Mathematical Aspects of Quantum Computing 2007Kinki University Series on Quantum ComputingEditor-in-Chief: Mikio Nakahara (Kinki University, Japan)ISSN: 1793-7299PublishedVol. 1 Mathematical Aspects of Quantum Computing 2007edited by Mikio Nakahara, Robabeh Rahimi (Kinki Univ., Japan) &Akira SaiToh (Osaka Univ., Japan)ZhangJi - Math'l Aspects of Quan.pmd 3/11/2008, 2:05 PM 2NE W J E RSE Y L ONDON SI NGAP ORE BE I J I NG SHANGHAI HONG KONG TAI P E I CHE NNAI World ScientifcKinki University Series on Quantum Computing Vol. 1Mathematical Aspects of Quantum Computing 2007editors Mikio NakaharaRobabeh Rahimi Kinki University, JapanAkira SaiTohOsaka University, JapanBritish Library Cataloguing-in-Publication DataA catalogue record for this book is available from the British Library.For photocopying of material in this volume, please pay a copying fee through the CopyrightClearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, USA. In this case permission tophotocopy is not required from the publisher.ISBN-13 978-981-281-447-0ISBN-10 981-281-447-7All rights reserved. This book, or parts thereof, may not be reproduced in any form or by any means,electronic or mechanical, including photocopying, recording or any information storage and retrievalsystem now known or to be invented, without written permission from the Publisher.Copyright 2008 by World Scientific Publishing Co. Pte. Ltd.Published byWorld Scientific Publishing Co. Pte. Ltd.5 Toh Tuck Link, Singapore 596224USA office: 27 Warren Street, Suite 401-402, Hackensack, NJ 07601UK office: 57 Shelton Street, Covent Garden, London WC2H 9HEPrinted in Singapore.MATHEMATICAL ASPECTS OF QUANTUM COMPUTING 2007Kinki University Series on Quantum Computing Vol. 1ZhangJi - Math'l Aspects of Quan.pmd 3/11/2008, 2:05 PM 1March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsvPREFACEThis volume contains lecture notes and poster contributions presentedat the summer school Mathematical Aspects of Quantum Computing,held from 27 to 29 August, 2007 at Kinki University in Osaka, Japan.The aims of this summer school were to exchange and share ideas amongresearchers working in various elds of quantum computing particularlyfrom a mathematical point of view, and to motivate students to tackle theadvanced subjects of quantum computing.Invited speakers to the summer school were asked to prepare their lec-ture notes in a self-contained way and therefore we expect that each contri-bution will be useful for students and researchers even with less backgroundto the topic.This summer school was supported by Open Research Center Projectfor Private Universities: matching fund subsidy from MEXT (Ministry ofEducation, Culture, Sports, Science and Technology).We yearn for continued outstanding success in this eld and wish toexpand summer schools and conferences further in the eld of quantumcomputing based at Kinki University.Finally, we would like to thank Ms Zhang Ji of World Scientic for herexcellent editorial work.Osaka, January 2008Mikio NakaharaRobabeh RahimiAkira SaiTohMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsThis page intentionally left blank This page intentionally left blankMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsviiSummer School onMathematical Aspects of Quantum ComputingKinki Univerisity, Osaka, Japan27 29 August 200727 AugustMikio Nakahara (Kinki Univerisity, Japan)Quantum Computing: An OverviewKazuhiro Sakuma (Kinki University, Japan)Braid Group and Topological Quantum Computing28 AugustDamian J. H. Markham (Tokyo University, Japan)An Introduction to Entanglement TheoryShogo Tanimura (Kyoto University, Japan)Holonomic Quantum Computing and Its Optimization29 AugustSahin Kaya Ozdemir (Osaka University, Japan)Playing Games in Quantum Mechanical Settings: Features of Quan-tum GamesManabu Hagiwara (National Institute of Advanced Industrial Science andTechnology, Japan)Quantum Error Correction CodesPoster PresentationsVahideh Ebrahimi (Kinki University, Japan)Controlled Teleportation of an Arbitrary Unknown Two-Qubit En-tangled StateMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsviiiYukihiro Ota (Kinki University, Japan)Notes on D ur-Cirac ClassicationRobabeh Rahimi (Kinki University, Japan)Bang-Bang Control of Entanglement in Spin-Bus-Boson ModelAkira SaiToh (Osaka University, Japan)Numerical Computation of Time-Dependent Multipartite Nonclas-sical CorrelationTakuya Yamano (Ochanomizu University, Japan)On Classical No-Cloning Theorem under Liouville Dynamics andDistancesMarch11,20088:59WSPC-ProceedingsTrimSize:9inx6inMathematicalAspectsixMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsxMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsxiMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsThis page intentionally left blank This page intentionally left blankMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsxiiiLIST OF PARTICIPANTSAoki, Takashi Kinki University, JapanChinen, Koji Osaka City University, JapanEbrahimi Bakhtavar, Vahideh Kinki University, JapanHagiwara, Manabu National Institute of AdvancedIndustrial Science andTechnology, JapanIkeda, Yasushi Kyoto University, JapanInoue, Kaiki Taro Kinki University, JapanKikuta, Toshiyuki Kinki University, JapanKobata, Kumi Kinki University, JapanKondo, Yasushi Kinki University, JapanMarkham, Damian James Harold Tokyo University, JapanMinami, Kaori Kinki University, JapanNagaoka, Shoyu Kinki University, JapanNakagawa, Nobuo Kinki University, JapanNakahara, Mikio Kinki University, JapanOotsuka, Takayoshi Kinki University, JapanOta, Yukihiro Kinki University, JapanOzdemir, Sahin Kaya Osaka University, JapanRahimi Darabad, Robabeh Kinki University, JapanSaiToh, Akira Osaka University, JapanSakuma, Kazuhiro Kinki University, JapanSasano, Hiroshi Kinki University, JapanSegawa, Etsuo Yokohama National University, JapanSugita, Ayumu Osaka City University, JapanTanimura, Shogo Kyoto University, JapanTomita, Hiroyuki Kinki University, JapanUnoki, Makoto Waseda University, JapanYamano, Takuya Ochanomizu University, JapanYamanouchi, Akiko ITOCHU Techno-Solutions Corpora-tions, JapanYoshida, Motoyuki Waseda University, JapanMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsThis page intentionally left blank This page intentionally left blankMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsxvCONTENTSPreface vQuantum Computing: An Overview 1M. NakaharaBraid Group and Topological Quantum Computing 55T. Ootsuka, K. SakumaAn Introduction to Entanglement Theory 91D. J. H. MarkhamHolonomic Quantum Computing and Its Optimization 115S. TanimuraPlaying Games in Quantum Mechanical Settings: Features ofQuantum Games 139S. K. Ozdemir, J. Shimamura, N. ImotoQuantum Error-Correcting Codes 181M. HagiwaraPoster SummariesControled Teleportation of an Arbitrary Unknown Two-QubitEntangled State 213V. Ebrahimi, R. Rahimi, M. NakaharaNotes on the D urCirac Classication 215Y. Ota, M. Yoshida, I. OhbaMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspectsxviBang-Bang Control of Entanglement in Spin-Bus-Boson Model 217R. Rahimi, A. SaiToh, M. NakaharaNumerical Computation of Time-Dependent MultipartiteNonclassical Correlation 219A. SaiToh, R. Rahimi, M. Nakahara, M. KitagawaOn Classical No-Cloning Theorem Under Liouville Dynamicsand Distances 221T. Yamano, O. IguchiMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects1QUANTUM COMPUTING: AN OVERVIEWMIKIO NAKAHARADepartment of Physics, Kinki University, Higashi-Osaka 577-8502, JapanE-mail: nakahara@math.kindai.ac.jpElements of quantum computing and quantum infromation processing are in-troduced for nonspecialists. Subjects inclulde quantum physics, qubits, quan-tum gates, quantum algorithms, decoherece, quantum error correcting codesand physical realizations. Presentations of these subjects are as pedagogical aspossible. Some sections are meant to be brief introductions to contributions byother lecturers.Keywords: Quantum Physics, Qubits, Quantum Gates, Quantum Algorithms.1. IntroductionQuantum computing and quantum information processing are emerging dis-ciplines in which the principles of quantum physics are employed to storeand process information. We use the classical digital technology at almostevery moment in our lives: computers, mobile phones, mp3 players, justto name a few. Even though quantum mechanics is used in the design ofdevices such as LSI, the logic is purely classical. This means that an ANDcircuit, for example, produces denitely 1 when the inputs are 1 and 1. Oneof the most remarkable aspects of the principles of quantum physics is thesuperposition principle by which a quantum system can take several dier-ent states simultaneously. The input for a quantum computing device maybe a superposition of many possible inputs, and accordingly the output isalso a superposition of the corresponding output states. Another aspect ofquantum physics, which is far beyond the classical description, is entangle-ment. Given several objects in a classical world, they can be described byspecifying each object separately. Given a group of ve people, for example,this group can be described by specifying the height, color of eyes, personal-ity and so on of each constituent person. In a quantum world, however, onlya very small subset of all possible states can be described by such individualspecications. In other words, most quantum states cannot be described byMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects2such individual specications, thereby being called entangled. Why andhow these two features give rise to the enormous computational power inquantum computing and quantum information processing will be explainedin this contribution.A part of this lecture note is based on our forthcoming book.1Generalreferences are [24].2. Quantum Physics2.1. Notation and conventionsWe will exclusively work with a nite-dimensional complex vector space Cnwith an inner product , (Hilbert spaces). A vector in Cnis called aket vector or a ket and is denoted as|x =x1...xn xi Cwhile a vector in the dual space Cn is called a bra vector or a bra anddenoted| = (1, . . . , n) i C.Index i sometimes runs from 0 to n1. The inner product of |x and | is|x =n

i=1ixi.This inner product naturally introduces a correspondence|x = (x1, . . . , xn)t x| = (x1, . . . , xn),by which an inner product of two vectors are dened as x|y =

ni=1xiyi.The inner product naturally denes the norm of a vector |x as |x =

x|x.Pauli matrices are generators of su(2) and denotedx =

0 11 0

, y =

0 ii 0

, z =

1 00 1

in the basis in which z is diagonalized. Symbols X = x, Y = iy andZ = z are also employed.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects3Let A be an mn matrix and B be a p q matrix. ThenA B =a11B, a12B, . . . , a1nBa21B, a22B, . . . , a2nB. . .am1B, am2B, . . . , amnBis an (mp) (nq) matrix called the tensor product of A and B.2.2. Axioms of quantum mechanicsQuantum mechanics was discovered roughly a century ago.510In spite ofits long history, the interpretation of the wave function remains an openquestion. Here we adopt the most popular one, called the Copenhagen in-terpretation.A 1 A pure state in quantum mechanics is represented by a normalized vec-tor | in a Hilbert space H associated with the system. If two states|1 and |2 are physical states of the system, their linear superpo-sition c1|1 + c2|2 (ck C), with 2i=1|ci|2= 1, is also a possiblestate of the same system (superposition principle).A 2 For any physical quantity (observable) a, there exists a correspond-ing Hermitian operator A acting on H. When a measurement of a ismade, the outcome is one of the eigenvalues j of A. Let 1 and 2be two eigenvalues of A: A|i = i|i. Consider a superposition statec1|1 + c2|2. If we measure a in this state, the state undergoes anabrupt change (wave function collapse) to one of the eigenstates |i

corresponding to the observed eigenvalue i. Suppose we prepare manycopies of the state c1|1 + c2|2. The probability of collapsing to thestate |i is given by |ci|2(i = 1, 2). The complex coecient ci is calledthe probability amplitude in this sense. It should be noted that a mea-surement produces one outcome i and the probability of obtainingit is experimentally evaluated only after repeating measurements withmany copies of the same state. These statements are easily generalizedto superposition states of more than two states.A 3 The time dependence of a state is governed by the Schr odinger equationi|t = H|, (1)where is a physical constant known as the Planck constant and H is aHermitian operator (matrix) corresponding to the energy of the systemand is called the Hamiltonian.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects4Several comments are in order. In Axiom A 1, the phase of the vector may be chosen arbitrarily; |in fact represents the ray {ei| | R}. This is called the rayrepresentation. The overall phase is not observable and has no physicalmeaning. Axiom A 2 may be formulated in a dierent but equivalent way asfollows. Suppose we would like to measure an observable a. Let thespectral decomposition of the corresponding operator A beA =

ii|ii|, where A|i = i|i.Then the expectation value A of a after measurements with respectto many copies of | isA = |A|. (2)Let us expand | in terms of |i as | = i ci|i. According toA 2, the probability of observing i upon measurement of a is |ci|2andtherefore the expectation value after many measurements is i i|ci|2.If, conversely, Eq. (2) is employed, we will obtain the same result since|A| =

i,jcjcij|A|i =

i,jicjciij =

ii|ci|2.This measurement is called the projective measurement. Any particularoutcome i will be found with the probability|ci|2= |Pi|, (3)where Pi = |ii| is the projection operator and the state immediatelyafter the measurement is |i or equivalentlyPi|/

|Pi|. (4) The Schr odinger equation (1) in Axiom A 3 is formally solved to yield|(t) = eiHt/|(0), (5)if the Hamiltonian H is time-independent, while|(t) = T exp

i

t0H(t)dt

|(0) (6)if H depends on t, where T is the time-ordering operator. The stateat t > 0 is |(t) = U(t)|(0). The operator U(t) : |(0) |(t),called the time-evolution operator, is unitary. Unitarity of U(t) guar-antees that the norm of |(t) is conserved: (0)|U(t)U(t)|(0) =(0)|(0) = 1 (t > 0).March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects5Two mutually commuting operators A and B have simultaneous eigen-states. If, in contrast, they do not commute, the measurement outcomes ofthese operators on any state | satisfy the following uncertainty relations.Let A = |A| and B = |B| be their respective expectationvalues and A = (A A)2 and B = (B B)2 be respectivestandard deviations. Then they satisfyAB 12||[A, B]||. (7)2.3. Simple exampleExamples to clarify the axioms introduced in the previous subsection aregiven. They are used to controll quantum states in physical realizations of aquantum computer. A spin-1/2 particle has two states, which we call spin-up state | and spin-down state | . It is common to assign components| = (1, 0)tand | = (0, 1)t. They form a basis of a vector space C2.Let us consider a time-independent HamiltonianH =

2x (8)acting on the spin Hilbert space C2. Suppose the system is in the eigenstateof z with the eigenvalue +1 at time t = 0; |(0) = | . The wave function|(t) (t > 0) is then found from Eq. (5) as|(t) = exp

i2xt

|(0) =

cos t/2 i sin t/2i sint/2 cos t/2

10

=

cos t/2i sin t/2

= cos 2t| +i sin 2t| . (9)Suppose we measure z in |(t). The spin is found spin-up with probabilityP(t) = cos2(t/2) and spin-down with probability P(t) = sin2(t/2).Consider a more general HamiltonianH =

2 n , (10)where n is a unit vector in R3. The time-evolution operator is readilyobtained, by making use of a well known formulaei( n)= cos I +i( n ) sin (11)asU(t) = exp(iHt/) = cos t/2 I +i( n ) sin t/2. (12)March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects6Suppose the initial state is |(0) = (1, 0)tfor example. Then we nd, at alater time t > 0,|(t) = U(t)|(0) =

cos(t/2) + inz sin(t/2)i(nx + iny) sin(t/2)

. (13)2.4. Multipartite system, tensor product and entangledstateSo far, we have implictly assumed that the system is made of a singlecomponent. Suppose a system is made of two components, one lives in aHilbert space H1 and the other in H2. A system composed of two separatecomponents is called bipartite. The system as a whole lives in a Hilbertspace H = H1 H2, whose general vector is written as| =

i,jcij|e1,i |e2,j, (14)where {|ea,i} (a = 1, 2) is an orthonormal basis in Ha and i,j |cij|2= 1.A state | H written as a tensor product of two vectors as | =|1|2, (|a Ha) is called a separable state or a tensor product state.A separable state admits a classical interpretation The rst system is inthe state |1 while the second system is in |2. It is clear that the setof separable state has dimension dimH1 +dimH2. Note, however, that thetotal space H has dierent dimension than this: dimH = dimH1 dimH2.This number is considerably larger than the dimension of the sparable stateswhen dimHa (a = 1, 2) are large. What are the missing states then? Letus consider a spin state| = 12 (| | + | | ) (15)of two electrons. Suppose | may be decomposed as| = (c1| + c2| ) (d1| + d2| )= c1d1| | + c1d2| | + c2d1| | + c2d2| | .However this decomposition is not possible since we must have c1d2 =c2d1 = 0, c1d1 = c2d2 = 1/2 simultaneously and it is clear that the aboveequations have no common solution, showing | is not separable.Such non-separable states are called entangled. Entangled states refuseclassical descriptions. Entanglement is used extensively as a powerful com-putational resource in the following.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects7Suppose a bipartite state (14) is given. We are interested in when thestate is separable and when entangled. The criterion is given by the Schmidtdecomposition of |.Theorem 2.1. Let H = H1H2 be the Hilbert space of a bipartite system.Then a vector | H admits the Schmidt decomposition| =r

i=1si|f1,i |f2,i, (16)where si > 0 are called the Schmidt coecients satisfying i si = 1 and{|fa,i} is an orthonormal set of Ha. The number r N is called theSchmidt number of |.It follows from the above theorem that a bipartite state | is separableif and only if its Schmidt number r is 1. See1for the proof.2.5. Mixed states and density matricesIt might happen in some cases that a quantum system under considertationis in the state |i with a probability pi. In other words, we cannot saydenitely which state the system is in. Therefore some random nature comesinto the description of the system. Such a system is said to be in a mixedstate while a system whose vector is uniquely specied is in a pure state.A pure state is a special case of a mixed state in which pi = 1 for some iand pj = 0 (j = i).A particular state |i H appears with probability pi in an ensembleof a mixed state, in which case the expectation value of the observable ais i|A|i. The mean value of a averaged over the ensemble is then givenbyA =N

i=1pii|A|i, (17)where N is the number of available states. Let us introduce the densitymatrix by =N

i=1pi|ii|. (18)Then Eq. (17) is rewritten in a compact form as A = Tr(A).Let A be a Hermitian matrix. A is called positive-semidenite if|A| 0 for any | H. It is easy to show all the eigenvalues ofMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects8a positive-semidenite Hermitian matrix are non-negative. Conversely, aHermitian matrix A whose every eigenvalue is non-negative is positive-semidenite.Properties which a density matrix satises are very much like axiomsfor pure states.A 1 A physical state of a system, whose Hilbert space is H, is completelyspecied by its associated density matrix : H H. A density matrixis a positive-semidenite Hermitian operator with tr = 1, see remarksbelow.A 2 The mean value of an observable a is given byA = tr (A). (19)A 3 The temporal evolution of the density matrix follows the Liouville-vonNeumann equationi ddt = [H, ] (20)where H is the system Hamiltonian, see remarks below.Several remarks are in order. The density matrix (18) is Hermitian since pi R. It is positive-semidenite since || =

i pi|i||2 0. Each |i follows the Schr odinger equation i ddt|i = H|i in a closedquantum system. Its Hermitian conjugate is i ddti| = i|H. Weprove the Liouville-von Neumann equation from these equations asi ddt = i ddt

ipi|ii| =

ipiH|ii|

ipi|ii|H = [H, ].We denote the set of all possible density matrices as S(H).Example 2.1. A pure state | is a special case in which the correspondingdensity matrix is = ||. Therefore is nothing but the projectionoperator onto the state. Observe that A = tr A = iei||A|ei =|A

i |eiei| = |A|, where {|ei} is an orthonormal set.Let us consider a beam of photons. We take a horizontally polarizedstate |e1 = | and a vertically polarized state |e2 = | as orthonormalbasis vectors. If the photons are a totally uniform mixture of two polarizedstates, the density matrix is given by = 12|e1e1| + 12|e2e2| = 12

1 00 1

= 12I.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects9This state is called a maximally mixed state.If photons are in a pure state | = (|e1 +|e2)/2, the density matrix,with {|ei} as basis, is = || = 12

1 11 1

.We are interested in when represents a pure state or a mixed state.Theorem 2.2. A state is pure if and only if tr 2= 1.Proof: Since is Hermitian, all its eigenvalues i (1 i dimH) are realand the corresponding eigenvectors {|i} are made orthonormal. Then2= i,j ij|ii|jj| = i 2i|ii|. Therefore tr 2= i 2i max

i i = max 1, where max is the largest eigenvalue of . There-fore tr 2= 1 implies max = 1 and all the other eigenvalues are zero. Theconverse is trivial.We classify mixed states into three classes, similarly to the classicationof pure states into separable states and entangled states. We use a bipartitesystem in the denition but generalization to multipartitle systems shouldbe obvious.Denition 2.1. A state is called separable if it is written in the form =

ipi1,i 2,i, (21)where 0 pi 1 and

i pi = 1. It is called inseparable, if does not admitthe decomposition (21).It is important to realize that only inseparable states have quantum cor-relations analogous to entangled pure states. It does not necessarily implythat a separable state has no non-classical correlation. It is pointed out thatuseful non-classical correlation exists in the subset of separable states.11In the next subsection, we discuss how to nd whether a given bipartitedensity matrix is separable or inseparable.2.6. NegativityLet be a bipartite state and dene the partial transpose ptof withrespect to the second Hilbert space asij,kl il,kj, (22)March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects10where ij,kl = (e1,i| e2,j|) (|e1,k|e2,l). Here {|e1,k} is the orthonor-mal basis of the rst system while {|e2,k} of the second system. Suppose takes a separable form (21). Then the partial transpose yieldspt=

ipi1,i t2,i. (23)Note here that tfor any density matrix is again a density matrix sinceit is still positive semi-denite Hermitian with unit trace. Therefore thepartial transposed density matrix (23) is another density matrix. It wasconjectured by Peres12and subsequently proven by the Hordecki family13that positivity of the partially transposed density matrix is necessary andsucient condition for to be separable in the cases of C2C2systems andC2C3systems. Conversely, if the partial transpose of of these systemsis not a density matrix, then is inseparable. Instead of giving the proof,we look at the following example.Example 2.2. Let us consider the Werner state =1p4 0 0 00 1+p4 p2 00 p21+p4 00 0 0 1p4, (24)where 0 p 1. Here the basis vectors are arranged in the order|e1,1|e2,1, |e1,1|e2,2, |e1,2|e2,1, |e1,2|e2,2.Partial transpose of yieldspt=1p4 0 0 p20 1+p4 0 00 0 1+p4 0p2 0 0 1p4.ptmust have non-negative eigenvalues to be a physically acceptable state.The characteristic equation of ptisD() = det(pt I) =

p + 14

3

1 3p4

= 0.There are threefold degenerate eigenvalue = (1+p)/4 and nondegenerateeigenvalue = (1 3p)/4. This shows that ptis an unphysical state for1/3 < p 1. If this is the case, is inseparable.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects11From the above observation, entangled states are characterized by non-vanishing negativity dened asN() 12(

i|i| 1). (25)Note that negativity vanishes if and only if all the eigenvalues of ptarenonnegative. However there is a class of inseparable states which are notcharacterized by negativity.2.7. Partial trace and puricationLet H = H1H2 be a Hilbert space of a bipartite system made of compo-nents 1 and 2 and let A be an arbitrary operator acting on H. The partialtrace of A over H2 generates an operator acting on H1 dened asA1 = tr2A

k(I k|)A(I |k). (26)We will be concerned with the partial trace of a density matrix in prac-tical applications. Let = || S(H) be a density matrix of a purestate |. Suppose we are interested only in the rst system and have noaccess to the second system. Then the partial trace allows us to forgetabout the second system. In other words, the partial trace quanties ourignarance on the second sytem.To be concrete, consider a pure state| = 12(|e1|e1 +|e2|e2),where {|ei} is an orthonormal basis of C2. The corresponding density ma-trix is = 121 0 0 10 0 0 00 0 0 01 0 0 1,where the basis vectors are ordered as {|e1|e1, |e1|e2, |e2|e1, |e2|e2}.The partial trace of is1 = tr2 =

i=1,2(I ei|)(I |ei) = 12

1 00 1

. (27)Note that a pure state | is mapped to a maximally mixed state 1.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects12We have seen above that the partial trace of a pure-state density matrixof a bipartite system over one of the constituent Hilbert spaces yields amixed state. How about the converse? Given a mixed state density matrix,is it always possible to nd a pure state density matrix whose partial traceover the extra Hilbert space yields the given density matrix? The answeris yes and the process to nd the pure state is called the purication. Let1 = k pk|kk| be a general density matrix of a system 1 with theHilbert space H1. Now let us introduce the second Hilbert space H2 whosedimension is the same as that of H1. Then formally introduce a normalizedvector| =

kpk|k |k, (28)where {|k} is an orthonormal basis of H2. We ndtr2|| =

i,j,k(I i|)

pjpk|j|jk|k|

(I |i)=

kpk|kk| = 1. (29)It is always possible to purify a mixed state by tensoring an extra Hilbertspace of the same dimension as that of the original Hilbert space. Purica-tion is far from unique.3. QubitsA (Boolean) bit assumes two distinct values, 0 and 1, and it constitutes thebuilding block of the classical information theory. Quantum informationtheory, on the other hand, is based on qubits.3.1. One qubitA qubit is a (unit) vector in the vector space C2, whose basis vectors aredenoted as|0 = (1, 0)tand |1 = (0, 1)t. (30)What these vectors physically mean depends on the physical realizationemployed for quantum information processing.They might represent spin states of an electron, |0 = | and |1 = | .Electrons are replaced by nuclei with spin 1/2 in NMR (Nuclear MagneticResonance).March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects13In some cases, |0 stands for a vertically polarized photon | while |1represents a horizontally polarized photon | . Alternatively they mightcorrespond to photons polarized in dierent directions. For example, |0may represent a polarization state | = 12(| +| ) while |1 representsa state | = 12(| | ).Truncated two states from many levels may be employed as a qubit. Wemay assign |0 to the ground state and |1 to the rst excited state of anatom or an ion.In any case, we have to x a set of basis vectors when we carry outquantum information processing. In the following, the basis is written inan abstract form as {|0, |1}, unless otherwise stated.It is convenient to assume the vector |0 corresponds to the classical bit0, while |1 to 1. Moreover a qubit may be in a superposition state:| = a|0 + b|1, with |a|2+ |b|2= 1. (31)If we measure | to see whether it is in |0 or |1, the outcome will be0 (1) with the probability |a|2(|b|2) and the state immediately after themeasurement is |0 (|1).Although a qubit may take innitely many dierent states, it should bekept in mind that we can extract from it as the same amount of informa-tion as that of a classical bit. Information can be extracted only throughmeasurements. When we measure a qubit, the state vector collapses to theeigenvector that corresponds to the eigenvalue observed. Suppose a spin isin the state a|0 + b|1. If we observe that the z-component of the spin is+1/2, the system immediately after the measurement is in |0. This hap-pens with probability |00| = |a|2. The measurement outcome of aqubit is always one of the eigenvalues, which we call abstractly 0 and 1.3.2. Bloch sphereIt is useful, for many purposes, to express a state of a single qubit graphi-cally. Let us parameterize a one-qubit pure state | with and as|(, ) = cos 2|0 + eisin 2|1. (32)The phase of | is xed in such a way that the coecient of |0 is real. It iseasy to verify that ( n(, ) )|(, ) = |(, ), where = (x, y, z)and n(, ) is a real unit vector called the Bloch vector with components n(, ) = (sin cos , sin sin , cos )t. It is therefore natural to assign n(, ) to a state vector |(, ) so that |(, ) is expressed as a unitMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects14vector n(, ) on the surface of the unit sphere, called the Bloch sphere.This correspondence is one-to-one if the ranges of and are restricted to0 and 0 < 2.It is veried that state (32) satises(, )||(, ) = n(, ). (33)A density matrix of a qubit can be represented as a point on a unitball. Since is a positive semi-denite Hermitian matrix with unit trace,its most general form is = 12I +

i=x,y,zuii, (34)where u R3satises |u| 1. The reality follows from the Hermitic-ity requirement and tr = 1 is obvious. The eigenvalues of are =12

1

|u|

/2 and therefore non-negative. The eigenvalue vanishes incase |u| = 1, for which rank = 1. Therefore the surface of the unit spherecorresponds to pure states. The converse is also shown easily. In contrast,all the points u inside a unit ball correspond to mixed states. The ball iscalled the Bloch ball and the vector u is also called the Bloch vector.It is easily veried that given by Eq. (34) satises = tr () = u. (35)3.3. Multi-qubit systems and entangled statesLet us consider a group of many (n) qubits next. Such a system behavesquite dierently from a classical one and this dierence gives a distinguish-ing aspect to quantum information theory. An n-qubit system is often calleda (quantum) register in the context of quantum computing.As an example, let us consider an n-qubit register. Suppose we specifythe state of each qubit separately like a classical case. Each of the qubit isthen described by a 2-d complex vector of the form ai|0+bi|1 and we need2n complex numbers {ai, bi}1in to specify the state. This corresponds thea tensor product state (a1|0 + b1|1) . . . (an|0 + bn|1) C2n. If thesystem is treated in a fully quantum-mechanical way, however, a generalstate vector of the register is represented as| =

ik=0,1ai1i2...in|i1 |i2 . . . |in C2n.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects15Note that 2n 2n for a large number n. The ratio 2n/2n is 10298forn = 1000. Most quantum states in a Hilbert space with large n are entan-gled having no classical analogues. Entanglement is an extremely powerfulresource for quantum computation and quantum communication.Let us consider a 2-qubit system for deniteness. The system has abinary basis {|00, |01, |10, |11}. More generally, a basis for a system ofn qubits may be {|bn1bn2. . . b0}, where bn1, bn2, . . . , b0 {0, 1}. It isalso possible to express the basis in terms of the decimal system. We write|x, instead of |bn1bn2. . . b0, where x = bn12n1+bn22n2+. . . +b0.The basis for a 2-qubit system may be written also as {|0, |1, |2, |3} withthis decimal notation.The set{|+ = 12(|00 +|11), | = 12(|00 |11),|+ = 12(|01 +|10), | = 12(|01 |10)}(36)is an orthonormal basis of a two-qubit system and is called the Bell basis.Each vector is called the Bell state or the Bell vector. Note that all the Bellstates are entangled.4. Quantum Gates, Quantum Circuit and QuantumComputation4.1. IntroductionNow that we have introduced qubits to store information, it is time toconsider operations acting on them. If they are simple, these operationsare called gates, or quantum gates, in analogy with those in classical logiccircuits. More complicated quantum circuits are composed of these simplegates. A collection of quantum circuits for executing a complicated algo-rithm, a quantum algorithm, is a part of a quantum computation.Denition 4.1. (Quantum Computation) A quantum computation is acollection of the following three elements:(1) A register or a set of registers,(2) A unitary matrix u, which is taylored to execute a given quantumalgorithm and(3) Measurements to extract information we need.More formally, a quantum computation is the set {H, U, {Mm}}, whereH = C2nis the Hilbert space of an n-qubit register, U U(2n) representsMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects16a quantum algorithm and {Mm} is the set of measurement operators. Thehardware (1) is called a quantum computer.Suppose the register is set to a ducial initial state, |in = |00 . . . 0for example. A unitary matrix Ualg is generated by an algorithm whichwe want to execute. Operation of Ualg on |in yields the output state|out = Ualg|in. Information is extracted from |out by appropriatemeasurements.4.2. Quantum gatesWe have so far studied the change of a state upon measurements. Whenmeasurements are not made, the time evolution of a state is describedby the Schr odinger equation. The time evolution operator U is unitary:UU = UU = I. We will be free from the Schrodinger equation in thefollowing and assume there always exist unitary matrices which we need.One of the important conclusions derived from the unitarity of gates isthat the computational process is reversible.4.2.1. Simple quantum gatesExamples of quantum gates which transform a one-qubit state are givenbelow. We call them one-qubit gates in the following. Linearity guaranteesthat the action of a gate is completely specied if its action on the basis{|0, |1} is given. Consider the gate I whose action on the basis vectors isI : |0 |0, |1 |1. The matrix expression of this gate isI = |00| + |11| =

1 00 1

. (37)Similarly we introduce X : |0 |1, |1 |0, Y : |0 |1, |1 |0and Z : |0 |0, |1 |1 byX = |10| + |01| =

0 11 0

= x, (38)Y = |01| |10| =

0 11 0

= iy, (39)Z = |00| |11| =

1 00 1

= z. (40)The transformation I is the identity transformation, while X is the negation(NOT), Z the phase shift and Y = XZ the combination thereof.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects17CNOT (controlled-NOT) gate is a 2-qubit gate, which plays an impor-tant role. The gate ips the second qubit (the target qubit) when the rstqubit (the control qubit) is |1, while leaving the second bit unchangedwhen the rst bit is |0. Let {|00, |01, |10, |11} be a basis for the 2-qubitsystem. We use the standard basis vectors with components|00 = (1, 0, 0, 0)t, |01 = (0, 1, 0, 0)t, |10 = (0, 0, 1, 0)t, |11 = (0, 0, 0, 1)t.The action of CNOT gate, whose matrix expression will be written asUCNOT, is UCNOT : |00 |00, |01 |01, |10 |11, |11 |10. Ithas two equivalent expressionsUCNOT = |0000| + |0101| + |1110| + |1011|= |00| I + |11| X, (41)having a matrix formUCNOT =1 0 0 00 1 0 00 0 0 10 0 1 0. (42)Let {|i} be the basis vectors, where i {0, 1}. The action of CNOT onthe input state |i, j is written as |i, i j, where i j is an addition mod2.A 1-qubit gate whose unitary matrix is U is graphically depicted asThe left horizontal line is the input qubit while the right horizontal line isthe output qubit: time ows from the left to the right.A CNOT gate is expressed aswhere denotes the control bit, while

denotes the conditional negation.There may be many control bits (see CCNOT gate below). More generally,we consider a controlled-U gate, V = |00| I + |11| U, in which thetarget bit is acted on by a unitary transformation U only when the controlbit is |1. This gate is denoted graphically asMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects18CCNOT (Controlled-Controlled-NOT) gate has three inputs and thethird qubit ips only when the rst two qubits are both in the state |1.The explicit form of the CCNOT gate isUCCNOT = (|0000| +|0101| +|1010|) I +|1111| X. (43)This gate is graphically expressed as4.2.2. Walsh-Hadamard transformationThe Hadamard gate or the Hadamard transformation H is an importantunitary transformation dened byUH : |0 12(|0 +|1): |1 12(|0 |1).(44)The matrix representation of H isUH = 12(|0 +|1)0| + 12(|0 |1)1| = 12

1 11 1

. (45)A Hadamard gate is depicted asThere are numerous important applications of the Hadamard transfor-mation. All possible 2nstates are generated when UH is applied on eachMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects19qubit of the state |00 . . . 0:(UHUH. . . UH)|00 . . . 0= 12(|0 +|1) 12(|0 +|1) . . . 12(|0 +|1)= 12n2n1

x=0|x. (46)Therefore, we produce a superposition of all the states |x with 0 x 2n 1 simultaneously. The transformation UnH is called the Walsh trans-formation, or Walsh-Hadamard transformation and denoted as Wn.The quantum circuitis used to generate Bell states from inputs |00, |01, |10 and |11.4.2.3. SWAP gate and Fredkin gateThe SWAP gate acts on a tensor product state asUSWAP|1, 2 = |2, 1. (47)The explict form of USWAP is given byUSWAP = |0000| +|0110| +|1001| +|1111| =1 0 0 00 0 1 00 1 0 00 0 0 1. (48)The SWAP gate is expressed asNote that the SWAP gate is a special gate which maps an arbitrary tensorproduct state to a tensor product state. In contrast, most 2-qubit gatesmap a tensor product state to an entangled state.The controlled-SWAP gateMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects20is also called the Fredkin gate. It ips the second (middle) and the third(bottom) qubits only when the rst (top) qubit is in the state |1. Its explicitform is UFredkin = |00| I4 + |11| USWAP.4.3. No-cloning theoremTheorem 4.1. (Wootters and Zurek14) An unknown quantum system can-not be cloned by unitary transformations.Proof: Suppose there would exist a unitary transformation U that makesa clone of a quantum system. Namely, suppose U acts, for any state |, asU : |0 |. Let | and | be two states that are linearly independent.Then we should have U|0 = | and U|0 = | by denition. Thenthe action of U on | = 12(| + |) yieldsU|0 = 12(U|0 +U|0) = 12(| + |).If U were a cloning transformation, we must also haveU|0 = | = 12(| + | + | + |),which contradicts the previous result. Therefore, there does not exist a uni-tary cloning transformation.Note however that the theorem does not apply if the states to be clonedare limited to |0 and |1. For these cases, the copying operator U shouldwork as U : |00 |00, : |10 |11. We can assign arbitrary action ofU on a state whose second input is |1 since this case will never happen.What we have to keep in mind is only that U be unitary. An example ofsuch U isU = (|0000| + |1110|) + (|0101| + |1011|). (49)where the rst set of operators renders U the cloning operator and thesecond set is added just to make U unitary. We immediately notice that Uis nothing but the CNOT gate.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects21Therefore, if the data under consideration is limited within |0 and |1,we can copy the qubit states even in a quantum computer. This fact is usedto construct quantum error correcting codes.4.4. Quantum teleportationThe purpose of quantum teleportation is to transmit an unknown quantumstate of a qubit using two classical bits in such a way that the recipientreproduces the same state as the original qubit state. Note that the qubititself is not transported but the information required to reproduce the quan-tum state is transmitted. The original state is destroyed such that quantumteleportation is not in contradiction with the no-cloning theorem.Fig. 1. In quantum teleportation, Alice sends Bob two classical bits so that Bob repro-duces a qubit state Alice initially had.Alice: Alice has a qubit, whose state she does not know. She wishes tosend Bob the quantum state of this qubit through a classical communicationchannel. Let | = a|0 + b|1 be the state of the qubit. Both of them havebeen given one of the qubits of the entangled pair|+ = 12(|00 +|11)in advance. They start with the state| |+ = 12 (a|000 + a|011 + b|100 + b|111) , (50)where Alice possesses the rst two qubits while Bob has the third. Aliceapplies UCNOTI followed by UHI I to this state, which results in(UHI I)(UCNOTI)(| |+)= 12[|00(a|0 + b|1) +|01(a|1 + b|0)+|10(a|0 b|1) +|11(a|1 b|0)]. (51)March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects22If Alice measures the 2 qubits in her hand, she will obtain one of the states|00, |01, |10 or |11 with equal probability 1/4. Bobs qubit (one of theEPR pair previously) collapses to a|0+b|1, a|1+b|0, a|0b|1 or a|1b|0, respectively, depending on the result of Alices measurement. Alicethen sends Bob her result of the measurement using two classical bits.Bob: After receiving two classical bits, Bob knows the state of the qubitin his hand;received bits Bobs state decoding00 a|0 + b|1 I01 a|1 + b|0 X10 a|0 b|1 Z11 a|1 b|0 Y(52)Bob reconstructs the intial state | by applying the decoding process shownabove. Suppose Alice sends Bob the classical bits 10, for example. Then Bobapplies Z on his qubit to reconstruct | as Z : (a|0b|1) (a|0+b|1) =|.Figure 2 shows the actual quantum circuit for quantum teleportation.Fig. 2. Quantum circuit implementation of quantum teleportation.4.5. Universal quantum gatesIt can be shown that any classical logic gate can be constructed by using asmall set of gates, AND, NOT and XOR for example. Such a set of gatesis called the universal set of gates. It can be shown that the CCNOT gatesimulates these classical gates, and hence quantum circuits simulate anyclassical circuits. The set of quantum gates is, however, much larger thanMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects23those classical gates. Thus we want to nd a universal set of quantum gatesfrom which any quantum circuits can be constructed.It can be shown that(1) the set of single qubit gates and(2) CNOT gateform a universal set of quantum circuits (universality theorem). The proofis highly technical and is not given here.1,2,16We, instead, sketch the proofin several lines.It can be shown that any U U(n) is written as a product of N two-levelunitary matrices, where N n(n 1)/2 and a two-level unitary matrix isa unit matrix In in which only four components Vaa, Vab, Vba and Vbb aredierent from In. Moreover V = (Vij) is an element of U(2). An exampleof a two-level unitary matrix isV = 0 0 0 1 0 00 0 1 0 0 0 , (||2+||2= 1)where a = 1 and b = 4.Now we need to prove the universality theorem for two-level unitarymatrices, which is certainly simpler than the general proof. By employingCNOT gates and their generalizations, it is possible to move the elementsVaa, Vab, Vba and Vbb so that they acts on a single qubit in the register. Weneed to implement the controlled-V gate whose target qubit is the one onwhich V acts. Implementation of the controlled-V gate requires generalizedCNOT gates and several U(2) gates.1,2,164.6. Quantum parallelism and entanglementGiven an input x, a typical quantum computer computes f(x) asUf : |x|0 |x|f(x), (53)where Uf is a unitary matrix which implements the function f.Suppose Uf acts on an input which is a superposition of many |x.Since Uf is a linear operator, it acts on all the constituent vectors of thesuperposition simultaneously. The output is also a superposition of all theresults;Uf :

x|x|0

x|x|f(x). (54)March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects24This feature, called the quantum parallelism, gives quantum computer anenormous power. A quantum computer is advantageous over a classicalcounterpart in that it makes use of this quantum parallelism and also en-tanglement.A unitary transformation acts on a superposition of all possible statesin most quantum algorithms. This superposition is prepared by the actionof the Walsh-Hadamard transformation on an n-qubit register in the initialstate |00 . . . 0 = |0 |0 . . . |0 resulting in

2n1x=0 |x/2n. This stateis a superposition of vectors encoding all the integers between 0 and 2n1.Then the linearlity of Uf leads toUf

12n2n1

x=0|x|0

= 12n2n1

x=0Uf|x|0 = 12n2n1

x=0|x|f(x). (55)Note that the superposition is made of 2n= enln2states, which makesquantum computation exponentially faster than classical counterpart in acertain kind of computation.What about the limitation of a quantum computer3? Let us considerthe CCNOT gate for example. This gate ips the third qubit if and onlyif the rst and the second qubits are both in the state |1 while it leavesthe third qubit unchanged otherwise. Let us x the third input qubit to|0. The third output qubit state is |x y, where |x and |y are the rstand the second input qubits respectively. Suppose the input state of therst and the second qubits is a superposition of all possible states while thethird qubit is xed to |0. This can be achieved by the Walsh-Hadamardtransformation asUH|0 UH|0 |0 = 12(|0 +|1) 12(|0 +|1) |0= 12(|000 +|010 +|100 +|110). (56)By operating CCNOT on this state, we obtainUCCNOT(UH|0 UH|0 |0) = 12(|000 +|010 +|100 +|111). (57)This output may be thought of as the truth table of AND: |x, y, xy. It isextremely important to note that the output is an entangled state and themeasurement projects the state to one line of the truth table, i.e., a singleterm in the RHS of Eq. (57).There is no advantage of quantum computation over classical one at thisstage. This is because only one result may be obtained by a single set ofMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects25measurements. What is worse, we cannot choose a specic vector |x, y, xyat our will! Thus any quantum algorithm should be programmed so thatthe particular vector we want to observe should have larger probability tobe measured compared to other vectors. The programming strategies todeal with this feature are(1) to amplify the amplitude, and hence the probability, of the vector thatwe want to observe. This strategy is employed in the Grovers databasesearch algorithm.(2) to nd a common property of all the f(x). This idea was employed inthe quantum Fourier transform to nd the orderaof f in the Shorsfactoring algorithm.Now we consider the power of entanglement. Suppose we have an n-qubit register, whose Hilbert space is 2n-dimensional. Since each qubit hastwo basis states {|0, |1}, there are 2n basis states, i.e., n |0s and n |1s,involved to span this Hilbert space. Imagine that we have a single quantumsystem, instead, which has the same Hilbert space. One might think thatthe system may do the same quantum computation as the n-qubit registerdoes. One possible problem is that one cannot measure the kth digitleaving other digits unaected. Even worse, consider how many dierentbasis vectors are required for this system. This single system must havean enormous number, 2n, of basis vectors! Multipartite implementationof a quantum algorithm requires exponentially smaller number of basisvectors than monopartite implementation since the former makes use ofentanglement as a computational resource.5. Simple Quantum AlgorithmsLet us introduce a few simple quantum algorithms which will be of helpto understand how quantum algorithms are dierent from and superior toclassical algorithms.5.1. Deutsch algorithmThe Deutsch algorithm is one of the rst quantum algorithms which showedquantum algorithms may be more ecient than their classical counterparts.aLet m, N N (m < N) be numbers coprime to each other. Then there exists P Nsuch that mP 1 (mod N). The smallest such number P is called the period or theorder. It is easily seen that mx+P mx(mod N), x N.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects26In spite of its simplicty, full usage of superposition principle and entangle-ment has been made here.Let f : {0, 1} {0, 1} be a binary function. Note that there are onlyfour possible f, namelyf1 : 0 0, 1 0, f2 : 0 1, 1 1,f3 : 0 0, 1 1, f4 : 0 1, 1 0.First two cases, f1 and f2, are called constant,while the rest, f3 and f4, arebalanced.If we only have classical resources, we need to evaluate f twice totell if f is constant or balanced. There is a quantum algorithm, in contrast,with which it is possible to tell if f is constant or balanced with a singleevaluation of f, as was shown by Deutsch.18Let |0 and |1 correspond to classical bits 0 and 1, respectively, andconsider the state |0 = 12(|00 |01 + |10 |11). We apply f on thisstate in terms of the unitary operator Uf : |x, y |x, y f(x), where is an addition mod 2. To be explicit, we obtain|1 = Uf|0 = 12(|0, f(0) |0, f(0) +|1, f(1) |1, f(1)),where stands for negation. Therefore this operation is nothing but theCNOT gate with the control bit f(x); the target bit y is ipped if andonly if f(x) = 1 and left unchanged otherwise. Subsequently we apply theHadamard gate on the rst qubit to obtain|2 = UH|1

= 122 [(|0 +|1)(|f(0) |f(0)) + (|0 |1)(|f(1) |f(1))]The wave function reduces to|2 = 12|0(|f(0) |f(0)) (58)in case f is constant, for which |f(0) = |f(1), and|2 = 12|1(|f(0) |f(1)) (59)if f is balanced, for which |f(0) = |f(1). Therefore the measurement ofthe rst qubit tells us whether f is constant or balanced.Let us consider a quantum circuit which implements the Deutsch algo-rithm. We rst apply the Walsh-Hadamard transformation W2 = UHUHon |01 to obtain |0. We need to introduce a conditional gate Uf,i.e., the controlled-NOT gate with the control bit f(x), whose action isMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects27Uf : |x, y |x, y f(x). Then the Hadamard gate is applied on the rstqubit before it is measured. Figure 3 depicts this implementation.Fig. 3. Implementation of the Deutsch algorithm.In the quauntum circuit, we assume the gate Uf is a black box for whichwe do not ask the explicit implementation. We might think it is a kind ofsubroutine. Such a black box is often called an oracle. The gate Uf is calledthe Deutsch oracle. Its implementation is given only after f is specied.Then what is the merit of the Deutsch algorithm? Suppose your friendgives you a unitary matrix Uf and asks you to tell if f is constant orbalanced. Instead of applying |0 and |1 separately, you may contruct thecircuit in Fig. 3 with the given matrix Uf and apply the circuit on the inputstate |01. Then you can tell your friend whether f is constant or balancedwith a single use of Uf.5.2. Deutsch-Jozsa algorithmThe Deutsch algorithm introduced in the previous section may be general-ized to the Deutsch-Jozsa algorithm.19Let us rst dene the Deutsch-Jozsaproblem. Suppose there is a binary functionf : Sn {0, 1, . . . , 2n1} {0, 1}. (60)We require f be either constantor balancedas before. When f is constant,it takes a constant value 0 or 1 irrespetive of the input value x. When it isbalanaced the value f(x) for a half of x Sn is 0 while it is 1 for the rest ofx. Although there are functions which are neither constant nor balanced, wewill not consider such cases here. Our task is to nd an algorithm which tellsif f is constant or balanced with the least possible number of evaluationsof f.It is clear that we need at least 2n1+ 1 steps, in the worst case withclassical manipulations, to make sure if f(x) is constant or balanced with100 % condence. It will be shown below that the number of steps reducesto a single step if we are allowed to use a quantum algorithm.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects28The algorithm is divided into the following steps:(1) Prepare an (n + 1)-qubit register in the state |0 = |0n|1. Firstn qubits work as input qubits while the (n + 1)st qubit serves as ascratch pad. Such qubits, which are neither input qubits nor outputqubits, but work as a scratch pad to store temporary information arecalled ancillas or ancillary qubits.(2) Apply the Walsh-Hadamard transforamtion to the register. Then wehave the state|1 = Un+1H |0 = 12n(|0 +|1)n 12(|0 |1)= 12n2n1

x=0|x 12(|0 |1). (61)(3) Apply the f(x)-controlled-NOT gate on the register, which ips the(n+1)st qubit if and only if f(x) = 1 for the input x. Therefore we needa Uf gate which evaluates f(x) and acts on the register as Uf|x|c =|x|c f(x), where |c is the one-qubit state of the (n + 1)st qubit.Observe that |c is ipped if and only if f(x) = 1 and left unchangedotherwise. We then obtain a state|2 = Uf|1 = 12n2n1

x=0|x 12(|f(x) |f(x))= 12n

x(1)f(x)|x 12(|0 |1). (62)Although the gate Uf is applied once for all, it is applied to all then-qubit states |x simultaneously.(4) The Walsh-Hadamard transformation (46) is applied on the rst nqubits next. We obtain|3 = (Wn I)|2 = 12n2n1

x=0(1)f(x)UnH |x 12(|0 |1). (63)It is instructive to write the action of the one-qubit Hadamard gate asUH|x = 12(|0 + (1)x|1) = 12

y{0,1}(1)xy|y,where x {0, 1}, to nd the resulting state. The action of the Walsh-March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects29Hadamard transformation on |x = |xn1. . . x1x0 yieldsWn|x = (UH|xn1)(UH|xn2) . . . (UH|x0)= 12n

yn1,yn2,...,y0{0,1}(1)xn1yn1+xn2yn2+...+x0y0|yn1yn2. . . y0 = 12n2n1

y=0(1)xy|y, (64)where x y = xn1yn1xn2yn2. . . x0y0. Substituting this resultinto Eq. (63), we obtain|3 = 12n

2n1

x,y=0(1)f(x)(1)xy|y

12(|0 |1). (65)(5) The rst n qubits are measured. Suppose f(x) is constant. Then |3

is put in the form|3 = 12n

x,y(1)xy|y 12(|0 |1)up to an overall phase. Let us consider the summation 12n

2n1x=0 (1)xyfor a xed y Sn. Clearly it vanishes since x y is 0 for half of x and1 for the other half of x unless y = 0. Therefore the summation yieldsy0. Now the state reduces to |3 = |0n 12(|0 |1) and the mea-surement outcome of the rst n qubits is always 00 . . . 0. Suppose f(x)is balanced next. The probability amplitude of |y = 0 in |3 is pro-portional to 2n1x=0 (1)f(x)(1)xy= 2n1x=0 (1)f(x)= 0. Thereforethe probability of obtaining measurement outcome 00 . . . 0 for the rstn qubits vanishes. In conclusion, the function f is constant if we obtain00 . . . 0 upon the meaurement of the rst n qubits in the state |3 andit is balanced otherwise.6. DecoherenceA quantum system is always in interaction with its environment. This in-teraction inevitably alter the state of the quantum system, which causesloss of information encoded in this system. The system under considerationis not a closed system when interaction with outside world is in action.We formulate the theory of open quantum system in this chapter by re-garding the combined system of the quantum system and its environmentas a closed system and subsequently trace out the environment degrees ofMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects30freedom. Let S and E be the initial density matrices of the system andthe environment, respectively. Even when the initial state is an uncorre-lated state SE, the system-environment interaction entangles the totalsystem so that the total state develops to an inseparable entangled statein general. Decoherence is a process in which environment causes variouschanges in the quantum system, which manifests itself as undesirable noise.6.1. Open quantum systemLet us start our exposition with some mathematical background materi-als.1,2,24We deal with general quantum states described by density matrices. Weare interested in a general evolution of a quantum system, which is describedby a powerful tool called a quantum operation. One of the simplest quantumoperations is a unitary time evolution of a closed system. Let S be a densitymatrix of a closed system at t = 0 and let U(t) be the time evolutionoperator. Then the corresponding quantum map E is dened asE(S) = U(t)SU(t). (66)One of our primary aims in this section is to generalize this map to casesof open quantum systems.6.1.1. Quantum operations and Kraus operatorsSuppose a system of interest is coupled with its environment. We mustspecify the details of the environment and the coupling between the systemand the environment to study the eect of the environment on the behav-ior of the system. Let HS, HE and HSE be the system Hamiltonian, theenvironment Hamiltonian and their interaction Hamiltonian, respectively.We assume the system-environment interaction is weak enough so that thisseparation into the system and its environment makes sense. To avoid con-fusion, we often call the system of interest the principal system. The totalHamiltonian HT is thenHT = HS + HE + HSE. (67)Correspondingly, we denote the system Hilbert space and the environmentHilbert space as HS and HE, respectively, and the total Hilbert space asHT = HS HE. The condition of weak system-environment interactionmay be lifted in some cases. Let us consider a qubit propagating througha noisy quantum channel, for example. Propagating does not necessarilyMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects31mean propagating in space. The qubit may be spatially xed and subjectto time-dependent noise. When the noise is localized in space and time, theinput and the output qubit states belong to a well dened Hilbert space HSand the above separation of the Hamiltonian is perfectly acceptable even forstrongly interacting cases. We consider, in the following, how the principalsystem state S at t = 0 evolves in time in the presence of its environment.A map which describes a general change of the state from S to E(S) iscalled a quantum operation. We have already noted that the unitary timeevolution is an example of a quantum operation. Other quantum operationsinclude state change associated with measurement and state change due tonoise. The latter quantum map is our primary interest in this chapter.The state of the total system is described by a density matrix . Suppose is uncorrelated initally at time t = 0,(0) = S E, (68)where S (E) is the initial density matrix of the principal system (envi-ronment). The total system is assumed to be closed and to evolve with aunitary matrix U(t) as(t) = U(t)(S E)U(t). (69)Note that the resulting state is not a tensor product state in general. Weare interested in extracting information on the state of the principal systemat some later time t > 0.Even under these circumstances, however, we may still dene the systemdensity matrix S(t) by taking partial trace of (t) over the environmentHilbert space asS(t) = trE[U(t)(S E)U(t)]. (70)We may forget about the environment by taking a trace over HE. This isan example of a quantum operation, E(S) = S(t). Let {|ej} be a basis ofthe system Hilbert space while {|a} be that of the environment Hilbertspace. We may take the basis of HT to be {|ej |a}. The initial densitymatrices may be written as S =

j pj|ejej|, E =

a ra|aa|.Action of the time evolution operator on a basis vector of HT is explicitlywritten asU(t)|ej, a =

k,bUkb;ja|ek, b, (71)March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects32where |ej, a = |ej |a for example. Using this expression, the densitymatrix (t) is written asU(t)(S E)U(t) =

j,apjraU(t)|ej, aej, a|U(t)=

j,a,k,b,l,cpjraUkb;ja|ek, bel, c|Ulc;ja. (72)The partial trace over HE is carried out to yieldS(t) = trE[U(t)(S E)U(t)] =

j,a,k,b,lpjraUkb;ja|ekel|Ulb;ja=

j,a,bpj

kraUkb;ja|ek

lrael|Ulb;ja

. (73)To write down the quantum operation in a closed form, we assume theinitial environment state is a pure state, which we take, without loss ofgenerality, E = |00|. Even when E is a mixed state, we may alwayscomplement HE with a ctitious Hilbert space to purify E, see 2.7.With this assumption, S(t) is written asS(t) = trE[U(t)(S |00|)U(t)]=

a(I a|)U(t)(S |00|)|U(t)(I |a)= a(I a|)U(t)(I |0)S(I 0|)U(t)(I |a).We will drop I from I a| hereafter, whenever it does not cause confu-sion. Let us dene the Kraus operator Ea(t) : HS HS byEa(t) = a|U(t)|0. (74)Then we may writeE(S) = S(t) =

aEa(t)SEa(t). (75)This is called the operator-sum representation (OSR) of a quantum opera-tion E. Note that {Ea} satises the completeness relation

aEa(t)Ea(t)

kl=

a0|U(t)|aa|U(t)|0

kl= kl, (76)where I is the unit matrix in HS. This is equivalent with the trace-preserving property of E as 1 = trSS(t) = trS(E(S)) = trS

a EaEaS

for any S S(HS). Completeness relation and trace-preserving propertyMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects33are satised since our total system is a closed system. A general quantummap does not necessarily satisfy these properties.25At this stage, it turns out to be useful to relax the condition that U(t) bea time evolution operator. Instead, we assume U be any operator includingan arbitrary unitary gate. Let us consider a two-qubit system on which theCNOT gate acts. Suppose the principal system is the control qubit whilethe environment is the target qubit. Then we ndE0 = (I 0|)UCNOT(I |0) = P0, E1 = (I 1|)UCNOT(I |0) = P1,where Pi = |ii|, and consequentlyE(S) = P0SP0 + P1SP1 = 00P0 + 11P1 =

00 00 11

, (77)where S =

00 0110 11

. Unitarity condition may be relaxed when mea-surements are included as quantum operations, for example.Tracing out the extra degrees of freedom makes it impossible to inverta quantum operation. Given an initial principal system state S, there areinnitely many U that yield the same E(S). Therefore even though it ispossible to compose two quantum operations, the set of quantum operationsis not a group but merely a semigroup. b6.1.2. Operator-sum representation and noisy quantum channelOperator-sum representation (OSR) introduced in the previous subsectionseems to be rather abstract. Here we give an interpretation of OSR as anoisy quantum channel. Suppose we have a set of unitary matrices {Ua}and a set of non-negative real numbers {pa} such that a pa = 1. Bychoosing Ua randomly with probability pa and applying it to S, we denethe expectation value of the resulting density matrix asM(S) =

apaUaSUa, (78)which we call a mixing process.26This occurs when a ying qubit is sentthrough a noisy quantum channel which transforms the density matrix byUa with probability pa, for example. Note that no enviroment has beenintroduced in the above denition, and hence no partial trace is involved.bA set S is called a semigroup if S is closed under a product satisfying associativity(ab)c = a(bc). If S has a unit element e, such that ea = ae = a, a S, it is called amonoid.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects34Now the correspondence between E(S) and M(S) should be clear. Letus dene Ea paUa. Then Eq. (78) is rewritten asM(S) =

aEaSEa (79)and the equivalence has been shown. Operators Ea are identied with theKraus operators. The system transforms, under the action of Ua, asS EaSEa/tr EaSEa

. (80)Conversely, given a noisy quantum channel {Ua, pa} we may intro-duce an environment with the Hilbert space HE as follows. Let HE =Span(|a) be a Hilbert space with the dimension equal to the number ofthe unitary matrices {Ua}, where {|a} is an orthonormal basis. Deneformally the environment density matrix E =

a pa|aa| andU

aUa |aa| (81)which acts on HS HE. It is easily veried from the orthonormality of{|a} that U is indeed a unitary matrix. Partial trace over HE then yieldsE(S) = trE[U(S E)U]=

a(I a|)

bUb |bb|

S

cpc|cc|

dUd |dd|

(I |a)=

apaUaSUa = M(S) (82)showing that the mixing process is also decribed by a quantum operationwith a ctitious environment.6.1.3. Completely positive mapsAll linear operators we have encountered so far map vectors to vectors.A quantum operation maps a density matrix to another density matrixlinearly.cA linear operator of this kind is called a superoperator. Let be a superoperator acting on the system density matrices, : S(HS) cOf course, the space of density matrices is not a linear vector space. What is meanthear is a linear operator, acting on the vector space of Hermitian matrices, also acts onthe space of density matrices and it maps a density matrix to another density matrix.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects35S(HS). The operator is easily extended to an operator acting on HT byT = IE, which acts on S(HS HE). Note, however, that T is notnecessarily a map S(HT) S(HT). It may happen that T() is not adensity matrix any more. We have already encountered this situation whenwe have introduced partial transpose operation in 2.7. Let HT = H1H2be a two-qubit Hilbert space, where Hk is the kth qubit Hilbert space.It is clear that the transpose operation t : 1 t1 on a single-qubitstate 1 preserves the density matrix properties. For a two-qubit densitymatrix 12, however, this is not always the case. In fact, we have seen thatt I : 12 pt12 dened by Eq. (22) maps a density matrix to a matrixwhich is not a density matrix when 12 is inseparable.A map which maps a positive operator acting on HS to anotherpositive operator on HS is said to be positive. Moreover, it is called acompletely positive map (CP map), if its extension T = In remains apositive operator for an arbitrary n N.Theorem 6.1. A linear map is CP if and only if there exists a set ofoperators {Ea} such that (S) can be written as(S) =

aEaSEa. (83)We require not only that be CP but also () be a density matrix:tr (S) = tr

aEaEa

= tr

aEaEa

= 1. (84)This condition is satised for any if and only if

aEaEa = IS. (85)Therefore, any quantum operation obtained by tracing out the environmentdegrees of freedom is CP and preserves trace.6.2. Measurements as quantum operationsWe have already seen that a unitary evoluation S USU and a mixingprocess S

i piUiSUi are quantum operations. We will see furtherexamples of quantum operations in this section and the next. This sectiondeals with measurements as quantum operations.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects366.2.1. Projective measurementsSuppose we measure an observable A =

i iPi, where Pi = |ii| is theprojection operator corresponding to the eigenvector |i. We have seen inChapter 2 that the probability of observing i upon a measurement of Ain a state isp(i) = i||i = tr (Pi) (86)and the state changes as PiPi/p(i). This process happens with a prob-ability p(i). Thus we may regard the measurement process as a quantumoperationS

ip(i)PiSPip(i) =

iPiSPi, (87)where the set {Pi} satisifes the completeness relation

i PiPi = I.The projective measurement is a special case of a quantum operation inwhich the Kraus operators are Ei = Pi.6.2.2. POVMWe have been concerned with projective measurements so far. However, itshould be noted that they are not unique type of measurements. Here wewill deal with the most general framework of measurement and show thatit is a quantum operation.Suppose a system and an environment, prepared initially in a productstate ||e0, are acted by a unitary operator U, which applies an operatorMi on the system and, at the same time, put the environment to |ei forvarious i. It is written explicitly as| = U||e0 =

iMi||ei. (88)The system and its environment are correlated in this way. This state mustsatisfy the normalization condition since U is unitary; |e0|UU||e0 =

i,j|ei|Mi Mj I||ej = |

i Mi Mi| = 1. Since | is arbitrary,we must have

iMi Mi = IS, (89)where IS is the unit matrix acting on the system Hilbert space HS. Opera-tors {Mi Mi} are said to form a POVM (positive operator-valued measure).March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects37Suppose we measure the environment with a measurement operatorO = IS

ii|eiei| =

ii (IS |eiei|) .We obtain a measurement outcome k with a probabilityp(k) = |(IS |ekek|)|=

i,j|ei|Mi (IS |ekek|)Mj||ej = |MkMk|, (90)where | = U||e0. The combined system immediately after the mea-surement is1

p(k)(IS |ekek|)U||e0 = 1

p(k)(IS |ekek|)

iMi||ei

= 1

p(k)Mk||ek. (91)Let S =

i pi|ii| be an arbitrary density matrix of the principalsystem. It follows from the above observation for a pure state || thatthe reduced density matrix immediately after the measurement is

kp(k)MkSMkp(k) =

kMkSMk. (92)This shows that POVM measurement is a quantum operation in which theKraus operators are given by the generalized measurement operators {Mi}.The projective measurement is a special class of POVM, in which {Mi} arethe projective operators.6.3. ExamplesNow we examine several important examples which have relevance in quan-tum information theory. Decoherence appears as an error in quantum infor-mation processing. The next chapter is devoted to strategies to ght againsterrors introduced in this section.6.3.1. Bit-ip channelConsider a closed two-qubit system with a Hilbert space C2C2. We callthe rst qubit the (principal) system while the second qubit the envi-ronment. A bit-ip channel is dened by a quantum operationE(S) = (1 p)S + pxSx, 0 p 1. (93)March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects38The input S is bit-ipped with a probability p while it remains in its inputstate with a probability 1 p. The Kraus operators are read o asE0 =

1 pI, E1 = px. (94)Fig. 4. Quantum circuit modelling a bit-ip channel. The gate is the inverted CNOTgate I |00| + x |11|.The circuit depicted in Fig. 4 models the bit-ip channel provided thatthe second qubit is in a mixed state (1 p)|00| + p|11|. The circuit isnothing but the inverted CNOT gate V = I |00| + x |11|. Theoutput of this circuit isV (S [(1 p)|00| + p|11|]) V = (1 p)S |00| + pxSx|11|, (95)from which we obtainE(S) = (1 p)S + pxSx (96)after tracing over the environment Hilbert space.The choice of the second qubit input state is far from unique and so isthe choice of the circuit. Suppose the initial state of the environment is apure state |E = 1 p|0 + p|1, for example. Then the output of thecircuit in Fig. 4 isE(S) = trE[V S |EE|V ] = (1 p)S + pxSx, (97)producing the same result as before.Let us see what transformation this quantum operation brings about inS. We parametrize S using the Bloch vector asS = 12I +

k=x,y,zckk, (ck R) (98)March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects39where k c2k 1. We obtainE(S) = (1 p)S + pxSx= 1 p2 (I + cxx + cyy + czz) + p2(I + cxx cyy czz)= 12

1 + (1 2p)cz cx i(1 2p)cycx + i(1 2p)cy 1 (1 2p)cz

. (99)Observe that the radius of the Bloch sphere is reduced along the y- and thez-axes so that the radius in these directions is |12p|. Equation (99) showsthat the quantum operation has produced a mixture of the Bloch vectorstates (cx, cy, cz) and (cx, cy, cz) with weights 1 p and p respectively.Figure 5 (a) shows the Bloch sphere which represents the input qubit states.The Bloch sphere shrinks along the y- and z-axes, which results in theellipsoid shown in Fig. 5 (b).-1-0.500.5-1-0.500.51-1-0.500.51-1-0.500.51-1-0.500.51-1-0.500.51Fig. 5. Bloch sphere of the input state S (a) and output states of (b) bit-ip channeland (c) phase-ip channel. The probability p = 0.2 is common to both channels.6.3.2. Phase-ip channelConsider again a closed two-qubit system with the (principal) system andits environment.The phase-ip channel is dened by a quantum operationE(S) = (1 p)S + pzSz, 0 p 1. (100)The input S is phase-ipped (|0 |0 and |1 |1) with a proba-bility p while it remains in its input state with a probability 1 p. Thecorresponding Kraus operators areE0 =

1 pI, E1 = pz. (101)March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects40Fig. 6. Quantum circuit modelling a phase-ip channel. The gate is the invertedcontrolled-z gate.A quantum circuit which models the phase-ip channel is shown inFig. 6. Let S be the rst qubit input state while (1 p)|00| + p|11| bethe second qubit input state. The circuit is the inverted controlled-z gateV = I |00| + z |11|.The output of this circuit isV (S [(1 p)|00| + p|11|]) V = (1 p)S |00| + pzSz |11|, (102)from which we obtainE(S) = (1 p)S + pzSz. (103)The second qubit input state may be a pure state|E =

1 p|0 + p|1, (104)for example. Then we ndE(S) = trE[V S |EE|V ] = E0SE0 + E1SE1, (105)where the Kraus operators areE0 = 0|V |E =

1 pI, E1 = 1|V |E = pz. (106)Let us work out the transformation this quantum operation brings aboutto S. We parametrize S using the Bloch vector as before. We obtainE(S) = (1 p)S + pzSz= 1 p2 (I + cxx + cyy + czz) + p2(I cxx cyy + czz)= 12

1 + cz (1 2p)(cx icy)(1 2p)(cx + icy) 1 cz

. (107)Observe that the o-diagonal components decay while the diagonal com-ponents remain the same. Equation (107) shows that the quantum oper-ation has produced a mixture of the Bloch vector states (cx, cy, cz) andMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects41(cx, cy, cz) with weights 1 p and p respectively. The initial state has adenite phase = tan1(cy/cx) in the o-diagonal components. The phaseafter the quantum operaition is applied is a mixture of states with and+. This process is called the phase relaxation process, or the T2 processin the context of NMR. The radius of the Bloch sphere is reduced alongthe x- and the y-axes as 1 |1 2p|. Figure 5 (c) shows the eect of thephase-ip channel on the Bloch sphere for p = 0.2.Other examples will be found in [1,2].7. Quantum Error Correcting Codes7.1. IntroductionIt has been shown in the previous chapter that interactions between a quan-tum system with environment cause undesirable changes in the state of thequantum system. In the case of qubits, they appear as bit-ip and phase-ip errors, for example. To reduce such errors, we must implement somesort of error correcting mechanism in the algorithm.Before we introduce quantum error correcting codes, we have a brief lookat the simplest version of error correcting code in classical bits. Supposewe transmit a serise of 0s and 1s through a noisy classical channel. Eachbit is assumed to ip independently with a probability p. Thus a bit 0 sentthrough the channel will be received as 0 with probability 1 p and as 1with probability p. To reduce channel errors, we may invoke to majorityvote. Namely, we encode logical 0 by 000 and 1 by 111, for example. When000 is sent through this channel, it will be received as 000 with probability(1p)3, as 100, 010 or 001 with probability 3p(1p)2, as 011, 101 or 110 withprobability 3p2(1 p) and nally as 111 with probability p3. By taking themajority vote, we correctly reproduce the desired result 0 with probabilityp0 = (1 p)3+ 3p(1 p)2= (1 p)2(1 + 2p) while fails with probabilityp1 = 3p2(1 p) +p3= (3 2p)p2. We obtain p0 p1 for suciently smallp 0. In fact, we nd p0 = 0.972 and p1 = 0.028 for p = 0.1. The successprobability p0 increases as p approaches to 0, or alternatively, if we usemore bits to encode 0 or 1.This method cannot be applicable to qubits, however, due to no-cloningtheorem. We have to somehow think out the way to overcome this theorem.7.2. Three-qubit bit-ip code: the simplest exampleIt is instructive to introduce a simple example of quantum error correctingcodes (QECC). We closely follow Steane30here.March 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects427.2.1. Bit-ip QECCSuppose Alice wants to send a qubit or a series of qubits to Bob through anoisy quantum channel. Let | = a|0+b|1 be the state she wants to send.If she is to transmit a serise of qubits, she sends them one by one and thefollowing argument applies to each of the qubits. Let p be the probabilitywith which a qubit is ipped and we assume there are no other types oferrors in the channel. In other words, the operator X is applied to the qubitwith probability p and consequently the state is mapped to| |

= X| = a|1 + b|0. (108)We have already seen in the previous section that this channel is describedby a quantum operation (93).7.2.2. EncodingTo reduce the error probability, we want to mimic somehow the classicalcounterpart without using a clone machine. Let us recall that the action ofa CNOT gate is CNOT : |j0 |jj, j {0, 1} and therefore it duplicatesthe control bit j {0, 1} when the target bit is initially set to |0. We usethis fact to triplicate the basis vectors as||00 = (a|0 + b|1)|00 |E = a|000 + b|111, (109)where |E denotes the encoded state. The state |E is called the logicalqubit while each constituent qubit is called the physical qubit. We borrowterminologies from classical error correcting code (ECC) and call the setC = {a|000 + b|111|a, b C, |a|2+ |b|2= 1} (110)the code and each member of C a codeword. It is important to note that thestate | is not triplicated but only the basis vectors are triplicated. Thisredundancy makes it possible to detect errors in |E and correct them aswe see below.A quantum circuit which implements the encoding (109) is easily foundfrom our experience in CNOT gate. Let us consider the circuit shown inFig. 7 (a) whose input state is ||00. It is immediately found that theoutput of this circuit is |E = a|000 + b|111 as promised.7.2.3. TransmissionNow the state |E is sent through a quantum channel which introducesbit-ip error with a rate p for each qubit independently. We assume p isMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects43Fig. 7. Quantum circuits to (a) encode, (b) detect bit-ip error syndrome, (c) makecorrection to a relevant qubit and (d) decode. The gate NX stands for the bit-ip noise.suciently small so that not many errors occur during qubit transmission.The received state depends on in which physical qubit(s) the bit-ip erroroccurred. Table 1 lists possible received states and the probabilities withwhich these states are received.Table 1. State Bob receives and theprobability which this may happen.State Bob receives Probabilitya|000 + b|111 (1 p)3a|100 + b|011 p(1 p)2a|010 + b|101 p(1 p)2a|001 + b|110 p(1 p)2a|110 + b|001 p2(1 p)a|101 + b|010 p2(1 p)a|011 + b|100 p2(1 p)a|111 + b|000 p37.2.4. Error syndrome dectection and correctionNow Bob has to extract from the received state which error occurred duringqubits transmission. For this purpose, Bob prepares two ancillary qubits inthe state |00 as depicted in Fig. 7 (b) and apply four CNOT operationswhose control bits are the encoded qubits while the target qubits are BobsMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects44two ancillary qubits. Let |x1x2x3 be a basis vectors Bob has received andlet A (B) be the output state of the rst (second) ancilla qubit. It is seenfrom Fig. 7 (b) that A = x1x2 and B = x1x3. Let a|100+b|011 be thereceived logical qubit for example. Note that the rst qubit state in bothof the basis vectors is dierent from the second and the third qubit states.These dierence are detected by the pairs of CNOT gates in Fig. 7 (b). Theerror extracting sequence transforms the anicillary qubits as(a|100 + b|011)|00 a|10011 + b|01111 = (a|100 + b|011)|11.Both of the ancillary qubits are ipped since x1 x2 = x1 x3 = 1 forboth |100 and |011. It is important to realize that (i) the syndrome isindependent of a and b and (ii) the received state a|100 + b|011 remainsintact; we have detected an error without measuring the received state!These features are common to all QECC.We list the result of other cases in Table 2. Note that among eightTable 2. States after error extraction is made and theprobabilities with which these states are produced.State after error syndrome extraction Probability(a|000 + b|111)|00 (1 p)3(a|100 + b|011)|11 p(1 p)2(a|010 + b|101)|10 p(1 p)2(a|001 + b|110)|01 p(1 p)2(a|110 + b|001)|01 p2(1 p)(a|101 + b|010)|10 p2(1 p)(a|011 + b|100)|11 p2(1 p)(a|111 + b|000)|00 p3possible states, there are exactly two states with the same ancilla state. Doesit mean this error extraction scheme does not work? Now let us compare theprobabilities associated with the same ancillary state. When the ancillarystate is |10, for example, there are two possible reseived states a|010 +b|101 and a|101+b|010. Note that the former is received with probabilityp(1 p)2while that latter with p2(1 p). Therefore the latter probabilityis negligible compared to the former for suciently small p.It is instructive to visualize what errors do to the encoded basis vectors.Consider a cube with the unit length. The vertices of the cube have coor-dinates (i, j, k) where i, j, k {0, 1}. We assign a vector |ijk to the vertex(i, j, k), under which the vectors |000 and |111 correspond to diagonallyseparated vertices. An action of Xi, the operator X = x acting on theith qubit, sends these basis vectors to the nearest neighbor vetices, whichMarch 11, 2008 8:59 WSPC - Proceedings Trim Size: 9in x 6in MathematicalAspects45dier from the correct basis vectors in the ith position. The intersectionof the sets of vectors obtained by a single action of Xi on |000 and |111is an empty set. Therefore an action of a single error operator X can becorrected with no ambiguity.Now Bob measures his ancillary qubits and obtains two bits of classicalinformation. The set of two bits is called the (error) syndrome and it tellsBob in which physical qubit the error occurred during transmission. Bobapplies correcting procedure to the received state according to the errorsydrome he has obtained. Ignoring extra error states with small probabili-ties, we immediately nd that the following action must be taken:error syndrome correction to be made00 identity operation (nothing is required)01 apply x to the third qubit10 apply x to the second qubit11 apply x to the rst qubitSuppose the syndrome is 01, for example. The state Bob received is likely tobe a|001 +b|110. Bob recovers the initial state Alice has sent by applyingI I x on the received state:(I I x)(a|001 + b|110) = a|000 + b|111.If Bob receives the state a|110 + b|001, unfortunately, he will obtain(I I x)(a|110 + b|001) = a|111 + b|000.In fact, for any error sydrome, Bob obtaines either a|000+b|111 or a|111+b|000. The latter case occurs if and only if more than one qubit are ipped,and hence it is less likely to happen for suciently small error rate p. Theprobability with which multiple error occurs is found from Table 1 asP(error) = 3p2(1 p) + p3= 3p2 2p3. (111)This error rate is less than p if p < 1/2. In contrast, success probability hasbeen enhanced from 1 p to 1 P(error) = 1 3p2+ 2p3. Let p = 0.1,for example. Then the error rate is lowered to P(error) = 0.028, while thesuccess probability is enhanced from 0.9 to 0.972.7.2.5