 1) + iη2

)· ... · sinN

(λj,α + iη

2 (−j + 1) + iη2

)sinN

(λj,α + iη

2 (j − 1)− iη2

)· ... · sinN

(λj,α + iη

2 (−j + 1)− iη2

)=

sinN(λj,α + j iη2

)· ... · sinN

(λj,α − j iη2 + iη

)sinN

(λj,α + j iη2 − iη

)· ... · sinN

(λj,α − j iη2

) =

sin(λj,α + j iη2

)sin(λj,α − j iη2

)N

. (3.38)

21


The last step is easy to see as soon as one realizes that the first term in the denominatorcancels with the second term in the nominator etc. such that only the last term of thedenominator and the first term of the nominator survive.We move on with the somewhat more tedious right hand side (RHS) and split the productinto three parts j∏

n=1

∏k 6=j,(β,m)

sin(λnj,α − λmk,β + iη)

sin(λnj,α − λmk,β − iη)

· j∏n=1

∏γ 6=α,m

sin(λnj,α − λmj,γ + iη)

sin(λnj,α − λmj,γ − iη)

·

j∏n=1

∏m6=n

sin(λnj,α − λmj,α + iη)

sin(λnj,α − λmj,α − iη)

. (3.39)

The last part of this expression reduces to a factor one

j∏n=1

∏m6=n

sin(λnj,α − λmj,α + iη)

sin(λnj,α − λmj,α − iη)=∏n,mn6=m

sin(iη(m− n+ 1))

sin(iη(m− n− 1))=∏n,mn6=m

sin(iη(m− n+ 1))

− sin(iη(n−m+ 1))= 1,

(3.40)

since the last product is taken over all n and m (with n 6= m) and there are thus j(j−1) =j2 − j factors which is an even amount regardless of the parity of j.The second part of (3.39) is written as

∏γ 6=α

∏n,m

sin(λnj,α − λmj,γ + iη)

sin(λnj,α − λmj,γ − iη)=∏γ 6=α

∏n,m

sin(λj,α − λj,γ + iη2 (2m− 2n+ 2))

sin(λj,α − λj,γ + iη2 (2m− 2n− 2))

. (3.41)

First, the product over m is performed. The term 2m− 2n+ 2 in the numerator takes thevalues 4−2n, 6−2n, . . . , 2j−2n, 2j+2−2n and the term 2m−2n−2 in the denominatorthe values −2n, 2 − 2n, . . . , 2j − 4 − 2n, 2j − 2 − 2n. Thus, the last two factors of thenominator and the first two of the denominator don’t cancel and we are left with

∏γ 6=α

∏n

sin(λj,α − λj,γ + iη2 (2j − 2n)) sin(λj,α − λj,γ + iη

2 (2j + 2− 2n))

sin(λj,α − λj,γ + iη2 (−2n)) sin(λj,α − λj,γ + iη

2 (2− 2n)). (3.42)

When performing the product over n in a similar fashion, only one factor cancels; thefactor with term 2j − 2n = 0 for n = j reduces against the factor with term 2 − 2n = 0for n = 1. With the following new notation

ej(λ) =sin(λ+ i jη2 )

sin(λ− i jη2 )(3.43)

this product is expressed as∏γ 6=α

(e2(λj,α − λj,γ)2 · e4(λj,α − λj,γ)2 · ... · e2j−2(λj,α − λj,γ)2 · e2j(λj,α − λj,γ)

). (3.44)

22


The very first part of the expression (3.39) works analogues to the second and becomeswith the new notation (3.43)∏k 6=j

(ek−j(λj,α − λk,β) · ek−j+2(λj,α − λk,β)2 · ... · ej+k−2(λj,α − λk,β)2 · ej+k(λj,α − λk,β)

).

(3.45)

However, if k < j, this expression can be further reduced. In fact, all factors eb−a withnegative b−a cancel with their positive counterparts ea−b. Keep in mind that ek−j occursjust once whereas e2

j−k is squared and, therefore, this product is in the general case∏k 6=j

(e|j−k|(λj,α − λk,β) · e|j−k|+2(λj,α − λk,β)2 · ... · ej+k−2(λj,α − λk,β)2 · ej+k(λj,α − λk,β)

).

(3.46)

The Bethe equations (3.28) for complex rapidities with the string hypothesis then become

ej(λj,α)N =∏

(k,β)6=(j,α)

Ejk(λj,α − λk,β), ∀j with Mj 6= 0 and α = 1, . . . ,Mj (3.47)

with Ejk = e(1−δjk)

|j−k| · e2|j−k|+2 · ... · e

2j+k−2 · ej+k (3.48)

and ej(λ) defined in (3.43). We see, that by using the string hypothesis the Bethe equationsare now only dependent on the real string centers λj,α.

String Hypothesis and Logarithmic Bethe Equations

After logarithmizing the Bethe equations in a similar fashion as done in section 3.1.1, theyare named Bethe-Takahashi equations and displayed below

Bethe-Takahashi Equations

2π

NIj,α =Θj(λj,α)− 1

N

M∑k=1(k,β)

Mk∑β=16=(j,α)

Θjk(λj,α − λk,β), (3.49)

∀ j with Mj 6= 0, and α = 1, . . . ,Mj ,

with Θj(λ) = 2 arctan

(tan(λ)

tanh( j·η2 )

)+ 2π

⌊λ

π+

1

2

⌋,

Θjk = (1− δjk)Θ|j−k| + 2Θ|j−k|+2 + · · ·+ 2Θj+k−2 + Θj+k,

and with Ij,α ∈

− N

2+

1

2, . . . ,

N

2− 1

2if Mj even

− N

2+ 1, . . . ,

N

2if Mj odd.

When searching for complex solutions to the Bethe-Takahashi equations, we draw Ns

quantum numbers for the Ns equations from N possible ones. Note, that the stringlength j for each equation decides whether the quantum number Ij,α is an integer or ahalf-odd integer. We call a specific choice of quantum numbers, i. e. a set of quantum

23


numbers {Ij,α} , a configuration. As mentioned before, not every possible configuration{Ij,α} generates a new eigenstate; some might coincide. This will be the main subject ofthe next chapter 4. Before turning to this, we seek the expressions for energy and totalmomentum in terms of complex rapidities.

Energy and Total Momentum for the String Hypothesis

We use the expression for the energy found for real rapidities (3.32) and replace the realrapidities with their complex form (3.35). Hence, we have to sum over all string lengthsj, centers α and rapidities within a string n and we derive

EM = −J∑j,α,n

sinh2(η)

cosh(η)− cos(2λnj,α)= −J

∑j,α

j∑n=1

sinh2(η)

cosh(η)− cos(2λj,α + iη(j + 1− 2n))

= −J∑j,α

sinh(η) sinh(jη)

cosh(jη)− cos(2λj,α). (3.50)

The sum over n reduces the expression to a compact form which is shown by induction.Similarly, the total momentum is derived by

P =∑j,α,n

−i ln

(sin(λnj,α + iη2 )

sin(λnj,α − iη2 )

)=∑j,α

Ij,απ −Θj(λj,α) mod (2π)

= πNs −2π

N

∑j,α

Ij,α mod (2π). (3.51)

24

Chapter 4

Classification of States

4.1 Possible Quantum Number Sets

Before we turn to the restrictions on the possible quantum number sets, let us first considerhow many quantum number configurations we are currently accounting for. When weintroduced the Bethe-Takahashi equations (3.49) in section 3.3 we allowed for N differentquantum numbers per level j. How many configurations are hence possible? To answerthis question we have to first look at the different bases. A base fulfills

∑j jMj = M .

This equals the partition of M , visualized by Young or Ferrers diagrams.The partition of M gives all the distinct ways to write the natural number M as a sumof other natural numbers. A Young diagram consists of M boxes placed in rows on top ofeach other, such that a lower row is never longer that an upper row, thus referring to oneway of expressing a natural number as sum of others. All Young diagrams with M boxesgive the complete partition. A diagram connects to a base as follows: The number ofboxes is the number of flipped spins M , the number of rows is the number of independentparameters Ns. The length of a row j is the length of a string or the level, the number ofrows with same length is Mj . Figure 4.11 depicts all Young diagrams up to 8 boxes andan example of how to connect a diagram to the configuration.

Example:

(# of boxes) ≡M

(# of rows) ≡ Ns

(length of row) ≡ j

(# of rows with length j) ≡Mj

M = 8

Ns = 3

M2 = 2

M4 = 1

Figure 4.1: Young diagrams up to eight boxes and connection to the different string bases.

We have now visualized the possible bases for a given magnetization - each base is rep-

1By R. A. Nonenmacher., CC BY-SA 4.0, https : //commons.wikimedia.org/w/index.php?curid =4766072, 15.08.18

25

Chapter 4 - Classification of States

resented by a Young diagram. For every sector with M flipped spins all possible bases{{Mj}|

∑j jMj = M

}are depicted by all Young diagrams with M boxes. In each base

the set of Bethe-Takahashi equations take a slightly different form in which we insertquantum numbers. How many configurations of quantum numbers exist for a given base?Recall, that two quantum numbers in the same level j cannot coincide as the rapiditieshave fermionic behavior but strings of different levels j are considered different excitations(’particles’) and can share a quantum number.On each level j one thus can choose Mj quantum numbers out of N possible quantumnumbers. This is visualized as follows:

The N circles stand for the possible quantum numbers that are given by the interval[Imin, Imax]. If Mj is even, Imin = −N/2 + 1/2 and Imax = N/2 − 1/2, and if Mj is oddImin = −N/2 + 1 and Imax = N/2. If a quantum number is chosen the circle is filled,thus there are Mj filled circles per line, representing a level j. The level j are sortedsuch that the shortest j is the upper most row. One configuration is depicted in such adiagram that could also be understood as particles occupying spots. Actually, there isa slightly different particle hole analogy to the quantum numbers to which we will comeback in section 4.7. For now, let us refer to strings with level j as j-particles. There areMj j-particles that can occupy N possible spots and thus there are

(NMj

)configurations.

That is why there are∏j

(NMj

)quantum number sets per base (per Young diagram) where

j runs over all levels (row lengths). Summing the number of configurations over all bases(Young diagrams) for a certain M we get∑

{Mj}∑jjMj=M

∏j

(N

Mj

)>

(N

M

)(4.1)

which is clearly more than the expected dimensionality of the Hilbert subspace dim(HM ) =(NM

): in the case of only real rapidities with M = M1 (Young diagrams with M boxes in a

column, or with M rows with one box each) we already have(NM1

)=(NM

)configurations

of quantum numbers (see section 3.1.1).It turns out that two quantum number sets may describe the same wave function. As hasbeen pointed out in section 3.1.1, the reason lies in the 2π-periodicity of the momenta k.The parametrization k(λ) in equation (3.26) shows that a ±2π-shift of the momentum kainduces a ±π-shift of the rapidity λa. A general consequence of the periodicity is thatthere exist infinitely many solutions to the Bethe equations, in most of which the rapidi-ties differ by multiples of π and then have the same wave function. This can of course be

26


remedied by restricting ourselves to the first Brillouin zone. We have already restrictedthe quantum numbers to N possible ones to account for that. However, this is seeminglynot strict enough − there are still too many sets of quantum numbers.Interestingly, this kind of over counting does not occur in the isotropic XXX chain. Itis easy to see that the kernel Θjk of the logarithmic Bethe-Takahashi equations in theXXX chain goes to a finite value for λ→∞. Therefore, it exists a quantum number Ij,∞

corresponding to an infinite rapidity λj . The quantum number Ij,∞ is thus an upper limitof quantum numbers. This imposes already enough restrictions on the choice of quan-tum numbers. Another aspect is that the solutions of the Bethe-equations in the XXXchain with only real rapidities are highest weight states of the irreducible representationsof SU(2). Therefore, one has to act with the global spin lowering operator to create theother states which is obtained by adding infinite rapidities to the state (for more detailssee [39], [64]).As firstly the gapped XXZ possesses only a U(1) symmetry and the irreducible represen-tations of U(1) are all one dimensional, and secondly the kernel Θ → ∞ for λ → ∞, wedo not have any natural restriction on configurations.Though what is the curse in the gapped XXZ chain could be its cure: The only chancewe have is to study how a ±π shift of a rapidity λ changes the quantum number configu-ration and based on this, formulate new restrictions on the sets of quantum numbers. Inother words, we seek a complete classification of quantum number sets, that generates thecorrect amount of states.The strategy is as follows: We start with defining and studying the symmetry transfor-mations linked to the ±π-jumps of rapidities in section 4.2. From there, we continue withfinding restrictions on the choice of quantum numbers in section 4.3. In the subsequentsection 4.4 it is argued why the resulting restricted set B of possible configurations isunique and complete. Furthermore in section 4.5, the possible configurations are counted,e.g. the cardinality of B is found. We can use the cardinality to show that the number ofconfigurations matches the dimensionality of the Hilbert subspace which is done in section4.6. Finally, the particle-hole excitations as an analogy to the quantum numbers and thephase diagram for some values of M and N in the anti-ferromagnetic regime are given insection 4.7.

4.2 Possible Symmetry Transformations

It has just been mentioned that k(λ ± π) = k(λ) ± 2π and therefore, shifting λ by ±πresults in the same eigenstate or wave function, a very consequence of the lattice structure.In general, the Bethe-(Takahashi) equations are coupled and if one rapidity is shifted, allthe other rapidities have to shift, too. Otherwise they would not give an Eigenstate. Sincethe momentum k(λ) is dependent only on one rapidity, shifts by π of single rapiditiesare justified and render the eigenstate unchanged. In the logarithmic Bethe equationshowever, shifting a single rapidity affects all quantum numbers in the coupled equations.To see this more explicitly, we study the resulting change of a quantum number set {Ij,α}after a single rapidity has been changed by ±π. Therefore, λ+π is plugged into the Bethe-Takahashi equations (3.49). Lets first look at the kernels Θ and how they transform under

27


the π-shift, starting with Θj [65]

Θj(λ± π) = 2 arctan

(tan(λ± π)

tanh( j·η2 )

)+ 2π

⌊λ± ππ

+1

2

⌋

= 2 arctan

(tan(λ)

tanh( j·η2 )

)+ 2π

⌊λ

π± 1 +

1

2

⌋= Θj(λ)± 2π (4.2)

The kernel Θjk is defined as a sum of Θj (see (3.49)) with

j + k − |j − k| − δjk = 2 min(j, k)− δjk (4.3)

summands. All terms apart from the edge terms occur twice (are multiplied by a factor 2)which accounts for the fact the indices are counted in steps of two. Thus, Θjk transformsas

Θjk(λ± π) = (1− δjk)(Θ|j−k|(λ)± 2π) + 2Θ|j−k|+2(λ)± 4π + . . .

· · ·+ 2Θj+k−2(λ)± 4π + Θj+k(λ)± 2π

= Θjk(λ)± (2 min(j, k)− δjk)2π. (4.4)

Going back to the Bethe-Takahashi equations, the change of quantum numbers dependson their relation to the shifted rapidity center, decrypted in their indices. The quantumnumber Ij,α corresponding to the shifted rapidity λj,α changes as

Ij,α =N

2π(Θj(λj,α)± 2π)− 1

2π

M∑k=1

(k,β) 6=

Mk∑β=1(j,α)

(Θjk(λj,α − λk,β)± (2 min(j, k)− δjk)2π)

= Ij,α ±N ∓M∑k=1

(k,β) 6=

Mk∑β=1(j,α)

(2 min(j, k)− δjk)

= Ij,α ±

N − 2∑k 6=j

min(j, k)Mk −∑β 6=α

(2j − 1)

= Ij,α ±

N − 2∑k 6=j

min(j, k)Mk − (Mj − 1)(2j − 1)

. (4.5)

If another rapidity λl,γ with (l, γ) 6= (j, α) is shifted, the quantum number Ij,α changes,too, since the kernel Θjk is summed over all rapidities

Ij,α =N

2πΘj(λj,α)− 1

2π

M∑k=1

(k,β) 6=

Mk∑β=1(j,α)

(Θjk(λj,α − λk,β)∓ δklδβγ(2 min(j, l)− δjl)2π)

= Ij,α ±M∑k=1

(k,β) 6=

Mk∑β=1(j,α)

δklδβγ(2 min(j, l)− δjl).

28


At this point we have to distinct to cases: either l = j with α 6= γ or l 6= j. The formercase gives

Ij,α = Ij,α ±M∑k=1

(k,β) 6=

Mk∑β=1(j,α)

δjkδβγ(2j − 1)

= Ij,α ± (2j − 1) (4.6)

and the latter

Ij,α = Ij,α ±M∑k=1

(k,β) 6=

Mk∑β=1(j,α)

δklδβγ2 min(j, l)

= Ij,α ± 2 min(j, l). (4.7)

We clearly see that the change of one rapidity center λjα by ±π causes a change of thewhole set of quantum numbers {Ik,β}. Summing over the transformed quantum numberset {Ik,β} gives∑

l,γ

{Ik,β} =

{Ik,β} ±N ∓ 2∑k 6=j

(min(j, k)Mk − (Mj − 1)(2j − 1))±∑k=j,β 6=α

(2j − 1)± 2∑k 6=j,β

min(j, l)

= {Ik,β} ±N. (4.8)

As expected, the quantum numbers change by N after a change of a single momentum kby 2π (or rapidity λ by π). Nevertheless, the total change of N splits over all quantumnumbers in the set and that is one of the reasons why restricting quantum numbers to Npossible ones is not enough.

Definition of the Symmetry Transformation Sj,αLets define the transformation Sj,α acting on a set of quantum numbers {Ik,β} by actingon each individual quantum number

Sj,α({Ik,β}) = {Sj,α(Ik1,β1), . . . ,Sj,α(Ikn,βn)} = {Ik,β} (4.9)

that describes the changes the set undergoes when λj,α → λj,α + π (see (4.5)-(4.7)).Therefore, Sj,α acts on a single quantum number depending on its indices as

Sj,α(Ij,α) = Ij,α +N − (2j − 1)(Mj − 1)− 2∑k 6=j

min(j, k)Mk = Ij,α (4.10)

Sj,α(Ij,β 6=α) = Ij,β + 2j − 1 = Ij,β (4.11)

Sj,α(Ik 6=j,β) = Ik,β + 2 min(j, k) = Ik,β. (4.12)

29


The inverse transformation S−1j,α describes λj,α → λj,α − π and acts on the individual

elements in the set as

S−1j,α(Ij,α) = Ij,α − (N − (2j − 1)(Mj − 1)− 2

∑k 6=j

min(j, k)Mk) = Ij,α (4.13)

S−1j,α(Ij,β 6=α) = Ij,β − (2j − 1) = Ij,β (4.14)

S−1j,α(Ik 6=j,α) = Ik,α − 2 min(j, k) = Ik,α. (4.15)

Before we will use the transformation Sj,α to restrict the sets of quantum numbers, someproperties of Sj,α are discussed that are useful for the subsequent argumentations.

Properties of the Symmetry Transformation Sj,αProperty 0. We start with a short remark on the notation: We will denote the set ofpossible configurations of quantum numbers for a given base (Young diagram) as A. Withthe set S we refer to all possible transformations Sj,α of a set in A. ’Possible’ meanshere, that obviously the indices of a transformation Sj,α have to coincide with the indicesof one quantum number in the set. Note, that there are collections of transformationsS◦j,α ◦ S◦k,β ◦ . . . included in the set S and the notation S◦j,α stands for either Sj,α or its

inverse S−1j,α .

Property 1. No transformation in S changes the indices of the quantum numbers. If westart with an ordered set of quantum numbers {Ij,α, Ij,β, Ik,γ , . . . Ik,α, Ik,β, Ij,γ . . . }, that isfirstly ordered in Latin indices and then ordered in Greek indices such that α < β impliesIj,α < Ij,β, a transformation in S applied on the set may only change the order in Greekindices Ij,α > Ij,β. We will ignore the order in Greek indices for now, e. g. α < β doesnot imply Ij,α < Ij,β and two unordered quantum number sets are considered equal. Thisimplies that transformations in S are applicable to all sets in A, in other words, there isone S per base (Young diagram).Property 2. It is now easy to see that all transformations in S commute

S◦j,α ◦ S◦k,β({Il,γ}) = S◦k,β ◦ S◦j,α({Il,γ}), (4.16)

since the action of transformations in S only adds or subtracts some constants to thequantum numbers and indices do not change. Therefore, a transformation Sj,α and itsinverse S−1

j,α are never considered in the same collection of transformations because theywould cancel each other anyway.Property 3. The next property is that quantum number shifts induced by Sj,α are alwayspositive and by S−1

j,α are always negative. Apart from (4.10) and (4.13), this is a trivialstatement. It can be shown to hold for (4.10) and (4.13), too:

N−(2j − 1)(Mj − 1)− 2∑k 6=j

min(j, k)Mk

= N + 2j − 1 +Mj − 2∑k

min(j, k)Mk. (4.17)

We can estimate the last sum as

2∑k

min(j, k)Mk ≤ 2∑k

kMk = 2M, (4.18)

30


in which the inequality changes to an equality only when j labels the longest string in theset, e.g. j > k ∀ k 6= j. Using this, we can show that

N + 2j − 1 +Mj − 2∑k

min(j, k)Mk

(4.18)

≥ N + 2j − 1 +Mj − 2M > 0, (4.19)

as we have chosen M ≤ N/2.Property 4. The quantum number Ij,α changes the most under S◦j,α as it picks up the

following summand that we denote Dsingj

S◦j,α(Ij,α) = Ij,α ± (N − (2j − 1)(Mj − 1)− 2∑k 6=j

min(j, k)Mk︸︷︷︸Dsingj

). (4.20)

This can be shown by subtracting the changes of the other quantum numbers Ij,β 6=α (4.11)and Ik 6=j,β (4.12), which is in both cases something positive:

Dsingj − (2j − 1) = N − (2j − 1)Mj − 2

∑k 6=j

min(j, k)Mk

= N +Mj − 2∑k

min(j, k)Mk

(4.18)

≥ N +Mj − 2M > 0 (4.21)

Dsingj − 2 min(j, k) = N + 2(j −min(j, k))− 1 +Mj − 2

∑k

min(j, k)Mk

(4.18)> N +Mj − 1− 2M ≥ 0, if j < k

(4.18)

≥ N + 2(j − k)− 1 +Mj − 2M > 0, if j > k. (4.22)

Property 5 : Finally, S defines an equivalence relation. If a set transforms by any transfor-mation in S into another set, they are equivalent. Reflexivity, symmetry and transitivityof this relation are obvious.Therefore, the problem of restricting quantum number sets can be seen as finding all thedistinct equivalence classes of the set of possible configurations A.

4.3 Restrictions on the Quantum Number Sets

We are searching for the subset B contained in A, in which every configuration is unique.It turns out that restricting the choice of quantum number sets such that the followingthree types of symmetry transformations are prohibited already solves the problem. Moreprecise, acting with one of these transformations on an element {Ij,α} ∈ B necessarilyimplies that {Ij,α} 6∈ B. The three transformations are:

(a) S◦j,α; jump by ±π of a rapidity

31


(b) Sj,α ◦ S−1j,β ; jump by π and jump by −π of two rapidities on the same level j

(c) Sj,α ◦ S−1k,β; jump by π and jump by −π of two rapidities on different levels j 6= k

It is later extensively discussed (see section 4.4), why the three resulting restrictions forbidany other collection of transformations as well.

Restrictions for Case (a)

How single ±π-shifts change a set of quantum numbers can directly be read off the defini-tion of the transformation S◦j,α (4.10)-(4.15). In virtue of property 4, the quantum numberof which the indices matches the single transformation changes the most by

Dsingj = N − (2j − 1)(Mj − 1)− 2

∑k 6=j

min(j, k)Mk, (4.23)

where Dsingj stands for the maximal difference after a single transformation. To prevent

single transformations S◦j,α, the interval [Imin, Imax] of N possible quantum numbers is

32


narrowed to the maximal difference Dsingj . As a consequence, no set containing Ij,α+Dsing

j

∀ j, α is considered anymore and therefore, no two sets are related by S◦j,α. We place the

new interval with width Dsingj in the middle of the initial interval [Imin, Imax] and recover

the first restriction on the sets of quantum numbers

First Condition on Quantum Number Sets

− 1

2Dsingj < Ij,α ≤

1

2Dsingj ∀j (4.24)

with Dsingj = N − (2j − 1)(Mj − 1)− 2

∑k 6=j

min(j, k)Mk.

By choice, equation (4.24) has a strict less-than sign as lower limit and a less-than-or-equalsign as upper limit to accommodate all situations of (half odd) integer quantum numberscombined with an even/odd interval width Dsing

j . This results in Dsingj possible quantum

numbers per j-string instead of N possible quantum numbers.The narrowing is dependent on the string level j, because Dsing

j is. We can make a

statement on the relationship between the widths Dsingj for different levels j.

For that we assume, every string with a certain string length l just occurs once, e.g. Ml = 1∀ l within the base. The width Dsing

l on every level l reduces to

Dsingl = N − 2

∑k 6=l

min(l, k). (4.25)

We can directly conclude Dsingk ≥ Dsing

l for k < l as the sum increases for increasing l.Actually, one can place a strict inequality with one exception: If k and l are the largestintegers in the configuration, Dsing

k = Dsingl .

Adding a l-string to the configuration gives an additional −(2l−1) to Dsingl , an additional

−2k to Dsingk and an additional −2l to Dsing

m with k < l < m. Because 2k < 2l − 1 < 2m,we directly conclude

Dsingk ≥ Dsing

l ≥ Dsingm with k < l < m (4.26)

in any base. Furthermore, the only case with an equality sign is the above mentioned.

Restrictions for Case (b)

If two or more rapidity centers λj,α/β on the same level j occur in a configuration, theircorresponding quantum numbers Ij,α and Ij,β cannot be drawn independently from thenarrowed interval. A transformation combined with an inverse transformation Sj,α ◦ S−1

j,β

encoding ±π-jumps of two string complexes on the same level j could still take place asthe net maximal change is smaller than the interval width Dsing

j .The distance of two rapidity centers λj,α and λj,β is either greater or smaller than π, sinceit only can be changed by jumps of π. It never equals π as this would imply that thereare transformed into each other by a symmetry transformation.

33


For that purpose, we are interested how quantum numbers change under Sj,α ◦ S−1j,β which

is given by

Sj,α ◦ S−1j,β (Ij,α) = Ij,α +N − (2j − 1)(Mj − 1)− 2

∑k 6=j

min(j, k)Mk − (2j − 1)

= Ij,α +N − (2j − 1)Mj − 2∑k 6=j

min(j, k)Mk︸︷︷︸Ddoubj

, (4.27)

where we have denoted the change as Ddoubj , and

Sj,α ◦ S−1j,β (Ij,β) = Ij,β −Ddoub

j .

Their width then changes as

Sj,α ◦ S−1j,β (|Ij,α − Ij,β|) = |Ij,α − Ij,β|+ 2Ddoub

j .

No other quantum number is moved as the imposed changes of both transformations canceleach other.As a consequence, we require that the relative distance between quantum numbers on thesame level j must be strictly smaller than Ddoub

j , which is the second restriction on thesets of quantum numbers [65]

Second Condition on Quantum Number Sets

|Ij,α − Ij,β| < Ddoubj ∀j, α, β (4.28)

with Ddoubj = N − (2j − 1)Mj − 2

∑k 6=j

min(j, k)Mk.

Again, the strict inequality ensures that the quantum numbers do not coincide after asingle transformation S◦j,α.It is easily seen that

Dsingj = Ddoub

j + 2j − 1, Dsingj > Ddoub

j (4.29)

and therefore, this criterion is needed, too.Furthermore, we have seen that there are quantum number sets that lead to invalid states,namely sets with quantum numbers on the same level j that, if transformed, are equal.The distance of these quantum numbers are multiples of Ddoub

j , e. g. |Ij,α−Ij,β| = n·Ddoubj

with n ∈ N. We explicitly exclude these sets from the set of possible quantum numbersets A because there are non physical and therefore not ’possible’.

Restrictions for Case (c)

The latest restriction is transferred to quantum numbers on different levels j and k. Trans-formations Sj,α ◦ S−1

k,β could still take place because, again, the maximal net change of

quantum numbers is smaller than the narrowed interval < Dsingj . For that purpose, we

have to examine the changes under such transformations of the quantum numbers Ik,max

34


and Ij,min that denote the maximal/minimal quantum number on a certain level in theset, e. g. Ik,max > Ik,β ∀ β 6= ‘ max ‘. This is

Sj,max ◦ S−1k,min(Ik,max − Ij,min)

= Ik,max − Ij,min −

N − (2j − 1)(Mj − 1)− 2∑l 6=j

min(j, l)Ml

−

N − (2k − 1)(Mk − 1)− 2∑l 6=k

min(k, l)Ml

+ 4 min(j, k)

= Ik,max − Ij,min −Dsingj −Dsing

k + 4 min(j, k) ≡ Ik,max − Ij,min − 2Ddoubj,k , (4.30)

where we have defined Ddoubj,k as

Ddoubj,k =

1

2(Dsing

j +Dsingk )− 2 min(j, k). (4.31)

With the same logic as in the previous section one could now formulate a third conditionby restricting their mutual difference. However, it is a valid state if two quantum numbersof different levels are equal. Therefore, we can be a bit more lax on the inequality signsin contrast to the last restriction, as

Ij,max − Ik,min =1

2(Dsing

j +Dsingk )− 2 min(j, k) = Ddoub

j,k (4.32)

is an allowed distance. Lets study this case in a bit more detail: Operating with bothtransformations S−1

j,max and Sk,min on each quantum number gives

S−1j,max ◦ Sk,min(Ij,max) = Ij,max −Dsing

j + 2 min(j, k) = Ij,min (4.33)

S−1j,max ◦ Sk,min(Ik,min) = Ik,min +Dsing

k − 2 min(j, k) = Ik,max. (4.34)

After the transformation the minimal/maximal quantum number of level j, k changesinto the maximal/minimal quantum number of some other set {Ij,α}, respectively. Thisstatement is true if the second condition is imposed on the set and by virtue of the fourthproperty of S. Looking at the width of the transformed quantum numbers Ij,min andIk,max of the set {Ij,α} and using equation (4.32) gives

Ik,max − Ij,min = Ik,min − Ij,max +Dsingk +Dsing

j − 4 min(j, k) (4.35)

=1

2(Dsing

j +Dsingk )− 2 min(j, k) = Ddoub

j,k

as expected.This shows that, if the difference between two quantum numbers on different levels isexactly 1

2(Dsingj + Dsing

k )− 2 min(j, k), there are two sets linked by a transformation thatdescribes the same eigenstate. In contrast to the last case this is a valid state and thus, onehas to choose which set is included in B and which not. Setting j < k and the inequalitysigns in exactly the fashion as done below is such a choice and we arrive at the thirdcondition on quantum number sets

35


Third Condition on Quantum Number Sets

Ij,max − Ik,min ≤ Ddoubj,k

Ik,max − Ij,min < Ddoubj,k

∀j, k if j < k (4.36)

with Ddoubj,k =

1

2(Dsing

j +Dsingk )− 2 min(j, k),

Dsingj = N − (2j − 1)(Mj − 1)− 2

∑k 6=j

min(j, k)Mk.

The found restrictions are depicted as follows:

On each level the number of available spots is narrowed. Furthermore, the distance be-tween every quantum number must either obey condition two (4.28) or three (4.36) sym-bolized by the orange lines.Now we are ready to discuss why the three conditions are sufficient. As mentioned before,each element in B must represent one element of all the equivalence classes of S to ensureuniqueness and completeness of quantum number sets. In other words, the elements ofthe subset B are unique in B, but every other element in the set A/B is equivalent to aquantum number set in B. This is to be shown in the next section.

4.4 Uniqueness and Completeness

This section is divided into two parts. First, we show uniqueness. More specifically,we show that no two sets in the subset B can transform into one each other by anytransformation in S. Second, completeness of sets in B is discussed. Specifically, we arguethat, indeed, there is a transformation in S that transforms every set in A/B into one ofthe sets in B. Therefore, every element in B is a representative of one equivalence class.

Uniqueness

Clearly, the conditions (4.24), (4.28) and (4.36) forbid any transformations of the formS◦j,α, Sj,α ◦ S−1

j,β and Sj,α ◦ S−1k,β, as they cause sets to leave the subset B. How can we

be sure that collections of three or more transformations in S do not transform sets backinside B again? We therefore will study ’higher order transformations’ by looking at their

36


action on a set {Is,σ} in the subset B. We will show, that the resulting set under suchtransformations {Is,σ} violates at least one of the three restrictions and thus, the resultingset is clearly 6∈ B. In order to do so, the higher order collections are categorized in threedifferent types.We start with all collections that include Sk,α and S−1

k,β that are generally written as

S◦j,γ ◦ · · · ◦ Sk,α ◦ · · · ◦ S−1k,β ◦ · · · ◦ S

◦l,δ({Is,σ}) = {Is,σ}. (4.37)

The question is if there exist any collection of such a type so that {Is,σ} is in the subsetB if {Is,σ} is in B? The answer is clearly no, because the distance between Ik,α and Ik,βis too large to agree with the second condition (4.28). This is not fixed by any othertransformation in the collection. Firstly, neither S−1

k,α nor Sk,β can be part of the collection

(see property 2) and additional Sk,α or S−1k,β enlarge the distance even more (see property

3). All other transformations S◦l,γ in the collection add either ±(2k − 1) (if l = k andγ 6= α, β) or ±2 min(j, k) (if j 6= k) to both quantum numbers. It means Ik,α and Ik,β areshifted together by other transformations so that their mutual distance stays invariant.Therefore, these collections of transformations (4.37) in S are ruled out by the secondrestriction (4.28).We turn to all uneven collections in S. For simplicity, we assume that the collectionconsists of transformations S◦j,α with only distinct Latin indices, e. g. that there is only onetransformation (or inverse) per level j in the collection. We can group an uneven collectionin pairs of transformations and we will stick with one unpaired transformation. Letspresume further, that apart from the unpaired transformation half of the transformationsare actually inverse transformations and the other half not.We have to distinguish two cases at this point. Either it is possible to build pairs of atransformation and an inverse, such that the Latin index of the unpaired transformationmatches with the smallest j (shortest string) in the set {Is,σ}

S±1j,α ◦ S

◦k,β ◦ S−◦l,γ︸︷︷︸ ◦ · · · ◦ S◦m,δ ◦ S−◦n,ε︸︷︷︸ ({Is,σ}) = {Is,σ} (4.38)

with j < l ∀ l 6= j in the collection. Then, {Is,σ} is clearly not in the subset B, as S±1j,α(Ij,α)

violates the first condition (4.24) and it is not affected from all the other pairs of trans-formations as 2 min(j, k)− 2 min(j, l) = 2j − 2j = 0.In the other case, this way of pairing is not possible as the transformation matching theshortest string length is of the underrepresented kind (transformation or inverse transfor-mation). The pairing done instead leaves the transformation S±1

k,β of the overrepresentedkind unpaired that matches the shortest possible string k. In particular, ∀ j < k in thecollection, S∓1

j,α has the opposite sign of the transformation S±1k,β. The collections are of the

form

S±1k,β ◦ S

◦j,α ◦ S−◦l,γ︸︷︷︸ ◦ · · · ◦ S◦m,δ ◦ S−◦n,ε︸︷︷︸ ({Is,σ}) = {Is,σ} (4.39)

with j < k ∀ l 6= j, k in the configuration.In such a case, the same argumentation holds in principle as for the last case. Hereadmittedly, Ik,β picks up additional summands of the form ±2k ∓ 2 min(j, k) as thereare transformations with j < k. The net shift of such a collection of transformations istherefore

Ik,β = Ik,β ±Dsingk ± 2

∑j<k

(k − j) (4.40)

37


which still conflicts with the first condition (4.24).By relaxing the assumptions made at the beginning of this argument, we can generalize thisfinding to all odd collections. Allowing for more unpaired transformations, e.g. changingthe ratio of transformations and inverse transformations, still enables us to write a part ofthe collection in the order above (either (4.38) or (4.39)). The rest of the collection onlyconsists of transformations with the same sign like the unpaired transformation S◦j,α/S◦k,β.

Consequently, Ij,α/Ik,β still violates the first condition (4.24), as due to property 3 theadditional transformations only enlarge the distances.Similarly, if we included more transformations S◦j,α, S

◦j,β, . . . with Latin index matching

the same level j, the same statement would hold when all transformations would havethe same sign. If they had different signs, the collection would be of the same type asdiscussed in (4.37) and therefore, would conflict with the second condition (4.28). Wethus realize, that no two sets in B are transformed into each other by an odd number oftransformations.It remains to show that even collections of transformations in S do not link two sets in Beither. We use the same strategy as in the afore discussed odd collections and start withhalf of the collection being inverse transformations and the other half not. Also here, onlyone transformation (or inverse transformation) per Latin index is considered. Then, wepair a transformation with an inverse such that S◦j,α corresponding to the smallest j is

paired with the transformation S−◦k,β with the smallest possible k

S±1j,α ◦ S

∓1k,β︸︷︷︸ ◦ · · · ◦ S◦l,γ ◦ S−◦m,δ︸︷︷︸ ({Is,σ}) = {Is,σ} (4.41)

with j < k in the collection.Since j is the shortest string transformed, Ij,α picks up a single ∓2j coming from S∓1

k,β

apart from the change induced by S±1j,α. All other changes induced by the other pairs of

transformations cancel. Ik,β picks up an additional ±2j from S±1j,α (we assumed j < k)

and furthermore ∓2k ± 2 min(k, l) from every pair of transformations (where l might besmaller than k).The difference between Ij,α and Ik,β was fulfilling the third condition (4.36) before thetransformation, but we will check their difference after such transformations

Ij,α − Ik,β = Ij,α ±Dsingj ∓ 2j − (Ik,β ∓Dsing

k ± 2j +∑

l, j<l<k

(∓2k ± 2 min(k, l)))

= Ij,α − Ik,β︸︷︷︸−Ddoub

j,k <Ij,α−Ik,β≤Ddoubj,k

±(2Ddoubj,k +

∑l, j<l<k

(2k − 2l)). (4.42)

To see, that the distance between the quantum numbers, indeed, does not match with thethird condition, we have to investigate two non obvious cases, namely

Ij,α − Ik,β

> Ddoub

j,k +∑

l, j<l<k

(2k − 2l)

≤ −(Ddoubj,k +

∑l, j<l<k

(2k − 2l)) ≤ −Ddoubj,k

(4.43)

Both cases are ruled out by the third condition (4.36) so that the transformed set {Is,σ}is not in the subset B. Note, that the second case conflicts, too, because of the strict

38


inequality < in the third condition (4.36).Again, choosing several transformations S◦j,α, S

◦j,β leads to similar conflicts as long as they

have the same sign and half of the transformations in the collection are inverse and theother half not. If they had different signs, the very first argument applies (see (4.37)).If we choose different ratios of transformations and inverse transformations, it is handyto divide such a collection in an odd collection of the form (4.38) or (4.39) and an oddcollection of only single transformations (or only single inverse transformations). In sucha case, the argumentation of odd collections applies.Therefore, all higher order transformations are excluded by the restrictions (4.24), (4.28)and (4.36), too, and uniqueness of the subset B is guaranteed.

Completeness

We still have to show that every set in A/B can be transformed inside the subset. We willargue it the other way around and motivate that by transforming sets in B any set in A/Bis obtained.First of all, applying the transformation Sj,α on every quantum number Ij,α in the interval

with width Dsingj recovers an interval with the same width next to the initial one. The same

is true for an interval to the left by applying S−1j,α . By applying the same transformations

over and over again all quantum numbers on Z (or Z + 1/2) are reached.Similarly, every arbitrary distance between quantum numbers Ij,α, Ij,β on the same levelj is created by applying Sj,α ◦ S−1

j,β ∀α, β. If the quantum numbers are for example nextto each other

Ij,α − Ij,β = 1 (4.44)

and a transformation is applied, the resulting distance is either

either |Sj,α ◦ S−1j,β (Ij,α − Ij,β)| = 2Ddoub

j + 1 (4.45)

or |S−1j,α ◦ Sj,β(Ij,α − Ij,β)| = 2Ddoub

j − 1. (4.46)

Otherwise, when their distance is the maximal allowed distance

Ij,α − Ij,β = Ddoubj − 1 (4.47)

it transforms to

either |Sj,α ◦ S−1j,β (Ij,α − Ij,β)| = 3Ddoub

j − 1 (4.48)

or |S−1j,α ◦ Sj,β(Ij,α − Ij,β)| = Ddoub

j + 1. (4.49)

In B there are as well all possible distances between quantum numbers in between the twocases (4.44) and (4.47). Hence, with similar transformations the distances change to allvalues in between Ddoub

j + 1 and 2Ddoubj − 1 and in between 2Ddoub

j + 1 and 3Ddoubj − 1.

Recall, that we have excluded sets where quantum numbers differ by multiples of Ddoubj .

Hence, by applying the same transformation more than once, any allowed distance betweenquantum numbers is created.A very similar argument can be given for distances of quantum numbers Ij,α, Ik,β ondifferent levels j 6= k. Thus, the argumentation is not repeated here.We conclude, that by applying transformations in S on the sets in B any set with arbitraryquantum numbers and distances between quantum numbers are obtained. Hence, everyset in A/B is connected by a transformation to a set in B.

39


4.5 Counting Sets of Quantum Numbers

We search for the cardinality of the subset B denoted as n(B), since we want to know if we,at least in principle, have enough eigenstates to span the Hilbert space. The elements inB generate the distinct states for one particular base (Young diagram). A distinct set B isassigned to every Young diagram. Summing the cardinality n(B) over all Young diagramswith M boxes should reveal

(NM

), the dimensionality of the Hilbert subspace dim(HM ).

Counting the sets in B for arbitrary bases is in general very messy. Luckily, there arecertain types of Young diagrams for which the counting is less challenging. They will bepresented first to give more insight into the way of counting. This is then generalizedto arbitrary Young diagrams and leads to a rather compact expression for the numberof states n(B) per base. This expression is then summed over all bases to show that therestrictions reveal the correct number of states matching the dimensionality of the Hilbertspace.

Rectangular Young Diagrams

The first type consists of all bases with only string complexes on the same level j. Inpictures, this describes the Young diagrams with rectangular shape as then all rows havethe same length.In such a case, only the first and second condition ((4.24) and (4.28)) have to be obeyed:We have to choose Mj quantum numbers out of Dsing

j possible quantum numbers such

that their mutual distance is always smaller than Ddoubj . The quantities Dsing

j and Ddoubj

truncate to

Dsingj = N − (2j − 1)(Mj − 1), Ddoub

j = N − (2j − 1)Mj . (4.50)

Using the particle picture for now, we count the ways of placing the Mj j-particles on

Ddoubj spots and additionally count how often Ddoub

j fits inside Dsingj . The latter counting

is achieved by subtracting Dsingj −Ddoub

j = 2j − 1 and hence, there are 2j different ways

of placing the smaller interval Ddoubj in the larger one Dsing

j . The overlap Ddoubj − 1 of

intervals with width Ddoubj has to be subtracted.

This is depicted for the example N = 20, j = 2, Mj = 3, to improve clarity:

Figure 4.2: Only j-particles. The way of combining j-particles is depicted for the exampleN = 20, j = 2, Mj = 3.

The filled circles indicate the possible choices within the interval Ddoubj and the darker

red filled circles represent one actual choice of three quantum numbers. The shifting of

40


the interval Ddoubj is visualized as well and it is clear from the graphic that the overlap of

shifted intervals must be subtracted to not count the same states multiple times.Putting all together, the number of sets n(B) is

n(B) = 2j ·(Ddoubj

Mj

)− (2j − 1) ·

(Ddoubj − 1

Mj

)=

N

N − (2j − 1)Mj

(N − (2j − 1)Mj

Mj

)=

N

Ddoubj

(Ddoubj

Mj

)(4.51)

This way of counting was done in [36] for the special case of only real rapidities (j = 1,Mj = M and thus Ddoub

j = N −M). Therefore, the above calculation (4.51) is a general-ization of what is stated in [36].Furthermore, we can compare this result to the anti-ferromagnetic ground state whosequantum numbers are rigorously known and it has been shown by Yang and Yang [54],that the corresponding solution to the Bethe equations is unique. In fact, this has not onlybeen proven for the ground state of the entire system but also for the ground state of everysubsystem with magnetization M . We will for now only be concerned with the former.The ground state is found at M = N/2, and indeed, the two sets with quantum num-bers matching the conditions are [−N−1

2 ,−N−32 , . . . , N−1

2 ] and [−N+12 ,−N−1

2 , . . . , N−32 ], in

which the former set is associated to the ground state and the latter to the quasi degenerateground state [64]. This was also realized in [36].

Young Diagrams with only Distinct Rows

Next, we consider those bases in which there is only one j-particle per level j, i. e. Mj =1, ∀ j. We thus can drop the Greek index when labeling the quantum numbers. Pictoriallyspeaking, this describes all Young diagrams with each row having a different length.The conditions to obey here are given by the first (4.24) and the third (4.36). The formercondition refines the interval of all possible values that the quantum number Ij on levelj can have; the latter restricts which quantum numbers Ij , Ik on different levels j, k arecombined.As for the previous type, the width Dsing

j of the interval with possible quantum numberssimplifies to

Dsingj = N − 2

∑k 6=j

min(j, k). (4.52)

It was discussed in section 4.3 that Dsingj > Dsing

k if j < k with the exception that

Dsingj = Dsing

k if k is the highest and j is the second highest level and the base is of thecurrently discussed type. Hence, we deal with this exception in the following discussion.In this discussion, we firstly look at only two particles, then three and then arbitrary manywith the given constraint that they are always on different level.

A. Two Particles

We start with a system of 2 strings with length j and k, or by using the particle picture,one j-particle and one k-particle. Here, Dsing

j = Dsingk as just mentioned. In particular,

they are

Dsingj = Dsing

k = N − 2j. (4.53)

41


The allowed distance Ddoubj,k between the two particles is

Ddoubj,k = N − 4j. (4.54)

Step 1. As a first approach, we can build sets by fixing the quantum number Ik andcombining it with the range Ddoub

j,k of quantum numbers Ij , so that automatically all

quantum numbers in these sets have the correct mutual distance. There are Dsingk spots

on which the quantum number Ik could have been fixed. Thus, there is an overall factorof Dsing

k . The graphic below demonstrates how the combination for each fixed k-levelquantum number is done.

Figure 4.3: A: Two Particles. Step 1 is depicted for the example N = 10, j = 1, k = 2,Mj = Mk = 1

To be combined with the subsequent quantum number Ik, the interval Ddoubj,k is right

shifted by one if it is possible.We thus have a total of

Dsingk ·Ddoub

j,k = (N − 2j)(N − 4j) (4.55)

distinct sets.There are still two ways of combining quantum numbers for that we have not accountedyet.Step 2. Firstly, recall that there is either < or ≤ in the third condition (4.36) dependingon how we subtract two quantum numbers from one each other. Therefore, we still cancombine a fixed quantum number Ik with Ij = Ik +Ddoub

j,k as we have chosen j < k. Sucha combination is possible in

Dsingj −Ddoub

j,k = N − 2j − (N − 4j) = 2j (4.56)

cases (see (a)), as otherwise the interval Dsingj is exceeded. In later discussions, we refer

to these combinations as ’new right sided combinations’.Step 3. Secondly, the allowed width between quantum numbers is mainly independentof their ordering, meaning that the quantum number Ij can differ by +Ddoub

j,k − 1 or by

−Ddoubj,k + 1 from the quantum number Ik. Therefore, there is in some cases a maximum

of 2 · Ddoubj,k − 2 combinations of Ij and a fixed Ik. In the following, we call all for this

reason missing combinations ’new left sided combinations’ (see (b)).

42


(a) ’New right sided combinations’(b) ’New left sided combinations’

Figure 4.4: A: Two Particles. Step 2 and 3 are depicted for the example N = 10, j =1, k = 2, Mj = Mk = 1

Equation (4.56) also tells us that there are 2j quantum numbers Ij , that could be combinedwith larger quantum numbers Ik > Ij in step 3. Thus, we pair each quantum number Ijwith (Ddoub

j,k − 1) quantum numbers Ik, so that there are in total

2j · (Ddoubj,k − 1) = 2j(N − 4j)− 2j (4.57)

more combinations. Note that upright combinations of Ij and Ik have already been countedin the step (4.55) and hence the −1.Adding the three ways of combining sets (4.55), (4.56) and (4.57) together, the totalnumber of combinations and thus the cardinality of B is found as

n(B) =Dsingk ·Ddoub

j,k +

2j∑l=1

(Ddoubj,k − 1) + (Dsing

j −Ddoubj,k )

= (N − 2j)(N − 4j) + 2j(N − 4j) = N(N − 4j)

= N(N − 2∑k

min(j, k)). (4.58)

B. Three Particles

Let us continue with three different particles, a j-particle, a k-particle and a l-particlewith j < k < l. First, the number of spots (interval widths) for each particle are

Dsingj = N − 2

∑k 6=j

min(j, k)Mk = N − 4j (4.59)

Dsingk = Dsing

l = N − 2j − 2k. (4.60)

The allowed distances between the particles are

Ddoubj,k = Ddoub

j,l = N − 5j − k (4.61)

Ddoubk,l = N − 2j − 4k. (4.62)

The smaller the string length, the more spots the particle has to occupy. Therefore, itis handy to start with picking the particles that have the least spots (particles on thehighest level, here l) and combine them with the others. From there, the general strategy

43


is essentially to transfer the three steps from 2 particles to 3 particles but step 2 and 3have to be repeated for every additional level. This strategy is easily generalized to moreparticles later on.Step 1. Let us define the intervals of the j- and k-particles that in every case can becombined with a l-particle fixed to one spot. For the k-particles, it is simply given byDdoubj,k . For the j-particle, there are more spots available, namely

Dsingj −Dsing

k = Dsingj −Dsing

l = 2(k − j) (4.63)

more than for k- or l-particles. Since we have organized the levels symmetrically on topof each other, there are k − j more quantum numbers on each side which we refer to asoffset. This offset together with Ddoub

j,k = Ddoubj,l gives the interval

Ddoubj,k +

1

2(Dsing

j −Dsingk ) = Ddoub

j,l +1

2(Dsing

j −Dsingl ) = Dsing

j − 2j ≡ Pj . (4.64)

We call this distance Pj and it is interestingly only dependent on j and not on k or lanymore. Therefore, we conclude that in general Pj ∀ j determines the least number ofspots that can be combined with every other k-particle with k 6= j occurring in the base.By looking at the differences of Pj for distinct j, one can check that such combinationswork in general without violating one of the restrictions (4.24), (4.28) or (4.36).For this type of Young diagrams for which the quantity Dsing

j is defined by (4.52), Pj takesthe form

Pj = N − 2∑k

min(j, k). (4.65)

In the case of three particles Pj and Pk are given by

Pj = N − 6j (4.66)

Pk = Dsingk − 2k = Ddoub

k,l = N − 2j − 4k. (4.67)

By right shifting the intervals Pj and Pk for subsequent quantum numbers Il we recoverthe combinations for the first step. This argumentation is again underlined with a graphic:

Figure 4.5: B: Three Particles. Step 1 is depicted for the example N = 14, j = 1, k =2, l = 3, Mj = Mk = Ml = 1. Not all combinations are marked but are recovered by rightshifting.

44


The right shifting is not included for the sake of visibility but it is done analogously tothe two particle case B. We thus start with the following number of combinations

Dsingl · Pj · Pk = (N − 2j − 2k) · (N − 2j − 4k) · (N − 6j). (4.68)

Step 2 and 3 for level k. The yet neglected combinations between k- and l-particles aresimply taken from the two particles discussion in A, (4.56) and (4.57). In contrast to A,this number is multiplied by Pj as we want to include a third particle.

(a) ’New right sided combinations’ (b) ’New right sided combinations’

Figure 4.6: B: Three Particles. Step 2 and 3 are depicted for the example N = 14,j = 1, k = 2, l = 3, Mj = Mk = Ml = 1. The new combinations are taken with respectto level k.

We will later comment on how the interval Pj is combined with the residual combinations.For now, we have

Pj(2k + 2k · (Pk − 1)) = 2k · Pj · Pk (4.69)

additional combinations.Step 2 for level j. First, all additional j-spots that lead to new right sided combinations(see also equation (4.56) and the accompanying explanation) are given by

Dsingj − Pj = 2j. (4.70)

Of course, we combine each of these spots with the usual Pk different k-particles (see (a)).Additionally, we can combine one new right sided k-particle with one new right sidedj-particle in 2j cases (see (b)).

(a) ’New right sided combinations’ (b) ’New right sided combinations’

Figure 4.7: B: Three Particles. Step 2 for level j is depicted for the example N = 14,j = 1, k = 2, l = 3, Mj = Mk = Ml = 1. In (a), there are only new right sided j-particlecombinations. In (b), new right sided j- and k-particles are combined.

45


Thus, we have additional

2j · Pk + 2j (4.71)

combinations. Lets shortly discuss why we have counted all ’new right sided combinations’already. The only missing combinations of ’new right sided’ j-particles in equation (4.71)are with ’new left sided’ k-particles. Going back to step 2 and 3 for level k, we did notmention how to combine the interval Pj with all combinations (see figure 4.6). In (a), it isplaced exactly as in the first step (4.68). In (b) however, we shifted it to the right by one incomparison to the first step, so that all ’new right sided’ j-particles are already combinedwith all possible ’new left sided’ k-particles. This was done on purpose to simplify thecombinations with ’new left sided’ j-particles.Step 3 for level j. At this point, only ’new left sided combinations’ for the j-particle aremissing. Again, there are also 2j spots on the left hand side of the j-line that are notcombined yet with possible spots to their right. The k-particle that is the furthest awayfrom one new left sided j-particle is at the spot

Ddoubj,k − (k − j). (4.72)

The same is true for the l-particle. Hence, (4.72) is an upper limit for combinations withthe ’new left sided’ j-particle (marked with a dashed line in the graphic below). The k-and l-particles under the limit are combined using the usual steps 1-3. Below, (a) showsstep 1 and (b) step 2 and 3.

(a) ’New left sided combinations’ (b) ’New left sided combinations’

Figure 4.8: B: Three Particles. Step 3 of level j is depicted for the example N = 14,j = 1, k = 2, l = 3, Mj = Mk = Ml = 1. In (a), step 1 between level k and l is shown.In (b), step 2 and 3 between level k and l are visualized.

In (a), the upright combination has already been accounted for. Thus, the number of yetto be combined l-particles is one less than in (4.72) and we have

2j · Pk · (Ddoubj,k − (k − j)− 1). (4.73)

In (b), step 2 and 3 are shown. The number of new left or right sided combinations is inthis case

Ddoubj,k − (k − j)− Pk = N − 5j − k − (k − j)− (N − 2j − 4k) = 2(k − j). (4.74)

46


The only difference to the usual argumentation is that combinations with the very firstl-particle on the left already have been considered. Therefore, we get one less new rightsided combination. Hence, we have

2j · ((2(k − j))(Pk − 1) + (2(k − j))− 1) = 2j · (Pk · 2(k − j)− 1). (4.75)

Putting all together, we have

2j · Pk + 2j + 2j · Pk · (Ddoubj,k − (k − j)− 1) + 2j · (Pk · 2(k − j)− 1)

= 2j · Pk · (Ddoubj,k + (k − j)) = 2j · Pj · Pk (4.76)

for the new left sided j-particles.Hence, we end up with a total of

n(B) = Dsingl · Pj · Pk + 2k · Pj · Pk + 2j · Pj · Pk = N · Pj · Pk

= N · (N − 2∑l

min(j, l)) · (N − 2∑k

min(k, l)) (4.77)

quantum number sets for three particles.

C. Arbitrary Number of Particles

Let us assume an arbitrary number of particles, but all on a different level. The z-particlecorresponds to the longest string in the configuration, i. e. the highest level. Let us defineall useful quantities as before

Dsingj = N − 2

∑k 6=j

min(j, k) (4.78)

Ddoubj,k = N −

∑l 6=j

min(j, l) +∑k 6=l

min(l, k)

− 2 min(j, k). (4.79)

Step 1 : The intervals Pj combined in the first step and the offset between the intervalwidths take the form

Pj = N − 2∑k

min(j, k) (4.80)

1

2

(Dsingj −Dsing

z

)=∑k 6=z

min(z, k)−∑k 6=j

min(j, k)

=∑k<z

k −∑k<j

k −∑k>j

j =∑k,

j≤k<z

(k − j)−∑k,

j<k<z

j

=∑k,

j<k<z

(k − j). (4.81)

As usually, the first step is

Dsingz ·

∏j<z

Pj . (4.82)

47


Step 2 and 3. The next two steps are combined to truncate some expressions. The way togo is mainly generalizing the logic of combining the j-particle in the last example. To doso, we split equation (4.71) and add the first summand to (4.73) and the second to (4.75).Both resulting equations are generalized and thus, all higher levels > j have to be takeninto account. The first becomes

∑j<z

2j ·

Ddoub

j,z −∑k,

j<k<z

(k − j)

∏k<z,k 6=j

Pk

. (4.83)

If j is the second largest integer, the inner sum is zero and we recover equation (4.69).To also generalize the second, we first need a general form of equation (4.74)

Ddoubj,k − 1

2(Dsing

j −Dsingk )− (Ddoub

k,z +1

2(Dsing

k −Dsingz )︸︷︷︸

Pk

) (4.84)

= N −∑l 6=j

min(j, l)−∑k 6=l

min(l, k)− 2j +∑l 6=j

min(j, l)−∑l 6=k

min(k, l)− Pk (4.85)

= N − 2∑k 6=l

min(l, k)− 2j −N + 2∑l

min(k, l) = 2k − 2j. (4.86)

Thus, the second resulting equation is

∑j<z

2j

2

∑k,

j<k<z

(k − j)

∏k<z,k 6=j

Pk

. (4.87)

Adding both terms (4.83) and (4.87) together gives

∑j<z

2j ·

Ddoub

j,z +∑k,

j<k<z

(k − j)

∏k<z,k 6=j

Pk

. (4.88)

Let us first massage only the inner most brackets

Ddoubj,z +

∑k,

j<k<z

(k − j) = N −

∑k 6=j

min(j, k) +∑k<z

k

− 2j +∑k,

j<k<z

(k − j)

= N −

2∑k<j

k +∑k>j

j +∑k,

j≤k<z

k

− 2j +∑k,

j<k<z

(k − j)

= N − 2∑k≤j

k −∑k>j

j − j −∑k,

j<k<z

k +∑k,

j<k<z

(k − j)

= N − 2∑k≤j

k − 2∑k>j

j = N − 2∑k

min(j, k) = Pj . (4.89)

48


Therefore, the expression truncates and summing all together gives

n(B) = Dsingz

∏j<z

Pj +

∑j<z

2j

∏j<z

Pj =

N − 2∑j<z

j +∑j<z

2j

∏j<z

Pj

= N ·∏j<z

Pj = N ·∏j<z

(N − 2

∑k

min(j, k)

). (4.90)

Arbitrarily Shaped Young Diagram

We can now find the cardinality n(B) for any base, in particular with more than oneparticle on each level. The interval Pj then takes the form

Pj = Dsingj − 2j = Ddoub

j − 1 = N − (2j − 1)Mj − 1− 2∑k 6=j

min(j, k)Mk. (4.91)

Lets define a new quantity

Qj = Pj − (Mj − 1) = N − 2∑l

min(l, j)Ml. (4.92)

Step 1. Clearly, Pj < Ddoubj , so that we just can replace all the intervals Pj by the

corresponding binomial coefficients(PjMj

)in our first approach. May z be the longest

string (highest level) in the base. Instead of fixing a single z-particle to a spot, we haveto fix a particular combination of all z-particles. Thus, Dsing

z is replaced by all possiblecombinations of z-particles. Similar to (4.51) we get

2z

(Ddoubz

Mz

)− (2z − 1)

(Ddoubz − 1

Mz

)=

(Ddoubz

Mz

)(Ddoubz + (2z − 1)Mz

Ddoubz

)=N − 2

∑l<z lMl

Ddoubz

(Ddoubz

Mz

). (4.93)

The first step becomes

N − 2∑

l<z lMl

Ddoubz

(Ddoubz

Mz

)·∏j<z

(PjMj

). (4.94)

We will use the information that we have learned from the last section. In particular, weknow in what cases combinations are possible that exceed the interval Pj . It implies thatin some cases we need to consider the binomial coefficient with Dj and not only with Pjin the top entry. Similar to the last section, we built up all missing combinations level bylevel starting with the second largest. In contrast, lower levels j (smaller j) are dependenton results from higher levels so that the derivation has some recursive character.Step 2 for level y. The second highest level describes y-particles. As mentioned, onlyintervals with width Py were considered, but the allowed distance between y-particles isactually Py + 1 = Ddoub

y . We know from before, new right sided combinations are possibleand there are 2y of them. We fix one z-particle and one new right sided y-particle suchthat in any case the conditions on the distances between quantum numbers are obeyed. To

49


count only the combinations, that haven’t been considered yet, we only distribute My − 1over Py and fix the last quantum number on the new Py + 1 site. This determines onwhich spot the z-particle with the largest possible distance has to be fixed. Comparingthis to the two particle sector in figure 4.4, the marked particles in (a) are now the fixedparticles. We then get

2y

(Py

My − 1

)((Dsingz

Mz

)−(Dsingz − 1

Mz

))= 2y

(Py

My − 1

)(Dsingz

Mz

)Mz

Ddoubz

. (4.95)

Step 3 for level y. In order to determine the new left sided combinations, we have to findout how often the interval Ddoub

z fits into the interval with all possible z-particles that canbe combined with a new left sided y-particle. The latter interval compares to the intervalmarked in figure 4.4 (b). Here, it is given by

Ddouby,z −

1

2(Dsing

y −Dsingz ) =

1

2(Dsing

y +Dsingz )− 2j − 1

2(Dsing

y −Dsingz )

= Dsingz − 2y = Ddoub

z + (2z − 2y)− 1 (4.96)

Therefore, Ddoubz fits (2z − 2y) inside (see figure 4.2) but keep in mind that the first one

was already counted in (4.94). Again, there are 2y new left sided y-particles which arefixed. The interval with z-particles is then right shifted and the overlap is subtractedanalogously to (4.51). New left and right combinations are then given by

2y

(Py

My − 1

)[(Ddoubz

Mz

)Mz

Ddoubz

+ (2z − 2y − 1)

(Ddoubz

Mz

)− (2z − 2y − 2)

(Ddoubz − 1

Mz

)]= 2y

(Py

My − 1

)(Ddoubz

Mz

)[Mz +Ddoub

z − 2Mz + (2z − 2y)Mz

Ddoubz

]= 2y

(Py

My − 1

)(Ddoubz

Mz

)[N − 2

∑l min(l, z)Ml + (2z − 2y)Mz

Ddoubz

]. (4.97)

The sum in the square brackets can be rewritten as sum over y

2∑l

min(l, z)Mz − (2z − 2y)Mz = 2∑l

min(l, y)Ml. (4.98)

The first binomial coefficient is also rewritten(Py

My − 1

)=

My

Py + 1−My

(PyMy

)=

My

N − 2∑

l min(l, y)Ml

(PyMy

)(4.99)

and thus, we end up with

2yMy

Ddoubz

(PyMy

)(Ddoubz

Mz

). (4.100)

Step 2 and 3 for level x. Moving on to the next level x and simply transferring equation(4.97), the sums do not cancel anymore and the number of combinations becomes

2xMx

Ddoubz

(PyMy

)(PxMx

)(Ddoubz

Mz

)[N − 2

∑l min(l, z)Ml + (2z − 2x)Mz

N − 2∑

l min(l, x)Ml

]. (4.101)

50


However, we need to count terms including new combinations from y- and x-particles, too.We see that Py+1 = Ddoub

y fits (2x−2y) times into the by equation (4.96) defined interval(for z → y and y → x) and we have the additional term

2xMx

Ddoubz

(PyMy

)(PxMx

)(Ddoubz

Mz

)[(2y − 2x)My

N − 2∑

l min(l, x)Ml

]. (4.102)

Adding the last two equations (4.101) and (4.102) together gives

2xMx

Ddoubz

(PxMx

)(PyMy

)(Ddoubz

Mz

). (4.103)

We can continue this procedure. For the next level, we would have a term similar to(4.97). Then, we would have to include all combinations with new sided x-particles andalso, combinations with new sided y-particles, etc.Thus, we get in general

∑j

(2jMj

Ddoubz

(Ddoubz

Mz

)∏l<z

(PlMl

))(4.104)

additional combinations to equation (4.94). Summing both leads to the total number ofcombinations

n(B) =

(N

Ddoubz

(Ddoubz

Mz

)∏l<z

(PlMl

))=

(N

Ddoubz

(Ddoubz

Mz

)∏l<z

Ql

Ddoubl

(Ddoubl

Ml

))(4.105)

If all Mj = 1 the binomial coefficient reduces with the denominator and we end up withthe above equation (4.90). Therefore, if only some particles occur once, the correspondingfactor reduces to Ql(= Pl) and the other factors keep some binomial coefficients. If wehave only one particle type as firstly discussed, the factor before the product sign is theonly thing what is left in agreement with (4.51).

4.6 Cardinality and Completeness of Sets

In an explicit notation the number of combinations for an arbitrary Young diagram isgiven by

Cardinality n(B) per Young Diagram

n(B) =M∏j=1

N − 2(1− δjM )∑

k min(j, k)Mk

N − (2j − 1)Mj − 2∑k 6=j

min(j, k)Mk

(N − (2j − 1)Mj − 2∑k 6=j

min(j, k)Mk

Mj

).

(4.106)

We see that for every string length/level, there is a factor. In this notation, j runs overall indices, also if Mj = 0. In such a case, the binomial coefficient is 1 and the numeratorand denominator reduces to 1. The exception is if MM = 0. Then the numerator is N andthe denominator cancels the numerator of the highest level in the Young diagram. Hence,the effective numerator of the highest level in a base is always N .

51


We have shown or motivated why the set B of quantum number sets is unique and completeand we have found the number of elements in B. However, we still need to show that weexceed the whole Hilbert space, e. g. that the number of quantum number sets matchesthe dimensionality of the Hilbert space. This is done in the following.We want to show that equation (4.106) if summed over the partition of M (all Youngdiagrams with M boxes) gives dim(H)M =

(NM

)∑{Mj}∑j jMj=M

n(B) =

(N

M

). (4.107)

Inspired by Bethe himself that proved for the numbers of string particles in the XXXspin chain ∑

{Mj}∑j jMj=M

∏j

(Ddoubj

Mj

)=N − 2M + 1

N −M + 1

(N

M

)(4.108)

we are able to do a similar proof for (4.107) by induction. We start by rewriting thequantity Qj (4.92) that is a function of j,N, {Mj} in a recursive fashion

Qj(N,M1,M2,M3, . . . ) = N − 2∑l

min(j, l)Ml = N − 2Ns − 2∑l

(min(j, l)− 1)Ml

= Qj−1(N − 2N2,M2,M3, . . . ) (4.109)

in which we used the number of string complexes Ns =∑

jMj . In particular, Qj becomesfor j = 1

Q1 = N − 2∑l

min(1, l)Ml = N − 2Ns. (4.110)

We use this notation to rewrite n(B) recursively as well. Therefore, let us express n(B) interms of Qj

n(B) = n(N,M1,M2, . . . ) =N

QM

∏j

QjQj +Mj

(Qj +Mj

Mj

), (4.111)

so that we can write

n(N,M1,M2, . . . ) =N

N − 2Ns

Q1

Q1 +M1

(Q1 +M1

M1

)n(N − 2Ns,M2,M3, . . . )

=N

N − 2Ns +M1

(N − 2Ns +M1

M1

)n(N − 2Ns,M2,M3, . . . ). (4.112)

The function n(N − 2Ns,M2,M3, . . . ) describes a spin chain with N − 2Ns sites, withM ′ =

∑jMj(j − 1) = M −Ns flipped spins and N ′s = Ns −M1 string complexes.

Let us furthermore define the number of combinations with fixed M and Ns as

n(N,M,Ns) =∑

{Mj} with∑jMj=Ns∑j jMj=Ns

n(N,M1,M2,M3, . . . ) (4.113)

=

Ns−1∑M1=0

N

N − 2Ns +M1

(N − 2Ns +M1

M1

)n(N − 2Ns,M

′, N ′s). (4.114)

52


In this equation it is summed over all Young diagrams of M boxes with the same numberof rows.It has to hold that

(N

M

)=

M∑Ns=1

n(N,M,Ns), (4.115)

which is another formulation of equation (4.107) we want to prove. With the conjecture

n(N,M,Ns) =N

N −M

(N −MNs

)(M − 1

Ns − 1

)(4.116)

and the Vandermode identity

k∑j=0

(m

j

)(n−mk − j

)=

(n

k

)(4.117)

we can quite simply show equation (4.115) to hold

M∑Ns=1

N

N −M

(N −MNs

)(M − 1

Ns − 1

)=

N

N −M

M∑Ns=1

(N − 1−M + 1

Ns

)(M − 1

M −Ns

)=

N

N −M

(N − 1

M

)=

(N

M

). (4.118)

It does not matter that the sum starts at Ns = 1 and not at Ns = 0 as for Ns = 0 thelatter binomial coefficient gives zero.The only part left to show is the conjecture (4.116) which is done by the promised inductionproof. We start with showing that (4.116) holds for some special cases. Firstly, there isonly one possibility for Ns = 1. Then, we only have a single M -string and the conjecturesimplifies to n(N,M, 1) = n(B) and holds as can be easily checked. Secondly, there isalso only one base for Ns = M . Then, we only have real rapidities and the similarly,the conjecture truncates to n(N,M,M) = n(B) and holds. Additionally, all cases up toM = 4 for arbitrary N are displayed below, where we use the notation ab with a the stringlength and b the number of a-strings for the bases:

53


M = 2 M = 3

base Ns n(B) = n(N, 2, Ns)

12 2 NN−2

(N−2

2

)21 1 N

base Ns n(B) = n(N, 3, Ns)

13 3 NN−3

(N−3

3

)11, 21 2 N · (N − 4)

31 1 N

M = 4 Check explicitly n(N, 4, 2):

base Ns n(B) = n(N, 4, Ns 6= 2)

14 4 NN−4

(N−4

4

)12, 21 3 N

N−4

(N−4

2

)· (N − 6)

22 2 NN−6

(N−6

2

)11, 31 2 N · (N − 4)

41 1 N

n(N, 4, 2)

= NN−6

(N−6

2

)+N · (N − 4)

= 3N2 (N − 5)

= 3NN−4

(N−4

2

)

In the next step we assume that the conjecture holds for N − 2Ns, M′ = M − Ns and

N ′s = Ns−M1 and we show that then the equation also holds for N , M , and Ns by usingthe recursive equation (4.114)

n(N,M,Ns) =

Ns−1∑M1=0

N

N − 2Ns +M1

(N − 2Ns +M1

M1

)n(N − 2Ns,M −Ns, Ns −M1)

=

Ns−1∑M1=0

N

N − 2Ns +M1

N − 2Ns

N −Ns −M

(N − 2Ns +M1

M1

)(N −Ns −MNs −M1

)(M −Ns − 1

Ns −M1 − 1

).

(4.119)

We can use the Vandermode identity again to rewrite the first binomial coefficient

N

N − 2Ns +M1

N − 2Ns

N −Ns −M

(N − 2Ns +M1

M1

)=

N

N −Ns −M

(N − 2Ns +M1 − 1

M1

)=

N

N −Ns −M

M1∑s=0

(M − 1

s

)(N − 2Ns +M1 −M

M1 − s

). (4.120)

54


We plug this back into (4.119)

Ns−1∑M1=0

M1∑s=0

N

N −Ns −M

(M − 1

s

)(N − 2Ns +M1 −M

M1 − s

)(N −Ns −MNs −M1

)(M −Ns − 1

Ns −M1 − 1

)=

Ns−1∑M1=0

M1∑s=0

N

N −Ns −M

(M − 1

s

)(M −Ns − 1

Ns −M1 − 1

)(N −Ns −M)!

(N − 2Ns −M + s)!(M1 − s)!(Ns −M1)!

=

Ns−1∑M1=0

M1∑s=0

N

N −Ns −M

(M − 1

s

)(M −Ns − 1

Ns −M1 − 1

)(N −Ns −M

Ns − s

)(Ns − sM1 − s

). (4.121)

We exchange the sums and use the Vandermode identity twice in the following

=

Ns−1∑s=0

Ns−1∑M1=s

N

N −Ns −M

(M − 1

s

)(N −Ns −M

Ns − s

)(M − s− 1− (Ns − s)Ns − s− 1− (M1 − s)

)(Ns − sM1 − s

)

=

Ns−1∑s=0

N

N −Ns −M

(M − 1

s

)(N −Ns −M

Ns − s

)(M − s− 1

Ns − s− 1

)

=

Ns−1∑s=0

N

N −Ns −M(M − 1)!

s!(M −NS)!(Ns − s− 1)!

(Ns − 1)!

(Ns − 1)!

(N −Ns −M

Ns − s

)

=

Ns−1∑s=0

N

N −Ns −M

(M − 1

Ns − 1

)(Ns − 1

s

)(N −Ns −M

Ns − s

)=

N

N −Ns −M

(M − 1

Ns − 1

)(N −M − 1

Ns

)=

N

N −M

(M − 1

Ns − 1

)(N −MNs

). (4.122)

We have arrived at the conjecture (4.116) and therefore, we conclude that if the conjectureholds for N − 2Ns, M

′ = M −Ns, N′s = Ns −M1, it holds also for N , M and Ns.

4.7 The Phase Space of the Anti-ferromagnet

We have mentioned before that there is a particle-hole excitation analogy to the quantumnumbers occupying spots. It provides an intuitive picture for the anti-ferromagnetic spec-trum - at least in the vicinity of the string hypothesis - and hence, it is explained first.Let us start with only considering quantum numbers corresponding to real rapidities (1-strings). Also the ground state at a given magnetization has always only real rapidities.The quantum numbers of the ground state are given by [54, 64]{

−M + 1

2,−M + 3

2, . . . ,

M − 1

2

}. (4.123)

Hence, all quantum numbers lie next to each other centered around zero and we can assignboundaries to the edges of these quantum numbers defining a bulk. Moving a quantumnumber outside the bulk creates thus a particle-hole excitation and the corresponding

55


eigenenergy will be higher. The second condition (4.28) tells how many particle-hole ex-citations are possible. Shifting all particles to the right or to the left but without creatingholes in between quantum numbers causes the energy to increase just slightly. The furtherthe particles are shifted from the ground state, the more the energy ascends.If a particle of the anti-ferromagnetic ground state at half filling is removed so that thenumber of flipped spins is M = N/2 − 1, the 2-spinon spectrum is recovered. TheM = N/2−1 sector has its own ground and also quasi degenerate ground state where thereare no holes in between quantum numbers. The conditions only allow for one particle holeexcitation on the (quasi degenerate) ground state giving the complete 2-spinon spectrum[36].Shifting two particles outside the bulk creates either two holes and two particles, the 4-spinon excitations, or they form a 2-string state. String states are still considered differenttypes of particles in this particle picture and thus, every base has to be visualized sepa-rately.If we for now consider bases with one type of particles only (corresponding to rectangularYoung diagrams) a ’ground state’ for this particle sector can still be defined. It turnsout that this ’ground state’ is given by generalizing the quantum number set of the trueground state for a given magnetization (4.123) as{

−Mj + 1

2,−Mj + 3

2, . . . ,

Mj − 1

2

}. (4.124)

Interestingly, there is also a ’half-filling’ analogy of these higher j-string particles, namelyif Mj = N/2j. The number of states is according to (4.106)

N

N − (2j − 1)N2j

(N − (2j − 1)N2j

N2j

)= 2j, (4.125)

which gives the number of (quasi degenerate) ’ground states’. The true ’ground state’ ofsuch a constellation is given by (4.124). According to the conditions (4.24) and (4.28), allother allowed configurations of particles are given by left or right shifting. Hence, thereare j quasi degenerate ground states given by right shifting the whole set by 1 each timeand j − 1 quasi degenerate ground states given by left shifting. Analogously to 1-stringsthe quasi degenerate energies converge if either N or ∆ is increased.If several types of particles are combined, the lowest energy is not necessarily given byplacing all quantum numbers around zero. Shifted sets combined may constitute thelowest energy of the base. Nevertheless, holes in between particles on any level result inan increase of the energy. Thus, the particle hole excitations analogy still holds.Furthermore, bound-states are more costly in energy than 1-strings. The more particlesare bound in one string, i.e. the longer the string, the higher is the eigenenergy. Therefore,the N M -string states for a given magnetization possess the highest eigenenergies and the1-strings the lowest. The energy range of any other base is determined by the number ofparticles which is equivalent to the number of rows in the corresponding Young diagram.The spectrum of two 1-strings and two 2-strings covers for instance the same energy rangeas the spectrum of three 1-strings and one 3-string.This is illustrated in figure 4.9 which shows the phase space for M = 6 flipped spins fortwo system sizes with N = 12 and N = 14. In particular, equation (3.50) and (3.51) areplotted.

56


Figure 4.9: Phase space for six flipped spins (M = 6) and N = 12/N = 14 lattice sites in(a)&(b)/(c)&(d), respectively. The energies are put in relation to the ground state energyE0, which is the energy at half filling described by the quantum numbers in (4.123). Theanisotropy is ∆ = 4 and the coupling is J = 1. All states corresponding to the same baseare depicted in the same color.

57


Figure 4.9 shows the energy spectrum for all bases in the sector M = 6 and each base(corresponding to a Young diagram) is depicted in a different color. It is clearly seen thatbases with the same number of particles (same number of rows) lie in the same energyinterval.The upper two graphs ((a) and (b)) have system size N = 12 and the lower to graphshave system size N = 14. Increasing the system size leads to a merging of energy sectorsin the phase space. Figure (a) shows in particular the (quasi degenerate) ground statesfor 1-strings, 2-strings and 3-strings that have been discussed earlier.The careful observer may have realized, that the amount of dots for a certain base may notalways match the number predicted by equation (4.106). This has two reasons: Firstly,the resolution in the phase space plots for some bases is not high enough and therefore,two dots appear as one. These plots are more meant to illustrate the above discussion thanto show quantitative results and hence, it is decided to plot the phase space in a rathercompact form to the disadvantage of the resolution. Secondly, two dots lie exactly on topof each other at total momentum 0 and π. In fact, a lot of eigenenergies are 2-fold degen-erate but differ in their total momentum. This is due to the fact, that a quantum numberset when multiplied by −1 may still give a valid set under the restrictions (4.24), (4.28)and (4.36). As a result, the corresponding rapidities of both sets only differ by a sign.The single momenta (3.29) and thus the wave function is still distinct and the eigenstatediffers, however the eigenenergy (3.50) does not. This can be easily read of equation (3.50)as the cosine of the rapidity is taken in order to calculate the eigenenergy and becausethe cosine is an even function, the sign of the rapidity does not matter. Furthermore, thetotal momentum (3.51) of two reflected sets only coincides at P = 0 or P = π due totaking the momentum modulo π.It has to be stressed again that these accidental degeneracies only occur because the totalmomentum, which is the sum of the single momenta corresponding to the rapidities, isplotted on the x-axis. The eigenstates defined by the single momenta are still distinct asjust argued. Because a multidimensional state is hardly visualized, the total momentumis taken instead which however may coincide for two states.Thus, we have explicitly shown one example for which all distinct eigenstates were ob-tained. However, it has to be taken with care as these states are calculated in the vicinityof the string hypothesis. It has been stressed throughout this thesis, that the actual eigen-states may differ because of string deviations. In fact, not only the eigenstates themselvesbut also the validity of the classification of states have to be discussed if string deviationsand violations to the string hypothesis are taken into account. This is to be done in thenext chapter.

58

Chapter 5

String Deviations

5.1 Completeness of Heisenberg spin chains

It was pointed out earlier that the completeness of Bethe eigenstates for the XXZ Heisen-berg spin chain is not proven by solely counting Bethe-Takahashi states. Nevertheless, asimilar argument of completeness based on combinatorial state counting has been done forthe XXX case by Hans Bethe himself [3] and for the gapless XXZ case by A. N. Kirillovet al. [66]. While Bethe’s combinatorial considerations are done without any specificationsof the string states, Kirillov makes use of the string hypothesis. The proof done in chapter4 completes these considerations by providing the counting for the gapped XXZ case.The reason why this is not a solid proof are string deviations [38]. The crucial point isthat because of large string deviations some string states may collapse onto the real axis.Then, additional real solutions are found as has been already realized by Bethe [3] andlater rediscovered by Essler et al. [67] in the XXX case. Just recently, a study of collapsed2-string states in the gapped XXZ case was published as well [68].Collapsed states are not only violating the string hypothesis but also render proofs of com-pleteness based on counting the string states questionable. Until now however, it couldalways be verified that additional solutions in one sector comply with missing solutionsof another sector and hence, are collapsed strings. It is therefore justified to believe, thatthe counting of strings states reveals the correct number of eigenstates, even though somestates are counted as a different type of strings than the type as which they appear in theBethe equations. Not only collapsed strings occur, but some string states are so deformedthat in order to find a numerical solution they have to be treated as separated strings inthe Bethe equations, for instance a 3-string as a 1-string and a 2-string. This is also stillconsistent with the counting argument.Another problem is encountered by so called singular configurations. If a string is orderedaround the origin, the Bethe equations may become singular. Furthermore, states then co-incide in the string hypothesis which is only lifted by taking string deviations into account[38]. Again, the counting is then not influenced, as the deviations lift the degeneracy.There is yet another reason why the Bethe equations are dubbed incomplete in the litera-ture. It has been explained in chapter 2 that the XXX model possesses a SU(2) symmetryand only highest weight states are found by the Bethe equations. Further solutions areobtained by acting with the lowering operator S− on these states which is equivalent toadding infinite rapidities [39] and thereby the complete set of eigenstates is found. As hasbeen mentioned in chapter 2, the gapless XXZ spin chain exhibits an additional sl2 loopalgebra symmetry when the anisotropy equals roots of unity [49]. Then again, only highestweight states are found and some more effort has to be made to determine the missingstates [69]. Therefore, the discussion of completeness has unfortunately been somewhatdeluded and the XXZ spin chain at roots of unity is repeatedly but unjustifiably calledincomplete in the literature [70].In the gapped phase no additional symmetry is encountered and thus, only string devia-tions have to be taken into account in order to get the complete set of proper eigenstates.

59

Chapter 5 - String Deviations

However, the deviations are not at all easy to handle as they are difficult to computenumerically and furthermore distinct types of deviations have to be treated differently.Calculating all deviations is an extreme complex and difficult task and hence, only someaspects are treated in the following: First, some features of string deviations in the gappedXXZ case are discussed. Second, the distinction between narrow and wide pairs of com-plex conjugate rapidities is explained [71], [72]. Last, it is derived how to calculate thestring deviations which are exponentially small in system size for bases consisting of one-and two-strings only.Thus, no exceptional cases like collapsed strings or singular states are treated and also,only the easiest bases are considered. This derivation is mainly done to illustrate how thedeviations are in principle calculated and to qualitatively show how to transfer the frame-work of obtaining string deviations [38] from the XXX case to the XXZ case. Finally, thedeviations in the two down spin sector are calculated and the eigenenergies are comparedwith exactly diagonalizing the Hamiltonian. This functions above all as a cross-check ofthe classification to see whether all eigenstates are correctly obtained.

String Deviations in the Gapped XXZ Case

In the recently published study [68] of the 2-down spin sector in the gapped XXZ Heisen-berg spin-1/2 chain, it was found that the system size N as well as the anisotropy ∆determine the stability of the 2-string state. That is, if the system size N and anisotropy∆ are relatively small, collapsed states are likely to occur. They were able to identify astable and an unstable region in the parameter space spanned by ∆ and N . Furthermore,it is shown that in the stable regime all string deviations are exponentially small in systemsize.This stable region probably extends to higher strings as well. Furthermore, N and ∆ bothinfluence the string deviations, which we may use to test the classification done in chapter4. In order to do so, the Bethe-Takahashi eigenenergies E calculated by using the classifi-cation of states and the string hypothesis are compared to eigenenergies Ed evaluated byexact diagonalization of the Hamiltonian using the python package provided in [73].The relative error between the eigenenergies is determined and either the system size N orthe anisotropy ∆ is varied. Both studies are done in a regime where no collapsed solutionsare expected (N < 21 [68]) as this is not accounted for in the Bethe-Takahashi equations.In figure 5.1, only the energy with th largest relative error is plotted. In both consid-ered cases string deviations become exponentially small and hence, the Bethe-Takahashieigenenergies then compare to the exact eigenenergies.Figure 5.1 underlines firstly, that string deviations are indeed dependent on the system sizeand the anisotropy, and secondly, that in the considered cases the classification of statesis done correctly. Hence, ramping up the anisotropy ∆ allows to check the classificationof states with exact diagonalization. The limit of this check is mainly given by the systemsize that the exact diagonalization method is still able to deal with.

60


(a) Relative error of eigenenergies for 2-strings (M = 2) depending on sthe ystemsize with fixed anisotropy ∆ = 4 and cou-pling constant J = 1.

(b) Largest relative error of all eigenener-gies for N = 12 lattices sites and six flippedspins (M = 6) and varying anisotropy ∆.The coupling constant is J = 1

Figure 5.1: The decrease of string deviations for increasing system size N or anisotropy∆ is shown by comparing eigenenergies. Bethe-Takahashi eigenenergies are compared toeigenenergies obtained by exact diagonalization. In both plots, only the largest relativeerror of all eigenenergies for each data point is plotted.

Narrow and Wide Pairs

In the following, string deviations for certain string states are derive following the methoddone in [38] for the isotropic case. Again, even though the method is very similar, [38] isnot repeatedly cited.As has been discussed in chapter 3, rapidities come in pairs of complex conjugate [59]. Letus denote the rapidity with positive (negative) imaginary part λ+

α,j (λ−α,j) such that

λ−j,α = (λ+j,α)?, (5.1)

where the star denotes the complex conjugate. We assume, that the real part of eachrapidity pair is distinct and that furthermore no pair of rapidities is centered around theorigin.The corresponding Bethe quantum numbers are denoted I+

j,α and I−j,α. Subtracting thelogarithmic Bethe equations (3.31) of such a pair of complex conjugate rapidities from oneeach other gives

2π

N(I+j,α − I

−j,α) =Θ1(λ+

j,α)−Θ1(λ−j,α)− 1

N

(Θ2(λ+

j,α − λ−j,α)−Θ2(λ−j,α − λ

+j,α))

− 1

N

∑(k,β),

λk,β 6=λ±j,α

(Θ2(λk,β − λ+

j,α)−Θ2(λk,β − λ−j,α))

mod (2π). (5.2)

The kernel Θj (see equation (3.31)) of the Bethe equations consists of an inverse tangentand a floor function, where the latter was introduced to guarantee the monotonicity of theformer. For this purpose, only the real part of the rapidity has to be inserted in the floor

61


function. Hence, it drops out in equation (5.2). The complex inverse tangent has branchcuts that are chosen to be

arctan(z?) =

(arctan z)? + π if z ∈ ]− i,−i∞[

(arctan z)? − π if z ∈ ]i, i∞[

(arctan z)? elsewhere.

(5.3)

Lets take the real part of the difference equation (5.2), recalling that we work modulo 2πand that <(z − z?) = 0

2π

N(I+j,α − I

−j,α) =

2

N

(arctan

(i tanh(2=(λ+

j,α))

tanh(η)

)− arctan

(−i tanh(2=(λ+

j,α))

tanh(η)

))mod (2π).

(5.4)

We thus find

I+j,α − I

−j,α =

− 1 if =(λ+

j,α) >η

2

0 if =(λ+j,α) <

η

2

(5.5)

in analogy to the XXX case [38]. We therefore see, that we have wide and narrow pairsof complex conjugate rapidities, the complex part of the former pair differs by more thaniη and the latter differs less [71], [72]. If the quantum numbers differ by exactly iη, theBethe equations are singular and this case need special attention but is left for futurework. Furthermore, the Bethe quantum numbers of narrow pairs coincide, rendering theclassification of states in terms of Bethe quantum numbers instead of Bethe-Takahashiquantum numbers almost impossible.

String Deviations for 2-Strings

In the following, expressions for string deviations are derived for bases, that only constituteof 1- and 2-strings. The strategy is as follows: a pair of complex conjugate rapidities isparametrized as a deviated 2-string and the corresponding logarithmic Bethe equationsare summed. Sending all deviations in this sum to 0 allows for a comparison to theBethe-Takahashi equations that connects Bethe quantum numbers with Bethe-Takahashiquantum numbers. Taking also the difference of the Bethe equations gives yet another setof equations that determines the deviations. Therefore, we have two independent sets ofequations obtained by summing and subtracting Bethe equations, the former reveals thestring centers and the latter the deviations.A similar, but more advanced strategy is needed to determine the deviations of higher orderstrings. If singular configurations are considered, a different approach is even needed. Bothtasks are not performed in this thesis but have to be transferred from the XXX case [38]to the gapped XXZ case in future work. The following derivation is thus only a first step.We start with parameterizing the complex conjugate rapidities of the two string as

λ±j,α = λj,α ± i(η

2+ δj,α

). (5.6)

62


For the sake of readability we will introduce the notation

λ±j,α ≡ λ± = λ± i(η

2+ δ), (5.7)

where we have suppressed all indices. The same notation is transfered to the Bethequantum numbers I. Taking the sum of the logarithmic Bethe equations then gives

2∑σ∈{+,−}

arctan

(tan(λσ)

tanh(η2 )

)=

2

N

∑σ∈{+,−}

πIσ +∑k=1,β

arctan

(tan(λσ − λk,β)

tanh(η)

)+

λσ 6=λk,β∑k=2,β

2 arctan

(tan(λσ − λ±k,β)

tanh(η)

)+ 2 arctan

(tan(λ+ − λ−)

tanh(η)

)+ 2 arctan

(tan(λ− − λ+)

tanh(η)

). (5.8)

We can rewrite the argument of the inverse tangent in terms of the real and imaginarypart which are

R(λ, µ, j) =sin(λ) cos(λ)

tanh(j·η2

)(cos2(λ) + sinh2(µ))

(5.9)

J(λ, µ, j) =sinh(µ) cosh(µ)

tanh(j·η2

)(cos2(λ) + sinh2(µ))

. (5.10)

We can use the relation

arctan(R+ i · J) + arctan(R− i · J) = ξ(R, 1 + J) + ξ(R, 1− J) (5.11)

where ξ(a, b) is the ’2-argument arctangent’ defined as

ξ(a, b) =1

2i(ln(b+ ia)− ln(b− ia)) = arctan

(ab

)+ πΘH(−b)sign(a) (5.12)

where ΘH is the Heaviside step function and sign() is the sign function. With the con-vention that sign(0) = 0 we get ξ(0, b) = 0. The branch cut of the complex logarithm ischosen to be such that −π < =(ln z) ≤ π. The deviations are taken to zero and using(5.11), equation (5.8) is then rewritten as

ξ[R(λ,η

2, 1), 1 + J(λ,

η

2, 1)] + ξ[R(λ,

η

2, 1), 1− J(λ,

η

2, 1)]︸︷︷︸

1.

=π

N(I+ + I−)

+1

N

∑k=1,β

ξ[R(λ− λk,β,η

2, 2), 1 + J(λ− λk,β,

η

2, 2)] + ξ[R(λ− λk,β,

η

2, 2), 1− J(λ− λk,β,

η

2, 2)]︸︷︷︸

2.

+1

N

λ 6=λk,β∑k=2,β

ξ[R(λ− λk,β, 0, 2), 1] + ξ[R(λ− λk, 0, 2), 1]︸︷︷︸3.

+ ξ[R(λ− λk, η, 2), 1 + J(λ− λk, η, 2)] + ξ[R(λ− λk, η, 2), 1− J(λ− λk, η, 2)]︸︷︷︸4.

+

1

N(ξ[0, 1 + J(0, η, 2)] + ξ[0, 1− J(0, η, 2)]︸︷︷︸

=0

). (5.13)

63


All four marked terms in this equation are treated separately. Taking care of the branchcuts and as long as a 6= 0, we can write

ξ(R, 1 + J) + ξ(R, 1− J) =1

2iln

((1 + J + i ·R) · (1− J + i ·R)

(1 + J − i ·R) · (1− J − i ·R)

)=

1

2iln

(1−R2 − J2 + 2i ·R1−R2 − J2 − 2i ·R

)= ξ(2R, 1−R2 − J2) (5.14)

which we use for the first term. It gives (see appendix B for details)

arctan

(2 sin(λ) cos(λ) tanh(η2 )(cos2(λ) + sinh2(η2 ))

tanh2(η2 )(cos2(λ) + sinh2(η2 ))2 − sin2(λ) cos2(λ)− sinh2(η2 ) cosh2(η2 )

)(B.1)= arctan

(−tanh(η)

tan(λ)

). (5.15)

Using the properties of the complex arctangent we get

−(π

2sign(tan(λ))− arctan

(tan(λ)

tanh(η)

))+

πΘH

(sinh2(η2 ) + cosh2(η2 )

sinh2(η2 )(sin2(λ))

)sign(sin(λ) cos(λ)). (5.16)

The tangent and the product of sine and cosine have always the same sign. Therefore, wereach at

arctan

(tan(λ)

tanh(η)

)+π

2sign (tan(λ)) . (5.17)

For the second term we have to again use the complex logarithm at first. Introducing thenotation λ− λk = λ and additionally

A = sin(λ) cos(λ) tanh(η)(cos2(λ) + sinh2(η

2)), (5.18)

we have (see appendix (B.2) - (B.8) for details)

1

2iln

(tanh2(η)(cos2(λ) + sinh2(η2 ))2 − sin2(λ) cos2(λ)− sinh2(η2 ) cosh2(η2 ) + 2iA

tanh2(η)(cos2(λ) + sinh2(η2 ))2 − sin2(λ) cos2(λ)− sinh2(η2 ) cosh2(η2 )− 2iA

)=

=1

2iln

(

tanh(η2 ) + i tan(λ))

(tanh(η2 )− i tan(λ)

)(

tanh(3η2 ) + i tan(λ)

)(

tanh(3η2 )− i tan(λ)

)

= arctan

(tan(λ)

tanh(η2 )

)+ arctan

(tan(λ)

tanh(3η2 )

). (5.19)

Because the imaginary part is always positive, the Heaviside step function ΘH gives alwayszero. The third term has b = 1 and µ = 0 and is therefore just exactly as in the BetheTakahashi equation. The forth term is treated like the first term with η → 2η everywhere.

64


We now are able to rewrite (5.8) which is given by

arctan

(tan(λ)

tanh(η)

)+π

2sign (tan(λ)) =

π

N(I+ + I−)

+1

N

∑k=1,β

(arctan

(tan(λ− λk,β)

tanh(η2 )

)+ arctan

(tan(λ− λk,β)

tanh(3η2 )

))

+1

N

∑k=2,βλ 6=λk,β

(2 arctan

(tan(λ− λk,β)

tanh(η)

)+ arctan

(tan(λ− λk,β)

tanh(2η)

)

+π

2sign (tan(λ− λk,β))

), (5.20)

and compare it with the Bethe Takahashi equations

arctan

(tan(λ)

tanh(η)

)+ π

⌊λ

π+

1

2

⌋=

π

NI2

+1

N

∑k=1,β

(arctan

(tan(λ− λk,β)

tanh(η2 )

)+ arctan

(tan(λ− λk,β)

tanh(3η2 )

)+ 2π

⌊λ− λk,β

π+

1

2

⌋)

+1

N

∑k=2,β,λ 6=λk,β

(2 arctan

(tan(λ− λk,β)

tanh(η)

)+ arctan

(tan(λ− λk,β)

tanh(2η)

)+ 3π

⌊λ− λk,β

π+

1

2

⌋).

(5.21)

Comparing gives⌊λ

π+

1

2

⌋− 1

2sign (tan(λ)) =

1

NI2 −

1

N(I+ + I−) +

2

N

∑k=1,β

⌊λ− λk,β

π+

1

2

⌋

+1

N


3

⌊λ− λk,β

π+

1

2

⌋− 1

2sign (tan(λ− λk,β)) mod (N), (5.22)

in which the Bethe quantum numbers can be rewritten using the distinction of narrowand wide pairs

2I+ + ΘH(δ) =N

2+ I2 + 2

∑k=1,β

⌊λ− λk,β

π+

1

2

⌋(5.23)

+∑k=2,β,λ 6=λk,β

3

⌊λ− λk,β

π+

1

2

⌋− 1

2sign (tan(λ− λk,β)) mod (N). (5.24)

Taking this equation modulo 2 reveals a condition for narrow and wide pairs, that giverise to the sign of the deviation

ΘH(δ) =N

2−M + 1 + I2 +


3

⌊λ− λk,β

π+

1

2

⌋− 1

2sign (tan(λ− λk,β)) mod (2)

(5.25)

65


The next step is to take the difference of the logarithmic Bethe equations that - whenexponentiated - give (see appendix B.9 )∣∣∣∣sinh(2η + 2δ)

sinh(2δ)

∣∣∣∣2 =

(∣∣∣∣sin(λ+ iη + iδ)

sin(λ+ iδ)

∣∣∣∣2)N ∏

k=1,β

∣∣∣∣∣sin(λ− λk,β + i12η − iδ)

sin(λ− λk,β + i32η + iδ)

∣∣∣∣∣2

∏k=2,βλ 6=λk,β

∣∣∣∣sin(λ− λk,β + iη − i(δ − δk,β))

sin(λ− λk,β + iη + i(δ − δk,β))

∣∣∣∣2 ∣∣∣∣ sin(λ− λk,β + i(δ + δk,β))

sin(λ− λk,β + 2iη + i(δ + δk,β))

∣∣∣∣2 . (5.26)

We are now able to calculate the string deviations numerically. The procedure is asfollows: The initial guess for the rapidity centers are the Bethe-Takahashi rapidities andall deviations are set to zero. These initial values are inserted into the RHS of equation(5.26), so that the LHS gives updated values for the deviations. The new deviationsare inserted into (5.8), where the sign of the deviation is given by (5.25) and the Bethequantum numbers are replaced by the Bethe-Takahashi quantum number using equation(5.24). Then, equation (5.8) reveals updated values for the string centers that are againplugged into equation (5.26) and so on. In such a recursive fashion, all deviations andrapidity centers are obtained. To ensure, that this recursion converges, it is of greatimportance that the RHS of (5.26) is not strongly dependent on the deviations. This isonly the case for deviations that are small compared to the rapidities and otherwise, thismethod fails.A cross check for small system sizes can be done by comparing the eigenenergy of theBethe states to the ones obtained by exactly diagonalizing the Hamiltonian. In order todo so, we need an expression for the eigenenergy that takes string deviations into account.Therefore, we use equation (3.32) and as a start, insert two complex conjugate rapiditieswhich gives(

sinh2(η)

cosh(η)− cos(2λ)+

sinh2(η)

cosh(η)− cos(2λ?)

)=

(2 sinh2(η)(cosh(η)−<(cos(2λ)))

| cosh(η)− cos(2λ)|2

).

(5.27)

Thus, the eigenenergy is given by the sum over all complex conjugated pairs

E = −J∑{λ+j,α}

(2 sinh2(η)(cosh(η)−<(cos(2λ+

j,α)))

| cosh(η)− cos(2λ+j,α)|2

). (5.28)

Eigenenergies of the 2-down spin sector calculated by the above presented method com-pare excellently to the energies given by the exact diagonalization method in most cases.If the Bethe-Takahashi quantum number equals zero, a singular configuration may beencountered and the above method does not converge.

66

Chapter 6

Conclusion

Summary and Conclusion

The main achievement of this thesis is the classification of all eigenstates for the anisotropicgapped XXZ Heisenberg spin-1/2 which is presented in chapter 4. In order to do so, theeigenstates were expressed in terms of rapidities (quasi momenta) within the frameworkof the coordinate Bethe ansatz (see chapter 3). The parametrization of the rapidities iscrucially dependent on the anisotropy ∆ of the spin chain and differs in the gapless orisotropic phase. This is one of the reasons why the in chapter 4 presented classification isonly valid if the anisotropy |∆| > 1.The Bethe equations - consistency relations acquired by imposing periodic boundary con-ditions - in their usual form are transcendental and furthermore very hard to solve nu-merically. Thus, a handier notation is obtained by taking the logarithm of the equationsthat moreover establishes a relation towards quantum numbers. By inserting differentsets of quantum numbers, all eigenstates are in principle found. A complicating feature isthat also complex rapidities lead to valid eigenstates - so called bound states. The corre-sponding quantum number sets are ambiguous and a classification of eigenstates is at firstonly possible when making use of the string hypothesis. This hypothesis purports thatthe complex rapidities are organized in regular patterns - strings - with only small or nodeviations. The string hypothesis has proven to give exact results in the thermodynamiclimit. However, there are some exceptions in which the string hypothesis fails. For thisparticular Heisenberg spin chain for instance, a study of collapsed 2-strings violating thestring hypothesis has just been published [68].Nevertheless, the string hypothesis still gives rise to most of the actual eigenstates as hasbeen partially shown in chapter 5. By using the string hypothesis, and again, by takingthe logarithm, the Bethe-Takahashi equations are obtained. In these equations, the ra-pidities are then connected to the Bethe-Takahashi quantum numbers in terms of whicha classification of states is possible.Only by using the symmetries of the rapidities, the full classification is obtained (see chap-ter 4). This classification of states is on the one hand consistent with known results, as itgives the same quantum numbers for the anti-ferromagnetic ground state with arbitrarynumber of flipped spins [47] and secondly the same quantum numbers for the 2-spinonspectrum [36]. On the other hand, uniqueness and completeness of quantum number setshave been proven and it has furthermore been shown by induction, that the total numberof quantum number sets matches the expected dimensionality of the Hilbert space. Withthe Bethe-Takahashi quantum numbers in hand, the corresponding sets of rapidities areobtained numerically by either using a recursive method or the Newton-Rhapson method.The system of Bethe-Takahashi equations is usually well behaved and a good numericalsolution can be found. The eigenstates and eigenenergies are expressed in terms of ra-pidities and are then easily calculated. Hence, we conclude that the full set of eigenstatesfor arbitrary system sizes and magnetization in the vicinity of the string hypothesis isprovided.As has been pointed out in chapter 5, counting the sets of Bethe-Takahashi quantum

67

Chapter 6 - Conclusion

numbers is not a proper proof of completeness of the Bethe equations. That is, the actualBethe eigenstates can vary very much from the Bethe-Takahashi ones. In particular, col-lapsed strings can occur that give extra solutions in a different base. The Bethe-Takahashieigenstates have been proven useful as initial guess to find the Bethe eigenstates in mostcases. The general path has been pointed out for one- and two-strings only in chapter 5.Only if a corresponding Bethe eigenstate to every Bethe-Takahashi quantum number setwas found, the combinatorial proof would give raise to the completeness of Bethe states.To find a general connection between these two is however very unlikely.The classification of states and the knowledge of its completeness within the string hypoth-esis is nevertheless very useful. For example, correlation functions and structure factorsmay now be calculated: First of all, most deviations are still exponentially small in systemsize and in the anisotropy ∆ (the larger ∆, the smaller the deviations). Secondly, only afraction of states violate the string hypothesis and may be neglected. The algorithm ABA-CUS designed for the calculation of these quantities does not account for largely deformedstring states [29], even though small deviations are taken into account. The classificationof states is thus especially useful for ABACUS which brings us directly to the discussionof future work.

Outlook

The very next step is to implement the restrictions on the set of quantum numbers intoABACUS and thereby solve its problem of double counting states which has been de-scribed in the introduction. By doing so, the dynamical structure factor is for examplecalculated by ABACUS which can be compared to experimental data.Not only correlation functions but for example also Quench or Floquet dynamics could bederived for the gapped XXZ Heisenberg spin-1/2 chain as now a good approximation tothe complete eigenbasis is accessible.Another yet open task is to continue the derivations for string deviations started in chap-ter 5 for arbitrary bases, similar to the derivation done for the isotropic XXX Heisenbergspin-1/2 chain in [38]. A numerical study of collapsed string solutions could be added, ex-tending the recently published study of collapsed 2-string solutions [68]. In this thesis, anequation determining the number of states for arbitrary bases within the string hypothesiswas derived (4.106). This equation is useful to identify the number of additional or missingsolutions. It would be interesting to find out whether an (un-)stable region determinedby the anisotropy ∆ and the system size N , which was discussed for the two flipped spinsector in [68], exists for higher string states. If so, further analytical or numerical studiescould reveal the boundary between the stable and the unstable region.Using the system’s symmetries to classify the eigenstates [65] is a novel approach to un-derstand the structure of eigenstates. This may be transfered to other integrable modelsthat lack a classification of eigenstates in terms of quantum numbers. It may even givesome more insight to the completeness problem of the Bethe equations.

Finally, it is only left to say that even though Bethe’s paper [3] was written almost acentury ago, there are still many interesting open questions, not only in condensed matterphysics but also in high energy physics, mathematical physics and in many experiments,that are yet to be explored by using the Bethe ansatz.

68

Appendix A

Bethe Equations for the XXX andthe Gapless XXZ Case

In [39] and [41], a more detailed derivation of the Bethe equations for both - the XXXHeisenberg spin chain and the gapless XXZ Heisenberg spin chain - is given. The notationused in chapter 3 for the gapped XXZ case is kept.

The Isotropic Case

The parametrization of the momenta k in terms of rapidities λ for the isotropic XXXHeisenberg spin chain is given by

k = π − 2 arctan(2λ), λ =1

2cot

(k

2

). (A.1)

The Bethe equations for the XXX model read(λj + i/2

λj − i/2

)N=

M∏k 6=j

(λj − λk + i

λj − λk − i

)with j = 1, . . . ,M, (A.2)

and after taking the logarithm

2π

NIj = Θ1(λj)−

1

N

M∑k=1

Θ2(λj − λk), with j = 1, . . . ,M (A.3)


(2λ

j

)

and with Ij ∈

− N

2+

1

2, . . . ,

N

2− 1

2if M even

− N

2+ 1, . . . ,

N

2if M odd.

The corresponding eigenenergy is

E = J

M∑j=1

−2

4λ2j + 1

(A.4)

and the total momentum

P =

M∑j=1

1

iln

(λj + i/2

λj − i/2

)= πM − 2π

N

M∑j=1

Ij mod (2π). (A.5)

Strings are parametrized as

λnj,α = λj,α +i

2(j + 1− 2n) + dnj,α. (A.6)

69

Appendix A - Bethe Equations for the XXX and the Gapless XXZ Case

The Bethe-Takahashi equations, in which the deviations dnj,α are neglected, read

2π

NIj,α = Θj(λj,α)− 1

N

Ns∑k=1

Mk∑β=1

Θjk(λj,α − λk,β), (A.7)

with Ij,α ∈

− N

2+

1

2, . . . ,

N

2− 1

2if Mj even

− N

2+ 1, . . . ,

N

2if Mj odd.

in which the kernel Θj is still defined as in (A.4) and Θjk is given by

Θjk = (1− δjk)Θ|j−k| + 2Θ|j−k|+2 + · · ·+ 2Θj+k−2 + Θj+k. (A.8)

The classification of states in terms of Bethe-Takahashi quantum numbers is known. Therestriction on quantum numbers is

Ij,α + j ≤ Ij,∞ =1

2

N − (2j − 1)(Mj − 1)− 2∑k 6=j

min(j, k)Mk

, (A.9)

which is in the case of only real rapidities (j = 1,Mj = M) also valid for the Betheequations.

The Gapped Case

The parametrization of the anisotropy ∆ for the gapless XXZ Heisenberg spin chain(|∆| < 1) reads η = arccos(∆) and η ∈]0, π/2[. The momentum k is expressed in terms ofrapidities as

k = π − 2 arctan

(tanh(λ)

tan(η/2)

). (A.10)

The Bethe equations are then given by(sinh(λj + iη/2)

sinh(λj − iη/2)

)N=

M∏k 6=j

(sinh(λj − λk + iη)

sinh(λj − λk − iη)

)with j = 1, . . . ,M. (A.11)

When taking the logarithm, they become

2π

NIj = Θ1(λj)−

1

N

M∑k=1

Θ2(λj − λk), with j = 1, . . . ,M (A.12)


(tanh(λ)

tan(jη/2)

)

and with Ij ∈

− N

2+

1

2, . . . ,

N

2− 1

2if M even

− N

2+ 1, . . . ,

N

2if M odd.

70

Appendix A - Bethe Equations for the XXX and the Gapless XXZ Case

The eigenenregy in terms of rapidities is

E = J

M∑j=1

− sin2(η)

cosh(2λj)− cos(η)(A.13)

and the total momentum is

P =M∑j=1

1

iln

(sinh(λj + iη/2)

sinh(λj − iη/2)

)= πM − 2π

N

M∑j=1

Ij mod (2π). (A.14)

String states for large nN are parametrized as

λmnj ,α = λnj ,α + iη

2(nj + 1− 2m) + i

π

4(1− vα) + iδmnj ,α, m = 1, . . . , nj , (A.15)

in which nj is the string length and vα = ±1 is the parity that have to be determined. Incomparison the gapped case, not all string lengths are allowed. A classification of statesis found in [39]. The Bethe-Takahashi equations read

2π

NIj,α = Θj(λj,α)− 1

N

Ns∑k=1

Mk∑β=1

Θjk(λj,α − λk,β), (A.16)

with Ij,α ∈

− N

2+

1

2, . . . ,

N

2− 1

2if Mj even

− N

2+ 1, . . . ,

N

2if Mj odd.

and the kernels are

Θj(λ) = 2vj arctan

(tanh(λ)

tan(njη/2)vj

)(A.17)

Θjk = (1− δnjnk)Θ|nj−nk| + 2Θ|nj−nk|+2 + · · ·+ 2Θnj+nk−2 + Θnj+nk . (A.18)

71

Appendix B

Supplemental Equations to Deter-mine String Deviations

To derive the expressions for string deviations, various relations of complex trigonometricand hyperbolic functions are used.The first term in equation (5.13) in chapter 5 gives

arctan


tanh2(η2 )(cos2(λ) + sinh2(η2 ))2 − sin2(λ) cos2(λ)− sinh2(η2 ) cosh2(η2 )

)(B.1)

= arctan


tanh2(η2 )(cos2(λ) + sinh2(η2 ))2 − (cos2(λ) + sinh2(η2 ))(1 + sinh2(η2 )− cos2(λ))

)

= arctan

2 sin(λ) cos(λ) tanh(η2 )

tanh2(η2 )(cos2(λ) + sinh2(η2 )− cosh4( η2

)

sinh2( η2

)+ 1

tanh(η/2) cos2(λ))

= arctan

2 sin(λ) cos(λ)

tanh(η2 )(cos2(λ)sinh2( η

2)+cosh2( η

2)

sinh2( η2

)− sinh2( η

2)+cosh2( η

2)

sinh2( η2

))

= arctan

(sin(λ) cos(λ) tanh(η)

− sin2(λ)

)= arctan

(−tanh(η)

tan(λ)

).

The second term of equation (5.13) is massaged

1

2iln

(tanh2(η)(cos2(λ) + sinh2(η2 ))2 − sin2(λ) cos2(λ)− sinh2(η2 ) cosh2(η2 ) + 2iA

tanh2(η)(cos2(λ) + sinh2(η2 ))2 − sin2(λ) cos2(λ)− sinh2(η2 ) cosh2(η2 )− 2iA

)(B.2)

=1

2iln

(tanh(η)(cos2(λ) + sinh2(η2 ))− 1

tanh(η)(1 + sinh2(η2 )− cos2(λ)) + 2i sin(λ) cos(λ)

tanh(η)(cos2(λ) + sinh2(η2 ))− 1tanh(η)(1 + sinh2(η2 )− cos2(λ))− 2i sin(λ) cos(λ)

)

=1

2iln

(sinh2(η)(cos2(λ) + sinh2(η2 ))− cosh2(η)(1 + sinh2(η2 )− cos2(λ)) + 2iA

sinh2(η)(cos2(λ) + sinh2(η2 ))− cosh2(η)(1 + sinh2(η2 )− cos2(λ))− 2iA

)

with the following notation for the imaginary part

A = sin(λ) cos(λ) sinh(η) cosh(η). (B.3)

73

Appendix B - Supplemental Equations to Determine String Deviations

We can furthermore write

1

2iln

(sinh2(η)(2 cos2(λ)− 1)− 1− sinh2(η2 ) + cos2(λ) + 2iA

sinh2(η)(2 cos2(λ)− 1)− 1− sinh2(η2 ) + cos2(λ)− 2iA

)(B.4)

=1

2iln

(cos2(λ)(2 sinh2(η) + 1)− 1− sinh2(η2 )− sinh2(η) + 2iA

cos2(λ)(2 sinh2(η) + 1)− 1− sinh2(η2 )− sinh2(η)− 2iA

)

=1

2iln

(sinh2(η)− sinh2(η2 )− tan2(λ)(sinh2(η2 ) + cosh2(η)) + 2i tan(λ) sinh(η) cosh(η)

sinh2(η)− sinh2(η2 )− tan2(λ)(sinh2(η2 ) + cosh2(η))− 2i tan(λ) sinh(η) cosh(η)

)

=1

2iln

(sinh2(η)− sinh2(η2 )− tan2(λ)(sinh2(η2 ) + cosh2(η)) + 2i tan(λ) sinh(η) cosh(η)

sinh2(η)− sinh2(η2 )− tan2(λ)(sinh2(η2 ) + cosh2(η))− 2i tan(λ) sinh(η) cosh(η)

)

Lets study some terms of (B.5) individually. We get

sinh2(η

2

)+ cosh2(η) = sinh2

(η2

)+ cosh(η)

(sinh2

(η2

)+ cosh2

(η2

))(B.5)

= 2 sinh2(η

2

)cosh2

(η2

)+ cosh(η) cosh2

(η2

)= cosh

(η2

)(sinh(η) sinh

(η2

)+ cosh(η) cosh

(η2

))= cosh

(η2

)cosh

(3η

2

)

and

sinh2(η)− sinh2(η

2

)=

1

2sinh2(η) + 2 sinh2

(η2

)cosh2

(η2

)− sinh2

(η2

)(B.6)

= sinh2(η) sinh(η

2

)cosh

(η2

)+ sinh2

(η2

)cosh2

(η2

)+ sinh4

(η2

)= sinh2(η) sinh

(η2

)cosh

(η2

)+ sinh2

(η2

)cosh(η)

= sinh(η

2

)sinh

(3η

2

).

Dividing the imaginary part of (B.5) by the first term (B.6) gives

2i tan(λ) sinh(η) cosh(η)

cosh(η2 ) cosh(3η2 )

= i tan(λ)sinh(η)(sinh2(η2 ) + cosh2(η2 )) + sinh(η) cosh(η)


(B.7)

= i tan(λ)sinh(η2 ) cosh(3η

2 ) + sinh(η2 ) cosh(η2 ) + sinh(η) cosh(η)


= i tan(λ)sinh(η2 ) cosh(3η

2 ) + cosh(η2 ) sinh(3η2 )


= i tan(λ)

(tanh(

η

2) + tanh(

3η

2)

).

74

Appendix B - Supplemental Equations to Determine String Deviations

Substituting all back into equation (B.5) gives

1

2iln

tanh(η2 ) tanh(3η2 )− tan2(λ) + i tan(λ)

(tanh(η2 ) + tanh(3η

2 ))

tanh(η2 ) tanh(3η2 )− tan2(λ)− i tan(λ)

(tanh(η2 ) + tanh(3η

2 )) (B.8)

=1

2iln

(

tanh(η2 ) + i tan(λ))

(tanh(η2 )− i tan(λ)

)(

tanh(3η2 ) + i tan(λ)

)(

tanh(3η2 )− i tan(λ)

)

= arctan

(tan(λ)

tanh(η2 )

)+ arctan

(tan(λ)

tanh(3η2 )

).

The difference of Bethe equations for complex conjugated rapidities after exponentiatinggive (

sin(λ+ iη + iδ)

sin(λ+ iδ)

)N (sin(λ− iη − iδ)sin(λ− iδ)

)N= (B.9)

∏k=1,β

(sin(λ− λk,β + i3

2η + iδ)

sin(λ− λk,β − i12η + iδ)

)(sin(λ− λk,β − i3

2η − iδ)sin(λ− λk,β + i1

2η − iδ)

)∏k=2,βλ 6=λk,β

(sin(λ− λk,β + iη + i(δ − δk,β)

sin(λ− λk,β − iη + i(δ − δk,β)

)(sin(λ− λk,β + 2iη + i(δ + δk,β))

sin(λ− λk,β + i(δ + δk,β))

)(

sin(λ− λk,β − iη − i(δ − δk,β)

sin(λ− λk,β + iη − i(δ − δk,β)

)(sin(λ− λk,β − 2iη − i(δ + δk,β))

sin(λ− λk,β − i(δ + δk,β))

)sin(2iη + 2iδ)

sin(2iδ)

sin(−2iη − 2iδ)

sin(−2iδ)

=

(∣∣∣∣sin(λ+ iη + iδ)

sin(λ+ iδ)

∣∣∣∣2)N ∏

k=1,β

∣∣∣∣∣sin(λ− λk,β + i12η − iδ)

sin(λ− λk,β + i32η + iδ)

∣∣∣∣∣2

.

75

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excitations of the gapped xxz heisenberg spin-1/2 chaincaux) that are designed to e ectively...

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