Lecture Notes on Quantum Information and Computation

Home , Raymond Laflamme

Draft for Internal Circulations: v1: Fall Semester, 2012, v2: Fall Semester, 2013 v3: Fall Semester, 2014, v4: Fall Semester, 2015 Lecture Notes on Quantum Information and Computation

Yong Zhang1

School of Physics and Technology, Wuhan University

(Fall 2015)

Abstract These lectures notes are written for both advanced undergraduate students and ﬁrst-year graduate students in the School of Physics and Technology, University Wuhan. They are mainly based on both online lecture notes of John Preskill from Caltech and the standard textbook of Michael Nielsen and Issac Chuang, so I do not claim any originality. These notes certainly have all kinds of typos or errors, so they will be updated from time to time. I do take the full responsibility for all kinds of typos or errors (besides errors in English writing), and please let me know of them.

* The second version of these notes are typeset by Yi Peng2 Undergraduate student (Id: 2010301020027), Participant in the Fall semester, 2013.

* The third version of these notes are typeset by Kun Zhang3 Graduate student (Id: 2013202020002), Participant in the Fall semester, 2012-2013-2014.

1 yong [email protected] 2 pengyi [email protected] 3 kun [email protected] Draft for Internal Circulations: v1: Fall Semester, 2012, v2: Fall Semester, 2013 v3: Fall Semester, 2014, v4: Fall Semester, 2015 Acknowledgements

I thank all participants in class including advanced undergraduate students, ﬁrst-year graduate students and French students for their patience and persis- tence and for their various enlightening questions. I especially thank students who are willing to devote their precious time to the typewriting of these lecture notes in Latex.

Main References to Lecture Notes

* [Preskill] John Preskill (Caltech): online lecture notes on QIC (1997-present), http://www.theory.caltech.edu/ preskill/ph229/ http://www.theory.caltech.edu/ preskill/ph229/ lecture ∼ * [Nilesen & Chuang] Michael A.∼ Nielsen and Isaac♯ L. Chuang, Quantum Computation and Quantum Information (Cambridge, 2000&2010)

* [KLM] Phillip Kaye, Raymond Laﬂamme and Michele Mosca, An Introduction to Quantum Computing (Oxford, 2007).

* [Zhang] Yong-De Zhang (University of Science & Technology of China), Principles of Quantum Information Physics (in Chinese) (Science Press, 2009).

Main References to Homeworks

* [Preskill] John Preskill (Caltech): online lecture notes on QIC (1997-present), http://www.theory.caltech.edu/ preskill/ph229/ lecture Research∼ Projects♯

* See Yong Zhang’s English and Chinese homepages.

2 To our parents and our teachers!

To be the best researcher is to be the best person ﬁrst of all: Respect and listen to our parents and our teachers always!

3 This course focuses on fundamental principles of quantum mechanics.

The aim of this course is to study the simulation of both quantum ﬁeld theory and quantum gravity on quantum computer.

4 Contents

I Introduction to Quantum Information and Computation 10

1 Overview 1 1.1 Reasons to learn Quantum Information and Computation ...... 1 1.2 What’s Quantum Information and Computation? ...... 1 1.3 Research topics ...... 2

2 Thermodynamics and Statistical Mechanics 4

3 Quantum Mechanics (I): Axioms 5 3.1 Axioms of quantum mechanics for closed system ...... 5 3.1.1 State ...... 5 3.1.2 Observable ...... 6 3.1.3 Projective measurement ...... 7 3.1.4 Schr¨odinger equation ...... 8 3.1.5 Composite system ...... 9

4 Qubit, Quantum Gates and Bell States 10 4.1 Overview ...... 10 4.2 Pure state formalism of a qubit ...... 11 4.2.1 Single-qubit gate SU 2 in the spin-1 2 case ...... 12 4.2.1.1 Spin-1 2 operator and Pauli matrices ...... 12 4.2.1.2 Spinor representation( ) of SU ~2 group ...... 12 4.2.2 Properties . . .~ ...... 13 4.2.3 Physical realization of qubit ...... ( .) ...... 16 4.3 Bell states ...... 16 4.3.1 Notation ...... 17 4.3.2 Parity-bit (i) and Phase-bit (j)...... 21 4.3.3 Orthonormal basis of the two-qubit Hilbert space ...... 22 4.3.4 How to distinguish (create) ψ i, j ...... 24 4.4 Quantum circuit model of Bell states ...... 25 4.4.1 Quantum circuit model . . .S .( . .)⟩ ...... 25 4.4.2 Hadamard gate: H ...... 25 4.4.3 CNOT gate ...... 28 4.4.4 Quantum circuit model of Bell states ...... 32

5 No-Cloning, Dense Coding, Teleportation and Cryptography 35 5.1 No-cloning theorem ...... 35 5.2 Dense coding ...... 38 5.3 Quantum teleportation ...... 40

5 5.4 The quantum teleportation using continuous variables ...... 44 5.5 Quantum cryptography (information security) ...... 47 5.5.1 Classical cryptography ...... 47 5.5.2 Quantum key distribution (QKD) ...... 48 5.5.3 BB84 quantum key distribution ...... 49

6 Bell Inequalities 51 6.1 Einstein’s quantum mechanics: local hidden variable theory (LHV) . . . . . 51 6.1.1 What hidden variable (HV) theory? ...... 51 6.1.2 What local theory? ...... 52 6.1.3 The rule to justify the rightful theory ...... 52 6.2 Bell’s inequality in the local hidden variable theory ...... 52 6.3 Bell’s inequality in quantum mechanics ...... 54 6.4 The CHSH inequality ...... 58 6.5 Hints for the violation of the Bell inequality ...... 60 6.6 Hardy’s theorem ...... 60 6.7 The GHZ theorem ...... 65

II Quantum Computing and Quantum Algorithm 67

7 Classical Circuit Model 68 7.1 Classical circuit ...... 68 7.1.1 Elementary logical gates ...... 69 7.1.2 Universal gate set ...... 70 7.2 Reversible classical computation ...... 70 7.2.1 Irreversible computation ...... 70 7.2.2 Classical reversible gate ...... 71 7.2.3 Three-bit Toffoli gate ...... 72 7.2.4 Three-bit Fredkin gate ...... 73 7.3 The construction of an n-bit Toffoli gate using the 3-bit Toffoli gate . . . . 74

8 Quantum Circuit Model 77 8.1 Deﬁnition of quantum circuit ...... 77 8.1.1 One-qubit gates ...... 77 8.1.2 Controlled two-qubit gates and controlled three-qubit gates . . . . . 81 8.1.2.1 Quantum Toﬀoli gate and Fredkin gate ...... 82 8.1.3 Quantum circuit model of GHZ states ...... 82 8.2 Universal quantum computation ...... 86 8.2.1 Quantum universal gate set ...... 86 8.2.2 Universal quantum gate set of two-qubit gates ...... 87 8.2.3 Deutsch’s gate is a universal quantum gate ...... 90

9 Physical Realization of Quantum Computers 92

10 Quantum Algorithms 93 10.1 Classical and quantum algorithm ...... 93 10.2 Oracle model ...... 94 10.3 Deutsch’s algorithm ...... 95 10.3.1 Deﬁnitions ...... 95

6 10.3.2 Classical algorithm ...... 96 10.3.3 Deutsch’s problem ...... 96 10.3.4 Phase kick-back ...... 96 10.3.5 Deutsch’s algorithm ...... 98 10.4 Deutsch-Jozsa’s algorithm ...... 99 10.4.1 Constant and balanced function in n-qubit ...... 99 10.4.2 Constant or balanced function? ...... 99 10.4.3 Notation and lemma ...... 100 10.4.4 Deutsch-Jozsa’s algorithm ...... 101 10.5 Bernstein-Vazirani’s algorithm ...... 103 10.6 Simon’s algorithm ...... 104 10.7 Grover’s algorithm ...... 104 10.7.1 Overview of the problem ...... 105 10.7.2 Grover’s algorithm ...... 105 10.7.3 Example: N 4 ...... 108 10.7.4 Quantum circuit model of Grover’s algorithm ...... 110 = 11 Quantum Circuit Complexity 113 11.1 Circuit complexity ...... 113 11.1.1 Deﬁnitions ...... 113 11.1.2 Complexity class ...... 114 11.1.3 Quantum complexity ...... 114 11.1.4 Accuracy ...... 115

12 Quantum Simulation 116

III Density Matrix and Quantum Entanglement 117

13 Quantum Mechanics (II): Density Matrix 118 13.1 Density matrix as state of quantum open system ...... 118 13.1.1 State ensemble formalism of density matrix ...... 119 13.1.2 Operator formalism of density matrix ...... 121 13.1.3 Reduced density matrix (State for subsystem) ...... 122 13.2 Mixed state formalism of a qubit ...... 124 13.2.1 Why polarization vector? ...... 125 13.2.2 Pure state and mixed state in two-dimensional Hilbert space H2 .. 126 13.3 Convexity of density matrix ...... 128 13.4 Two-qubit system and its subsystem ...... 129 13.4.1 Example: EPR pair (Bell state) ...... 129 13.4.2 Maximally entangled two-qubit pure states ...... 130 13.4.3 Monogamy of maximal entanglement ...... 131

14 Schmidt Decomposition, Puriﬁcation and GHJW Theorem 133 14.1 Introduction ...... 133 14.2 Schmidt decomposition and quantum entanglement ...... 133 14.2.1 Schmidt decomposition ...... 133 14.2.2 Quantum entanglement ...... 134 14.2.3 Proof for the theorem of the Schmidt decomposition ...... 135 14.3 Example for the Schmidt decomposition ...... 136

7 14.4 The puriﬁcation theorem and GHJW theorem ...... 140 14.4.1 Puriﬁcation ...... 140 14.4.2 The GHJW theorem ...... 141 14.5 Information is physics ...... 142

15 Mixed State Entanglement and Multi-partite Entanglement 145 15.1 Bipartite mixed state entanglement ...... 145 15.1.1 Separability ...... 145 15.1.1.1 Bipartite pure states ...... 145 15.1.1.2 Bipartite mixed state ...... 146 15.1.2 Quantum bipartite entanglement ...... 146 15.1.3 Positive-partial transpose (PPT) criterion for quantum bipartite separability ...... 147 15.1.4 Example: the Werner state and the PPT criterion ...... 147 15.1.5 Example: the Werner state and the CHSH inequality ...... 151 15.2 Multi-partite entanglement ...... 152 15.2.1 Deﬁnition ...... 152 15.2.2 The GHZ state ...... 153 15.2.3 Properties of GHZ states ...... 156

16 Entanglement Measures and Entropy: Bi-Partite System 158 16.1 Pure states and mixed states ...... 158 16.2 Bipartite entangled pure states ...... 159 16.3 Example for two-qubit pure entangled state ...... 162 16.4 Axioms of entanglement measure ...... 163 16.5 Example: entanglement of formation ...... 164 16.6 PPT criterion for quantum separability ...... 165 16.7 Example: Werner state in d-dimension ...... 165 16.7.1 ρ λ is a density matrix ...... 166 16.7.2 Partial transpose of ρ λ ...... 166 16.7.3 Positivity( ) of partial transpose ρ λ ...... 167 16.8 Example: the other form of the( ) Werner state ...... 167 16.8.1 ρanti is a density matrix . . . .( . .) ...... 167 16.8.2 Partial transpose of ρanti λ ...... 168

( ) IV Quantum Open System and Quantum Error Correction Codes 169

17 Quantum Mechanics (III): Quantum Open System 170 17.1 Introduction ...... 171 17.1.1 Why we talk about quantum open system ...... 171 17.1.2 Closed system and open system ...... 171 17.2 Projective measurement ...... 171 17.3 General measurement theory ...... 173 17.4 Deﬁnition of POVM ...... 174 17.5 More on POVM ...... 177 17.5.1 Dimensional analysis ...... 177 17.5.2 POVM on subsystem can be viewed as projective measurements on the entire system ...... 177 17.5.3 Tensor product realization of POVM ...... 177

8 17.5.3.1 Operator formula of F a ...... 178 17.5.3.2 Matrix formalism of Fa ...... 178 A 17.5.3.3 The properties of F a ...... 178 17.5.3.4 Example ...... 179 17.5.4 Direct-sum realization of POVM ...... 181 17.5.5 POVM as quantum operation (superoperator) ...... 181 17.5.6 Realization of a POVM ...... 182 17.6 Quantum operation (superoperator) ...... 184 17.6.1 Definition of the superoperator ...... 184 17.6.2 The wonderful theorem ...... 184 17.6.3 The CPTP mapping ...... 185 17.6.4 The Kraus representation ...... 186 17.6.5 The Stinespring representation ...... 187 17.6.6 Remarks on quantum operation ...... 187 17.7 Quantum channel ...... 188 17.7.1 The bit-flip channel ...... 188 17.7.2 The phase-flip channel ...... 190 17.7.3 Depolarizing channel ...... 190 17.7.4 The phase-damping channel ...... 191 17.7.5 The amplitude-damping channel ...... 193 17.7.6 The two-Pauli channel ...... 194 17.8 The master equation ...... 194

18 Notes on Finite Group Theory 196

19 Notes on Stabilizer Formalism of Quantum Error Correction Codes 197

V Fault-tolerant Quantum Computation 198

20 Measurement-Based Quantum Computation 199

21 Topological Quantum Computation 200

22 Integrable Quantum Computation 201

VI Research Projects 202

23 Research Projects 203

9 Part I

Introduction to Quantum Information and Computation

10 Chapter 1

Overview

References: [Preskill] Chapter 1: Introduction and overview;

● [Nielsen & Chuang] Chapter 1: Introduction and overview.

● 1.1 Reasons to learn Quantum Information and Computa- tion

For an advanced undergraduate major in physics, he or she has to learn Quantum Infor- mation and Computation, because

• Quantum Information and Computation can be seen as a new type of advanced Quantum Mechanics between Quantum Mechanics and Quantum Field Theory;

• Quantum Information and Computation represents a further development of Quan- tum Mechanics;

• what Quantum Information and Computation focuses is the logic of the Quantum Mechanics. The reason for a graduate student major in physics to learn Quantum Information and Computation is that

• if a graduate want to do great in modern theoretical physics (especially in Quantum Field Theory or High Energy Physics or Condensed Matter Physics), he or she has to understand Quantum Information and Computation very well, because he or she must understand Modern Quantum Mechanics which is represented by Quantum Information and Computation.

1.2 What’s Quantum Information and Computation?

There are many diﬀerent opinions about this question:

• In Michael A. Nielsen and Isaac L. Chuang’s opinion, Quantum computation and quantum information is the study of the information process tasks that can be accomplished using quantum mechanical⋯ systems (or using fundamentals of quantum mechanics)

1 ⋯ • According to Rolf Landauer (1961), Information is physical

which⋯ says that information is⋯ something that is encoded in the states of physical systems.

• According to David Deutsch (1985), computation is a physical process and is a task that can be performed on an actual physically realized device. What computers can or can not compute is determined by the law of physics alone, and not by mathematics. We indeed can see some similarities between computers and physical systems as shown in Table 1.1. Table 1.1: Computers vs. Physical Systems

Computer Physical System Computation Motion Input Initial State Rules Laws of Motion Output Final State

1.3 Research topics

See Issac Chuang’s homepage at MIT physics department. There are two main research topics in Quantum Information and Computation:

• How can physical system represent and process information?

• Can nature be better understood in terms of information or computation? In the book of Nielson & Chuang [NC] P.P. 203: “A detailed examination and attempted justiﬁcation of the physics underlying quantum computer (the quantum circuit model) is outside the scope of the present discussions and indeed outside the scope present knowledge.” Nowadays, there are even more radical ideas

– Quantum Mechanics + Special Relativity = Quantum Field Theory. Quantum Information and Computation + Special Relativity = ? – Physics is information. – Physics is computation. – The universe is a computer.

Gift (open problem) to fresh students in Quantum Information and Computation: P versus NP problem (P NP or P NP )

• One of seven Millenium= problem≠ by Clay Mathematical Institute.

2 • Experts intend to believe P NP or P NP , but no proof up to now.

• P NP means there may exist≠ problems⊂ which can not be solved eﬃciently, and it may put a new constraint on Nature like the light speed or the uncertainty principle. ≠ • Google & Wiki for details

3 Chapter 2

Thermodynamics and Statistical Mechanics

4 Chapter 3

Quantum Mechanics (I): Axioms

For those who are not shocked when they ﬁrst come across quantum theory can not possibly have understood it. —Niels Bohr

I think I can safely say that nobody understands quantum mechanics. —Richard Feynman

Quantum mechanics: Real black magic calculus —Albert Einstein

References:

[Preskill] Chapter 2: Foundations I: states and ensembles;

● [Nielsen & Chuang] Chapter 2: Introduction to quantum mechanics.

● 3.1 Axioms of quantum mechanics for closed system

Principles of quantum mechanics can be classiﬁed into two parts: the static part includes “States” and “Observables, and the dynamic part includes “Evolution” and “Measure- ment”. When we talk about axioms of quantum mechanics, we introduce axioms for quantum closed systems and axioms for quantum open systems respectively, see the following table. The axioms for quantum open system will be discussed in detail in Chapter 17.

3.1.1 State Axiom 3.1.1. A state ψ or a ray eiα ψ , from the Hilbert space H , can make a complete description of a physical system (with no hidden variable). S ⟩ S ⟩ Ray is an equivalent class of vectors, where global phase has no physical meaning. But, notice that the relative phase however is of physical significance. For example, the s- tate vector 0 1 is physically different from 0 eiα 1 with eiα 1. Note that the superposition principle is defined only for state vectors, not for state rays. S ⟩ + S ⟩ S ⟩ + S ⟩ ≠

5 Axioms of Quantum Mechanics

Closed Systems Open Systems Eg: the entire universe Eg: subsystems of a composite system Space Hilbert space H pure state vector Sφ⟩ ∈ H State density matrix (operator) ρ pure state ray {eiα Sφ⟩} ∈ H called mixed state self-adjoint operator A = ∑ a P Observable n n n with {an} ⊂ R and {P n} being orthogonal projections projective measurement general measurement Measurement (orthogonal) (non-orthogonal) ⟨A⟩ = ⟨φS A Sφ⟩ ⟨A⟩ = tr(Aρ) unitary evolution non-unitary evolution via superoperator Dynamics ̵ d S ( )⟩ = S ( )⟩ ̵ d ( ) = [ ( )] Eg: ih dt φ t H φ t Eg: ih dt ρ t H, ρ t

3.1.2 Observable An observable in physics is a property of a physical system that can be measured. Axiom 3.1.2. With every observable, there exists an associated linear, self-adjoint operator A, which acts in the Hilbert space H , A† A ψ Aφ Aψ φ , with ψ , φ H . (3.1.1)

Let an be one of the eigenvalues= ⇐⇒ ⟨ ofS A⟩and= ⟨ anS ⟩is the associatedS ⟩ S eigenvector,⟩ ∈ A an S a⟩n an . (3.1.2) All the eigenvalues an of A are real, whileS ⟩ = theS eigenvectors⟩ an form a complete orthogonal basis of the Hilbert space H . It would be easy to prove that the eigenvalues of the observable S ⟩ A is real, and the its eigenvectors are orthogonal.

Proof. (i) The eigenvalues of the observable A are real.

A an an an ∗ ∗ † an an, (3.1.3) an A an an A an an A an S ⟩ = S ⟩ ¡ ⇒ = i.e., an is real. ⟨ S S ⟩ = ⟨ S S ⟩ = ⟨ S S ⟩ (ii) The eigenvectors of the observable A are orthogonal.

• In the case of two eigenvectors with diﬀerent eigenvalues, namely A a a a k k k (3.1.4) A a` a` a` S ⟩ = S ⟩ with k ` and ak an. Therefore, S ⟩ = S ⟩

ak A a` a` ak a` ≠ ≠ ak a` ak a` 0 ak a` 0. (3.1.5) ak A a` ak ak a` ⟨ S S ⟩ = ⟨ S ⟩ ¡ ⇒ ( − )⟨ S ⟩ = ⇒ ⟨ S ⟩ = ⟨ S S ⟩ = ⟨ S ⟩ 6 • In the circumstance of degeneration, namely, there are at least two mutually independent eigenvectors of A, but with the same eigenvalue. The set of al- l these eigenvectors of A associated with the speciﬁc eigenvalue would span a subspace of the Hilbert space H . Thus, we can employ the Schmidt orthogonal- ization process, to make an orthogonal basis for such a subspace, with the basis vectors still being the eigenvectors of the observable A and the corresponding eigenvalue unchanged.

The spectral theorem. We can deﬁne the orthogonal projection operators P n onto the subspace associated with the eigenvalue an of the Hilbert space H , namely

Han A φ an φ φ H , (3.1.6) as ∶= S ⟩ = S ⟩ T ∀ S ⟩ ∈ Degeneracy (k) (k) P n an an , (3.1.7) k=1 (k) = Q U g b U with an k 1,..., Degeneracy be an orthonormal basis for the subspace Han . It’s easy to see that U g T = † P n P n, 2 P n P n, (3.1.8) ⎧ = ⎪ P nP m 0, if n m. ⎨ = Hence, we can expand the operator⎪ A as ⎩⎪ = ≠

A anP n, (3.1.9) n = which is the spectral theorem. Q

3.1.3 Projective measurement

Axiom 3.1.3. For an observable A with eigenvalues an and eigenvectors an , given the system is in the state ψ , the probability of obtaining an (in the non-degeneration case) as the outcome of the measurement of A is S ⟩ S ⟩ 2 Prob an an ψ . (3.1.10)

And the expectation value (mean value)( of) the= S⟨ observableS ⟩S A would be

ψ A ψ A . (3.1.11) ψ ψ ⟨ S S ⟩ ⟨ ⟩ = After the measurement, the system is left in⟨ theS state⟩ within the subspace corresponding to the eigenvalue an (the so called wavepacket collapse). The name “ projection measurement” is evident as we shall see. For example, we can see that the operator P n an an , itself is a self-adjoint operator, which has the mean value, = S ⟩⟨ S ψ P n ψ ψ an an ψ

ψ an an ψ ⟨ S S ⟩ = ⟨ S S ⟩⟨2 S S ⟩ an ψ Prob an , (3.1.12) = ⟨ S ⟩⟨ S ⟩ = S⟨ S ⟩S = ( ) 7 where ψ is the normalized state vector of the physical system.

S ⟩ P n=San⟩⟨anS P n ψ an ψ ψ an , (3.1.13) ψ P ψ an ψ S n⟩ ⟨ S ⟩ S ⟩ ÐÐÐÐÐÐÐ→» = S ⟩ which is the post-measurement state. ⟨ S S ⟩ T ⟨ S ⟩ T

Remark: The measurement of an observable is an irreversible process, i.e., we cannot get the original state back from the measurement results. In this process, we acquire information from the system through measurement. The wave package collapse is a truly random process, and we would lose information, namely P n would kill the information encoded in the state ak with k n, in the same time.

S ⟩ ≠ ψ - (1). Information acquirement: Pn ψ .

(3). Ture randomness: S ⟩Projector S ⟩ ? ψ Pn ψ with probability. Pn ψ - (2). Information loss // wavepacket collapse: ⟨ S S ⟩ P ψ killed (m n). S ⟩ m (4). Irreversible, non-unitary process: S ⟩ ≠ ψ Pn ψ , Pn ψ ψ .

S ⟩ Ð→ S ⟩ S ⟩ Ð→~ S ⟩

3.1.4 Schr¨odingerequation Axiom 3.1.4. The time evolution of a closed state is unitary. Equivalently, the time evolution of the state vector ψ t is governed by the Schr¨odingerequation:

∂ S (ih)⟩ ψ t H t ψ t , (3.1.14) ∂t ̵ where H t is the Hamiltonian operatorS ( )⟩ of= the( physical) S ( )⟩ system.

From the( Schr¨odingerequation) we can see that the time evolution operator U t ∆t, t should be iHdt ( + ) U t dt, t 1 . (3.1.15) h As we shall see that ( + ) = − ̵ U † t dt, t U t dt, t U t dt, t U † t dt, t I O dt2 . which means that( +U t )dt,( t +is unitary) = ( up+ to the) second( + order) = of+ dt.( And) we can infer that ψ t( + U t,) t dt U t dt, t 2dt U dt, 0 ψ 0 . (3.1.16)

In the special caseS ( of)⟩ time-independent= ( − ) ( Hamiltonian− − )H⋯ , we( can)S write( )⟩ down the explicit expression of the time evolution operator U t, 0 , for simplicity which we denote as U t , ̵ U t (e−iH) t~h. (3.1.17)( )

( ) =

8 In the case of time-dependent Hamiltonian H t , we have to use the Dyson’s formula

∞ 1 i n t t ( ) t U t I dtn dtn−1 dt1 H tn H tn−1 H t1 , (3.1.18) n=1 n! h 0 0 0 tn>tn−1>⋯>t1 ( ) = + Q − ̵ S S ⋯ S ( ) ( )⋯ ( ) see Google & Wiki for more details. ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

Remark: Why is the Schr¨odingerequation linear? Why is the time evolution unitary, but diﬀerent with measurement which is a non-unitary process? Why we have two distinct evolutions in quantum mechanics?

3.1.5 Composite system

For system A, ψ A HA, and system B, ϕ B HB, the composite system of system A and system B is described by the tensor product of Hilbert spaces, i.e., ψ A ϕ B HA HB. S ⟩ ∈ S ⟩ ∈ S ⟩ ⊗S ⟩ ∈ ⊗

9 Chapter 4

Qubit, Quantum Gates and Bell States

References: [Preskill]Chapter 2: Foundations I: states and ensembles;

● [Preskill] Chapter 4: Quantum entanglement;

● [Nielsen & Chuang] Chapter 1: Introduction and overview;

● [Nielsen & Chuang] Chapter 2: Introduction to quantum mechanics.

● 4.1 Overview

Classical computation vs. Quantum Computation

Classical Computation Quantum Computation Information unit bit qubit Operation gate quantum gate

For Classical Computation, the basic units that store the information and are manipulated are the bits. The “tool” that can manipulate bits are the so-called classical logic gate.

• bit: the short name of “binary digit”, which can only take the value of 0 or 1.

• gate: one-bit gate NOT gate two-bit gate AND, OR, ...

NOT gate: a in mod 2. AND gate: a b a b, OR gate: a b a b in mod 2.

While for Quantum computation, the counterpart∧ ≡ for⋅ bit should be∨ qubit,≡ ⊕ and for classical logic gates are quantum gates.

• qubit: the short name of “quantum bit”, which can be considered as a state vector in the two-dimensional Hilbert space H2: α α 0 β 1 , with α 2 β 2 1, α, β , (4.1.1) β C = S ⟩ + S ⟩ S S + S S = ∈ 10 where 0 could mean “spin up” and for 1 mean “spin down”, i.e., this is a two-level 1 physical system, and the spin- 2 system is a typical example. S ⟩ S ⟩ • quantum gate:

α SU(2) α′ single-qubit gate: , β β′ quantum gate ⎪⎧ ⎪ ( ) ′ ′ ⎪ α1 α2 ÐÐÐ→SU 4 α1 α2 ⎪ two-qubit gate: ′ ′ . ⎨ β1 β2 β1 β2 ⎪ ⎪ ⊗ ÐÐÐ→ ⊗ 4.2 Pure state formalism⎩⎪ of a qubit

The Hilbert space H2 can be spanned by basis 0 , 1 , i.e.,

α S ⟩ S ⟩ 2 2 ψ α 0 β 1 , ψ H2, with α, β and α β 1. β C

We may noticeS ⟩ = thatS ⟩ +α andS ⟩ =βbeing complex∀ S ⟩ ∈ numbers means∈ four realS numbersS + S S (four= degrees of freedom). With the constraint α 2 β 2 1, we can cut down the degrees of freedom into three. And, if we ignore the global phase of the state vector (qubit), which has no physical meaning, then we can reduceS S + theS S degrees= of freedom to be two. Therefore, the state vector ψ can be described by two real numbers θ, ϕ , for example

θ S ⟩ θ iϕ θ cos 2 ( ) ψ+ θ, ϕ cos 0 e sin 1 iϕ θ , with 0 θ π, and 0 ϕ 2π. (4.2.1) 2 2 e sin 2 TheS ( two real)⟩ ∶= variablesS ⟩ +θ, ϕ canS actually⟩ = determine a unit≤ vector< in the three-dimensional≤ < 3 Euclidean R , namelyn ˆ sin θ cos ϕ, sin θ sin ϕ, cos θ , as shown in Figure 4.1. And we ( ) = ( z )

nˆ sin θ cos ϕ, sin θ sin ϕ, cos θ θ = ( )

ϕ y

Figure 4.1: Bloch sphere shall also see that withn ˆ pointing along different directions, namely different values of θ and ϕ, the state vectors ψ+ θ, ϕ are different: π π 1 (1)ˆn eˆ 1, 0, 0 , θ , ϕ 0, ψ+ , 0 √ 0 1 ; x S 2 ( )⟩ 2 2 π π π π 1 (2)ˆn = eˆ = (0, 1, 0), θ = , ϕ = , ⇒ Tψ+( , )f = √ S 0⟩ + Si⟩1 ; y 2 2 2 2 2

(3)ˆn = eˆz = (0, 0, 1), θ = 0, ϕ =0, 2π , ⇒ Tψ+ (0, ϕ )f =0 . S ⟩ + S ⟩

The= vector= (n ˆ in three) = dimensional∈[ ) space⇒ S is( called)⟩ = BlochS ⟩ vector, and the unit boundary of the vector set is called Bloch sphere, i.e., nˆ 1.

11S S = 4.2.1 Single-qubit gate SU(2) in the spin-1~2 case 4.2.1.1 Spin-1 2 operator and Pauli matrices

The spin-1 2 operator~ can be expressed as 1 ~ J hσ 2 or ⃗ = ̵ ⃗ 1 J σ, (4.2.2) 2 if we set the reduced Planck constant h to⃗ be= one.⃗ And the σ is the so-called Pauli vector, deﬁned as ̵ σ σxeˆx σyeˆy σzeˆz ⃗ (4.2.3) where σx, σy and σz are the Pauli⃗ ∶= matrices,+ which+ have the standard form 0 1 0 i 1 0 σ σ , σ σ , σ σ . x 1 1 0 x 2 i 0 z 3 0 1 − ∶= ∶= ∶= ∶= ∶= ∶= The Pauli matrices satisfy the following properties, −

• the anticommutative relation σi, σj 2δij; (4.2.4)

• the commutative relation =

σi, σj 2iεijkσk, with i, j, k 1, 2, 3 and k i, j . (4.2.5)

4.2.1.2 Spinor representation = of SU 2 group∈ { } ∉ { }

We can deﬁne a unitary operator D 1 θ, n for the spin-1 2 system which is induced by 2 ( ) 3 the rotation in the Euclidean space R along the direction n through an angle of θ: ( ⃗) ~

D 1 θ, n exp iθn J ⃗ (4.2.6) 2 which is equivalent to ( ⃗) ∶= − ⃗⋅⃗ θ θ θ D 1 θ, n exp i n σ cos in σ sin . (4.2.7) 2 2 2 2 ( ⃗) = − ⃗⋅⃗ = − ⃗⋅⃗ D 1 θ, n is a single-qubit gate. And it has some interesting features, for instance 2

( ⃗) D 1 2π, n 1, D 1 4π, n 1. (4.2.8) 2 2 Though this would be meaningless( if⃗ it) only= − gives the( global⃗) = phase, there can be signiﬁcant eﬀect if the relative phase is changed because of this.

12 4.2.2 Properties

Thm 4.2.2.1. If we deﬁne σn σ n, then

σn ψ+ θ, ϕ ψ+ θ, ϕ= ⃗,⋅⃗ with n sin θ cos ϕ, sin θ sin ϕ, cos θ . (4.2.9)

Proof. Let’sS ﬁrstly( express)⟩ = S (σn with)⟩ the Pauli⃗ ∶= matrices,( )

σn sin θ cos ϕσσ1 sin θ sin ϕσσ2 cos θσσ3 0 1 0 i 1 0 = sin θ cos ϕ + sin θ sin+ϕ cos θ 1 0 i 0 0 1 − = + + cos θ sin θ cos ϕ i sin θ sin ϕ − sin θ cos ϕ i sin θ sin ϕ cos θ − = cos θ sin θ cos ϕ i sin ϕ + − , sin θ cos ϕ i sin ϕ cos θ ( − ) = namely ( + ) − cos θ sin θe−iϕ σ . (4.2.10) n sin θeiϕ cos θ = Therefore, we can calculate σn ψ+ θ, ϕ in the following− way −iϕ θ S ( )⟩cos θ sin θe cos 2 σn ψ+ θ, ϕ iϕ iϕ θ sin θe cos θ e sin 2 S ( )⟩ = cos θ cos θ sin θe−iϕ eiϕ sin θ 2 − 2 sin θeiϕ cos θ cos θeiϕ sin θ ⎛ + 2 2 ⎞ = θ ⎝ cos 2 − ⎠ iϕ θ e sin 2 = ψ+ θ, ϕ . (4.2.11)

There we get E.Q. (4.2.9) proved. = S ( )⟩

Thm 4.2.2.2. ψ+ θ, ϕ σ m ψ+ θ, ϕ n m, (4.2.12) 3 with n sin θ cos ϕ, sin θ sin ϕ,⟨ cos(θ and)S ⃗⋅m⃗ S R( , )⟩m = ⃗1⋅ ⃗. Proof.⃗ ∶In= ( analogy to E.Q.(4.2.10), we) can⃗ get∈ the expressionY ⃗ Y = for σ m:

′ ′ −iϕ′ cos θ sin θ e ′ ′ ′ ⃗⋅ ⃗′ ′ σ m ′ , with m sin θ cos ϕ , sin θ sin ϕ , cos θ . (4.2.13) sin θ′eiϕ cos θ′ ⃗⋅ ⃗ = ⃗ ∶= ( ) −

13 Therefore, we can get

ψ+ θ, ϕ σ m ψ+ θ, ϕ ′ ′ −iϕ′ θ θ −iϕ θ cos θ sin θ e cos 2 ⟨cos(2 e)S ⃗⋅ sin⃗ S 2 ( )⟩′ iϕ′ ′ iϕ θ sin θ e cos θ e sin 2 = θ ′ θ ′ i(ϕ−ϕ′) θ −iϕ θ cos 2 cos θ −sin 2 sin θ e cos 2 e sin 2 θ ′ iϕ′ θ ′ iϕ cos 2 sin θ e sin 2 cos θ e + = θ θ θ ′ cos cos cos θ′ sin sin θ′ei(ϕ−−ϕ ) 2 2 2 θ θ ′ θ = e−iϕ sin cos sin+θ′eiϕ sin cos θ′eiϕ 2 2 2 θ θ θ θ ′ ′ +cos2 sin 2 cos θ′ sin− cos sin θ′ ei(ϕ−ϕ ) e−i(ϕ−ϕ ) 2 2 2 2 ′ ′ ′ = cos θ cos−θ sin θsin θ cos+ ϕ ϕ + = n m, + ( − ) (4.2.14) where we have= used⃗⋅ ⃗ the fact that n m sin θ cos ϕ, sin θ sin ϕ, cos θ sin θ′ cos ϕ′, sin θ′ sin ϕ′, cos θ′ T ⃗⋅ ⃗ sin θ sin θ′ cos ϕ cos ϕ′ sin θ sin θ′ sin ϕ sin ϕ′ cos θ cos θ′ = ( )( ) sin θ sin θ′ cos ϕ cos ϕ′ sin ϕ sin ϕ′ cos θ cos θ′ = + + sin θ sin θ′ cos ϕ ϕ′ cos θ cos θ′. (4.2.15) = ( + ) + There, E.Q. (4.2.12)= is veriﬁed,( too.− ) + Remark: For the expression in terms of density matrix, we have

tr ρ σ m n m, (4.2.16) where ρ ψ+ θ, ϕ ψ+ θ, ϕ . ( (⃗⋅ ⃗ )) = ⃗⋅ ⃗

Thm 4.2.2.3.= S ( As)⟩⟨ we shall( know)S from the deﬁnition of ψ+ θ, ϕ that

0 z ψ+ 0, ϕ , S ( )⟩ (4.2.17) then S ⟩ = S↑ ⟩ = S ( )⟩ ψ+ θ, ϕ D eˆz n 0 , with n sin θ cos ϕ, sin θ sin ϕ, cosθ , (4.2.18) where S ( )⟩ = ( →⃗)S ⟩ ⃗ = (θ −iϕ θ ) cos 2 e sin 2 D eˆz n iϕ θ θ . (4.2.19) e sin 2 cos 2 − Proof. Firstly, we should realize( → that⃗) ∶=

′ θ ′ D 1 eˆz n D 1 θ, nxy exp i σ nxy , (4.2.20) 2 2 2 where we deﬁne ( →⃗) = ( ⃗ ) = − ⃗⋅⃗

nxy cos ϕ, sin ϕ, 0 , ′ π π (4.2.21) ⎧ nxy cos ϕ 2 , sin ϕ 2 , 0 sin ϕ, cos ϕ, 0 , ⎪ ⃗ ∶= ( ) ⎨ ⎪ ⃗ ∶= ( + ) ( + ) = (− ) ⎩⎪ 14 z

n sin θ cos ϕ, sin θ sin ϕ, cos θ θ ⃗ = ( ) ′ nxy sin ϕ, cos ϕ, 0 ϕ ⃗ = (− ) ϕ y x nxy cos ϕ, sin ϕ, 0

Figure 4.2: Rotation⃗ = ( in the Euclidean) space

′ i.e., nxy is a unit vector that is orthogonal to both n ande ˆz, and they can form a right- hand coordinate system, see Figure 4.2, which means that the corresponding rotation ⃗ ⃗ ′ taken place in the Euclidean space should be a rotation around nxy through an angle θ. And, now it would be easy to show that E.Q. (4.2.18) is right. E.Q. (4.2.20) can be rewritten as ⃗

θ ′ θ D 1 ez n cos iσ nxy sin 2 2 2 θ θ (⃗ →⃗) = cos − i⃗sin⋅⃗ sin ϕσσ cos ϕσσ 2 2 1 2 = θ− (− + ) cos 2 0 θ 0 cos ϕ i sin ϕ θ sin 0 cos 2 2 cos ϕ i sin ϕ 0 − + = + cos θ sin θ e−iϕ + 2 2 . (4.2.22) sin θ eiϕ cos θ ⎛ 2 − 2 ⎞ = And we can verify E.Q. (4.2.18)⎝ directly, ⎠ θ θ −iϕ θ cos 2 sin 2 e 1 cos 2 D 1 ez n 0 ψ+ θ, ϕ . (4.2.23) 2 sin θ eiϕ cos θ 0 sin θ eiϕ ⎛ 2 − 2 ⎞ ⎛ 2 ⎞ (⃗ →⃗)S ⟩ = = = S ( )⟩ Remarks: ⎝ ⎠ ⎝ ⎠ (1) As we have shown above

σ n ψ+ θ, ϕ ψ+ θ, ϕ , (4.2.24) ⎧ ψ+ θ, ϕ D 1 ez n 0 . ⎪ ⃗⋅⃗ S ( )⟩ = S 2( )⟩ ⎨ If we deﬁne ⎪ S ( )⟩ = (⃗ →⃗)S ⟩ ⎩⎪ ψ− θ, ϕ D 1 ez n 1 , (4.2.25) 2 then, we could verify that S ( )⟩ = (⃗ →⃗)S ⟩

σ n ψ− θ, ϕ ψ− θ, ϕ . (4.2.26)

Firstly, ⃗⋅⃗ S ( )⟩ = − S ( )⟩ θ θ −iϕ θ −iϕ cos 2 sin 2 e 0 sin 2 e ψ− θ, ϕ . (4.2.27) sin θ eiϕ cos θ 1 cos θ ⎛ 2 − 2 ⎞ ⎛− 2 ⎞ S ( )⟩ = = ⎝ 15 ⎠ ⎝ ⎠ Then,

−iϕ θ −iϕ cos θ sin θe sin 2 e σ n ψ− θ, ϕ iϕ sin θe cos θ cos θ ⎛− 2 ⎞ ⃗⋅⃗ S ( )⟩ = θ −iϕ− θ −iϕ sin 2 e cos θ cos⎝ 2 sin θe ⎠ sin θ e−iϕ sin θeiϕ cos θ cos θ ⎛− 2 + 2 ⎞ = θ −iϕ ⎝sin− 2 e − ⎠ cos θ ⎛ 2 ⎞ = ⎝ ψ−− θ, ϕ ⎠ . (4.2.28)

(2) The spinor representation of the= SO− S 3 ( group,)⟩ D R , satisﬁes

D R x( σσD) † R x(′ σσ,) (4.2.29)

′ ′ in which the vector x is rotated( under)⃗ ⋅ ⃗x (Rx),= i.e.,⃗ ⋅ ⃗xj Rijxj.

Example: D R D 1 ez n gives ⃗2 ⃗ = ⃗ =

( ) ≡ (⃗ →⃗) † D R σ3D R σ n. (4.2.30)

It implies that ( ) ( ) = ⃗ ⋅ ⃗

† σ n n D R σ3D R D R ez

D R σ3 0 ⃗ ⋅ ⃗S⃗⟩ = ( ) ( ) ( )S⃗ ⟩ n . (4.2.31) = ( ) S ⟩ 4.2.3 Physical realization of qubit= S⃗⟩

Electron Photon 1 Spin 2 1 Mass 0.5Mev 0 Qubit Spin-state Photon-polarization

1 Note: Though the two-level quantum system is equivalent with the 2 -spin system, not every two-level system, like photon-polarization state, is transformed as a spinor, due to the fact that the photon has spin 1.

4.3 Bell states

• Bell states are maximally entangled two-qubit pure states, also named as EPR pair states.

• Bell states are widely used in quantum information and computation: Bell’s inequalities, dense coding, teleportation, cryptography, etc.

• Y.Z., “Braid Group, Temperley–Lieb Algebra, and Quantum Information and Com- putation”, arXiv: quant-ph/0601050.

16 • Y.Z., Jinglong Pang, “Space-Time Topology in Teleportation-Based Quantum Com- putation”, arXiv:1309.0955.

• Y.Z., Kun Zhang, “Bell Transform, Teleportation Operator and Teleportation-Based Quantum Computation”, arXiv:1401.7009.

4.3.1 Notation 1 Pauli matrices are unitary matrices deﬁned in spin- 2 space: 0 1 σ X ; x 1 0 ⎧ ⎪ ⎪ ∶= ∶= 0 i ⎪ σy iYY ; (4.3.1) ⎪ i 0 ⎪ − ⎨ ∶= − ∶= 1 0 ⎪ σz Z . ⎪ 0 1 ⎪ ⎪ ∶ ∶ There we get three quantum⎪ gates: quantum= gate= X , quantum gate Z and quantum gate ⎩⎪ − Y ZX.

Def= 4.3.1 (Bell states). The following four double-qubit states

φ+ ψ 0, 0 √1 00 11 , 2 ⎧ ⎪ Sφ−⟩ ∶= Sψ(0, 1)⟩ ∶= √1 S00⟩ + S11⟩ , ⎪ 2 ⎪ (4.3.2) ⎪ ⎪ S +⟩ ∶= S ( )⟩ ∶= √1 S ⟩ − S ⟩ ⎪ ψ ψ 1, 0 01 10 , ⎨ 2 ⎪ − 1 ⎪ Sψ ⟩ ∶= Sψ(1, 1)⟩ ∶= √ S01⟩ + S10⟩ , ⎪ 2 ⎪ ⎪ S ⟩ ∶= S ( )⟩ ∶= S ⟩ − S ⟩ are the so called Bell states.⎩⎪ Lemma 4.3.1.1. All the four Bell states deﬁned in E.Q. (4.3.2) can be expressed as

i j ψ i, j I 2 X Z ψ 0, 0 , with i, j 0, 1. (4.3.3)

Proof. With the deﬁnitionS ( )⟩ of= the Bell⊗ states S (4.3.2),( )⟩ we can check= E.Q. (4.3.3) one by one for all the four cases of i, j 0, 1.

(1) For the case of i j =0, E.Q (4.3.3) is absolutely right.

(2) For the case of i = 0,=j 1, the right-hand-side (RHS) of E.Q. (4.3.3) should be

= = RHS I 2 Z ψ 0, 0 1 = I 2⊗Z S ( 00)⟩ 11 2 1 = ⊗ 00√ 11(S ⟩ + S ⟩) 2 = √ψ 0(S, 1 ⟩,− S ⟩) (4.3.4)

which means E.Q (4.3.3) is correct= S in( this)⟩ case.

17 (3) For the case of i 1, j 0, we evaluate the right-hand-side (RHS) of E.Q. (4.3.3)

= =RHS I 2 X ψ 0, 0 1 = I 2⊗X S ( 00)⟩ 11 2 1 = ⊗ 01√ 10(S ⟩ + S ⟩) 2 = √ψ 1(S, 0 ⟩,+ S ⟩) (4.3.5)

which also shows the legitimation= ofS E.Q( )⟩ (4.3.3) in this circumstance.

(4) For the case of i j 1, we can get the right-hand-side (RHS) of E.Q. (4.3.3)

= = RHS I 2 XXZZ ψ 0, 0 1 = I 2⊗XXZZ S ( 00)⟩ 11 2 = ⊗ 1√ (S ⟩ + S ⟩) I 2 X 00 11 2 1 = ⊗ 01√ 10(S ⟩ − S ⟩) 2 = √ψ 1(S, 1 ⟩,− S ⟩) (4.3.6)

which says E.Q (4.3.3) for the last= S situation.( )⟩

Now, we can conclude that for all the four cases i, j 0, 1, E.Q (4.3.3) is valid. Remark: For four Bell states (4.3.2), we have the following geometric representations, =

ψ 00 φ+ √1 00 11 (4.3.7) 2 S ( )⟩ = S ⟩ = (S ⟩ + S ⟩) = + 1 ψ 10 ψ I2 X ψ 00 √ 01 10 X (4.3.8) 2 2 r S ( )⟩ = S ⟩ = ⊗ S ( )⟩ = (S ⟩ + S ⟩) = − 1 ψ 01 φ I2 Z ψ 00 √ 00 11 Z (4.3.9) 2 2 r S ( )⟩ = S ⟩ = ⊗ S ( )⟩ = (S ⟩ − S ⟩) = − 1 ψ 11 ψ I2 XZ ψ 00 √ 01 10 XZ (4.3.10) 2 2 r S ( )⟩ = S ⟩ = ⊗ S ( )⟩ = (S ⟩ − S ⟩) = The vertical line denotes one Hilbert space H2. Lemma 4.3.1.2. The four Bell states can also be expressed as 1 ψ i, j 0i 1 j 1¯i , (4.3.11) 2 S ( )⟩ = √ S ⟩ + (− ) S ⟩ where ¯i i 1 mod 2 and i 0, 1.

We can= show( + in) the following= that E.Q. (4.3.11) is consistent with E.Q. (4.3.2).

18 Proof. From E.Q. (4.3.3) we can get

1 i j ψ i, j I 2 X Z 00 11 2 1 S ( )⟩ = √ 0⊗ X iZj 0 S ⟩ 1+ S X⟩ iZ j 1 2 1 = √ S0⟩ ⊗X i 0 S ⟩ +1S ⟩j ⊗1 X iS 1⟩ 2 1 = √ S0⟩ ⊗ i S ⟩ +1(−j 1) S ⟩¯i⊗ , S ⟩ 2 = √ S ⟩ ⊗ S ⟩ + (− ) S ⟩ ⊗ S ⟩ which is what exactly E.Q. (4.3.11) shows.

Remark: All the four Bell states deﬁned in E.Q. (4.3.2) are normalized:

(i) ψ 0, 0 ψ 0, 0 1, since 1 1 1 ⟨ ( )S00( 11)⟩ = 00 11 00 00 00 11 11 00 11 11 2 2 2 1 √ ⟨ S + ⟨ S √ S ⟩ + S ⟩ = (⟨1 0S ⟩0+ ⟨1 S ⟩ + ⟨ S ⟩ + ⟨ S ⟩) 2 1. = ( + + + )

(ii) ψ i, j ψ i, j 1, because =

j i i j ⟨ ( )S ( ψ )⟩i, j= ψ i, j ψ 0, 0 I 2 Z X I 2 X Z ψ 0, 0 j i i j ⟨ ( )S ( )⟩ = ψ ⟨ 0(, 0 )SI 2 ⊗Z X XZ ⊗ψ 0, 0 S ( )⟩ ψ 0, 0 ψ 0, 0 = a ( ) T ⊗ T ( )f 1. = ⟨ ( )S ( )⟩ Lemma 4.3.1.3. Let’s M denotes= an arbitrary SU 2 matrix, namely single-qubit gate, and M T is the transpose of M . Then ( ) T I 2 M ψ 0, 0 M I 2 ψ 0, 0 , (4.3.12) with the diagrammatical representation ⊗ S ( )⟩ = ⊗ S ( )⟩

M M T (4.3.13) r r =

19 Proof. Firstly, evaluate the left-hand-side (LHS) of E.Q. (4.3.12)

LHS I 2 M ψ 0, 0 I M 1 = 2 ⊗ S ii( )⟩ 2 i=0 ⊗ = 1√ 1 Q S ⟩ i M i 2 i=0 = √1 Q S ⟩ ⊗ S ⟩ 0 M00 0 M10 1 1 M01 0 M11 1 2 = √1 S ⟩ ⊗ S ⟩ + S ⟩ + S ⟩ ⊗ S ⟩ + S ⟩ M00 0 M01 1 0 M10 0 M11 1 1 2 1 = √ M T 0S ⟩ +0 MS T⟩ 1⊗ S ⟩1+ S ⟩ + S ⟩ ⊗ S ⟩ 2 = M√ T I S ⟩ ⊗ S ⟩ + S ⟩ ⊗ S ⟩ 2 0 0 1 1 , 2 ⊗ i.e., = √ S ⟩ ⊗ S ⟩ + S ⟩ ⊗ S ⟩ T LHS M I 2 ψ 00 , (4.3.14) which happens to be equal to the right-hand-side= ⊗ S of( E.Q.)⟩ (4.3.12). In the derivation we have assumed that M has the matrix form M M M 00 01 , M10 M11 ∶= namely M M M T 00 10 . M01 M11 We may notice that = X T XX, Z T ZZ, (4.3.15) ⎧ T = ⎪ XZ ZXZX. ⎨ = Therefore, we can get the conclusion⎪ that ⎩⎪ ( ) = + ψ I 2 X ψ 00 X I 2 ψ 00 , − I Z Z I ⎪⎧ Sφ ⟩ = I 2⊗Z Sψ (00 )⟩ = Z ⊗I 2 Sψ (00 )⟩, (4.3.16) ⎪ ⎪ − I XZ ZX I ⎨ Sψ ⟩ = I 2⊗XZ S (ψ 00)⟩ = ZX⊗ I2 S (ψ 00)⟩ . ⎪ ⎪ S ⟩ = ⊗ S ( )⟩ = ⊗ S ( )⟩ ⎩ + ψ 10 ψ X X (4.3.17) r r S ( )⟩ = S ⟩ = = − ψ 01 φ Z Z (4.3.18) r r S ( )⟩ = S ⟩ = = − ψ 11 ψ XZ ZX (4.3.19) r r S ( )⟩ = S ⟩ = = 20 4.3.2 Parity-bit (i) and Phase-bit (j) • Parity-bit (i) – i 0, two spins are aligned, denoted by “φ”; – i 1, two spins are anti-aligned, denoted by “ψ”. = • Phase-bit= (j) – j 0, superposition with “ ”, i.e., with equal phase; – j 1, superposition with “ ”, i.e., with opposite phase. = + = − Table 4.1: Parity-bit (i) and Phase-bit (j)

PP PP j PP 0 1 i PPP 0 φ+ φ− 1 ψ+ ψ− S ⟩ S ⟩ S ⟩ S ⟩ Lemma 4.3.2.1. Bell states are eigenstates of the two commutative operators:

1. parity-bit operator: Z Z,

2. phase-bit operator: X⊗X, namely ⊗ Z Z ψ i, j 1 i ψ i, j , (4.3.20) ⎧ X X ψ i, j 1 j ψ i, j . ⎪ ⊗ S ( )⟩ = (− ) S ( )⟩ ⎨ Proof. The parity-bit operator⎪ ⊗ and S the( phase-bit)⟩ = ( operator− ) S ( are)⟩ commutative, because ⎩⎪ X X Z Z XZ XZ ( ⊗ )( ⊗ ) ZX ZX = ( ) ⊗ ( ) ZX ZX = (− ) ⊗ (− ) Z Z X X , (4.3.21) = ( ) ⊗ ( ) namely = ( ⊗ )( ⊗ ) X XX,ZZ Z 0. (4.3.22)

We may examine the two operators[ separately.⊗ ⊗ ] =

21 (a) Parity-bit operator Z Z.

⊗Z Z ψ i, j 1 Z⊗Z S ( 0)⟩i 1 j 1¯i 2 1 = ⊗Z0√ Z S i⟩ + (−1)jZS 1⟩ Z ¯i 2 1 = √ 0S ⟩ ⊗ 1S i⟩i+ (− )1 j S ⟩1⊗ 1S ⟩ 1 i−1 ¯i 2 1 = √ S ⟩1⊗i(−0 ) Si⟩ + (−1)j(−1)Si 1⟩ ⊗(−¯i ) S ⟩ 2 1 = √1(i − ) S0⟩ ⊗ Si⟩ + (−1)j(1− ) ¯Si ⟩ ⊗, S ⟩ 2 i.e., = (− ) √ S ⟩ ⊗ S ⟩ + (− ) S ⟩ ⊗ S ⟩ Z Z ψ i, j 1 i ψ i, j . (4.3.23)

In the derivation, we have used⊗ the S ( facts)⟩ that= (− ) S ( )⟩ Z i 1 i i , (4.3.24) − ⎧ Z ¯i 1 i 1 ¯i . ⎪ S ⟩ = (− ) S ⟩ ⎨ ⎪ (b) Phase-bit operator X X. ⎩⎪ S ⟩ = (− ) S ⟩

⊗ X X ψ i, j 1 X⊗X S ( 0)⟩i 1 j 1¯i 2 1 = ⊗X 0√ X S i⟩ + (−1)jXS 1⟩ X ¯i 2 1 = √ 1 S ⟩ ⊗¯i S ⟩ +1 (j−0) iS ⟩ ⊗ S ⟩ 2 1 = √1j S ⟩ ⊗ S 0⟩ + (−i ) S ⟩1⊗jS 1⟩ ¯i , 2 namely = (− ) √ S ⟩ ⊗ S ⟩ + (− ) S ⟩ ⊗ S ⟩ X X ψ i, j 1 j ψ i, j . (4.3.25)

And we have used the facts in⊗ the S following( )⟩ = ( to− ) goS through( )⟩ the above derivation, X i ¯i , (4.3.26) X ¯i i . S ⟩ = S ⟩ With E.Q. (4.3.23) and E.Q. (4.3.25), we getS ⟩ E.Q.= (4.3.20)S ⟩ proved.

4.3.3 Orthonormal basis of the two-qubit Hilbert space Thm 4.3.3.1. Bell states form an orhonormal basis of two-qubit Hilber space, namely

φ+ , ψ+ , φ− , ψ− is an orthonormal basis of two-qubit{S Hilbert⟩ S ⟩ space.S ⟩ S ⟩}

22 Proof. Firstly, we would prove that all the Bell states are mutually orthogonal and normalized. Secondly, the completeness of the set of the four Bell state vectors would be veriﬁed. (a) The set of the four Bell state vectors make an orthonormal vector set. From E.Q. (4.3.3), we can derive that ′ ′ j i i′ j′ ψ i, j ψ i , j ψ 0, 0 I 2 Z X I 2 X Z ψ 0, 0

1 j i+i′ j′ a T f 00 11 I 2 Z X Z 00 11 ( ) ( ) = ⟨2 ( )S ⊗ ⊗ S ( )⟩ 1 = 1 ⟨ S + ⟨ S j⊗ i+i′ j′ S ⟩ + S ⟩ kk I 2 Z X Z `` 2 k,`=0 1 = 1 Q ⟨ S ⊗ ′ ′ S ⟩ k ` k Z jX i+i Z j ` 2 k,`=0 = Q1 ⟨ S ⟩ ⊗ ⟨ S S ⟩ 1 j i+i′ j′ δk` k Z X Z ` 2 k,`=0 1 = 1 Q ⟨ S ′ ′ S ⟩ k Z jX i+i Zj k 2 k=0 = 1 ⟨ S + ′ ′ S ⟩ trQZ jX i i Z j 2 1 ′ ′ = trZ j +jX i+i . 2 Because = 2 2 Z X I 2, trZ trX tr ZX 0, and i, j, i′, j′ 0, 1 , then we can obtain = = = = ( ) = ′ ′ ∈ { } ψ i, j ψ i , j δjj′ δii′ , (4.3.27) which means that the four Bella ( states) T ( are mutually)f = orthogonal and are all of unit length. In diagrammatical representation, we use the cup conﬁguration for ket state, and cap conﬁguration for bra state, shown as

′ ′ i′ j′ j i ψ i j X Z ψ ij Z X (4.3.28) r r Therefore, the orhonormalS ( relation)⟩ = has the diagrammatical⟨ ( )S = representation

Zj Xi ψ ij ψ i′j′ (4.3.29) r ′ ′ Xi Zj ⟨ ( )S ( )⟩ = r From the diagrammatical rules, we would have the normalized trace of the single- qubit gates on the loop,

1 ′ ′ ψ ij ψ i′j′ tr Z jX iX i Z j 2 1 ′ ′ ⟨ ( )S ( )⟩ = trZ j +jX i+i 2 = δjj′ δii′ . (4.3.30) = 23 (b) The vector set consisted of the four Bell states is complete. By utilizing E.Q. (4.3.11), we can get

1 ψ i, j ψ i, j i,j=0 Q1 S 1( )⟩ ⟨ ( )S 1 0i 1 j 1¯i 0i 1 j 1¯i i,j=0 2 2 = 1Q 1√ S ⟩ + (− ) S ⟩ ⟨ S + (− ) ⟨ S √ 0i 0i 1 j 1¯i 0i 1 j 0i 1¯i 1¯i 1¯i . 2 i,j=0 As we see that= Q S ⟩⟨ S + (− ) S ⟩⟨ S + (− ) S ⟩⟨ S + S ⟩⟨ S 1 1 1 j 1¯i 0i 0, 1 j 0i 1¯i 0, (4.3.31) i,j=0 i,j=0 therefore Q (− ) S ⟩⟨ S = Q (− ) S ⟩⟨ S = 1 1 1 ψ i, j ψ i, j 0i 0i 1¯i 1¯i i,j=0 2 i,j=0 Q S ( )⟩ ⟨ ( )S = 1 Q S ⟩⟨ S + S ⟩⟨ S 0i 0i 1i 1i i=0 = Q1 S ⟩⟨ S + S ⟩⟨ S ji ji i,j=0

= IQ4, S ⟩⟨ S

i.e., = 1 ψ i, j ψ i, j I 4. (4.3.32) i,j=0 S ( )⟩ ⟨ ( )S = Remark: Q

• ψ i, j i, j 0, 1 is the Bell basis of H2 H2;

• {Si,( j )⟩i, j S 0,=1 is} the product basis of H2⊗ H2.

4.3.4{S How⟩S to= distinguish} (create) Sψ(i, j⊗)⟩. There are two diﬀerent cases (i) Alice and Bob are in the same lab, which means that the distance between them is very close. Therefore, they can do the measurement jointly, e.g. X A X B and Z A Z B. ⊗ (ii) Alice⊗ and Bob are far away from each other, which leaves them two choices:

• perform local measurement, e.g. X A I B, Z A I B, I A X B, and I A Z B. • classical communication (phone call). ⊗ ⊗ ⊗ ⊗ But, with only local operation (LO) and classical communication (CC), it is impossible to distinguish/create ψi,j . The reason is that local operation changes ψ i, j , since S ⟩ X I 2,ZZ Z 0, S ( )⟩ (4.3.33) ⎧ Z I 2,XX X 0. ⎪ [ ⊗ ⊗ ] ≠ ⎨ ⎪ [ ⊗ ⊗ ] ≠ ⎩ 24 Remark: Quantum entanglement cannot be created by remote pairs by using local operation and classical communication.

4.4 Quantum circuit model of Bell states

4.4.1 Quantum circuit model • A Quantum Circuit model is composed of a series of single-qubit-gates SU 2 U 2 and two-qubit-gates SU 4 U 4 . ( )~ ( ) • Diagrammatic representation( )~ as( ) shown in Figure 4.3.

x1 y1 x y S 2⟩ U S 2⟩ x1 y1 U x y S ⟩ S ⟩ U Sx2⟩ Sy2⟩ xn yn ⋮ ⋮ (a) single-qubit gate (b) two-qubit gate (c) n-qubit gate S ⟩ S ⟩ S ⟩ S ⟩ S ⟩ S ⟩ Figure 4.3: Quantum Circuit Model of single-qubit, two-qubit,and n-qubit gates

4.4.2 Hadamard gate: H Def 4.4.1 (Hadamard gate).

1 1 1 1 H H X Z , (4.4.1) 2 2 1 1 ∶= √ ( + ) = √ is named as Hadamard gate. − We can easily verify the following relations by utilizing the properties of the Pauli matrices,

HXH Z; (4.4.2) HZH X; (4.4.3) = HYH Y ; (4.4.4) = H † H; (4.4.5) 2 = − H I 2. (4.4.6) = As we can see that Hadamard gate H is not= only a unitary transformation but also self- adjoint.

Proof. From the deﬁnition (4.4.1) of the Hadamard gate H, and the properties of the Pauli matrices, we can derive the properties (4.4.2) (4.4.6) of the Hadamard gate.

∼

25 (1) For E.Q. (4.4.2), we can evaluate its left-hand-side, 1 1 HXH X Z X X Z 2 2 1 = √ X( Z+ X) X√ Z( + ) 2 1 = (XXX+ ) ZXX( + )XXZ ZXZ 2 1 = (X Z +Z XZZ+ + ) 2 1 = (X + Z + Z − X ) 2 Z = ZZ,( + + − ) which is equivalent to the corresponding= right-hand-side of E.Q. (4.4.2). (2) As for E.Q. (4.4.3), we can get in the following manner 1 1 HZH X Z Z X Z 2 2 1 = √ X( Z+ Z) X√ Z( + ) 2 1 = (XZX+ ) ZZX( + XZZ) ZZZ 2 1 = ( ZXX+ X X+ Z + ) 2 1 = (−Z X + X + Z + ) 2 X = XX.(− + + + ) (3) We do the same trick as in the= former two cases to derive E.Q. (4.4.4), 1 1 HYH X Z Y X Z 2 2 1 = √ X( Z+ Y) X√ Z( + ) 2 1 = (XYX+ ) ZYX( + )XYZ ZYZ 2 1 = ( Y i+ i Y + + ) 2 Y = YY.(− − + − ) (4) Because the Pauli matrices X= and− Z are self-adjoint, therefore should be the Hadamard operator H, 1 H † X Z † 2 1 = √ (X †+ Z)† 2 1 = √ (X +Z ) 2 = HH.√ ( + )

= 26 (5) Still from the deﬁnition (4.4.1) of the Hadamard gate H, we can calculate the square of the Hadamard operator H, 1 H 2 X Z 2 2 1 = (X 2+ Z)2 XX,ZZ 2 1 = I +I +0 2 2 2 I = I 2(. + + ) Therefore, we have shown that H= gate is a self-adjoint unitary 2 2 matrix. In the view point of rotation operator, we have × 1 H σx σz 2 = √1 ( + ) ex ez σ, 2 namely = √ (⃗ + ⃗ ) ⋅⃗ 1 3 H n σσ, with n eˆx eˆy R . (4.4.7) 2 Thus, = ⃗⋅⃗ ⃗ ∶= √ ( + ) ∈ θ iHH exp i n σ , 2 θ=π which we can put in another way = − ⃗⋅⃗V iHH D n, π , (4.4.8) namely a unitary transformation induced= by( the⃗ rotation) along n through angle π in the three-dimensional Euclidean space. As a matter of fact, the Hadmard gate H can make a unitary basis transformations between Z basis and X basis: ⃗ H 0 , (4.4.9) H 1 , S ⟩ = S+⟩ and S ⟩ = S−⟩ H 0 , (4.4.10) H 1 . S+⟩ = S ⟩ E.Q. (4.4.9) can be easily derived by utilizing the deﬁnition (4.4.1) of the Hadamard gate S−⟩ = S ⟩ H as shown in the following 1 H 0 X Z 0 2 1 S ⟩ = √ ( 1 + 0)S ⟩ 2 = √ ,(S ⟩ + S ⟩) and = S+⟩ 1 H 1 X Z 1 2 1 S ⟩ = √ ( 0 + 1)S ⟩ 2 = √ .(S ⟩ − S ⟩)

= S−27⟩ While E.Q. (4.4.10) can be deduced by applying H on both sides of E.Q. (4.4.9) and utilizing E.Q. (4.4.6). Furthermore, We can reformulate E.Q. (4.4.9) into one index form, expressed as 1 H i 1 i i ¯i , (4.4.11) 2 where i 0, 1. S ⟩ = √ (− ) S ⟩ + S ⟩ On the other hand we can also get = 1+i H y y , 2 (4.4.12) 1−i ⎧ H y y , ⎪ S↑ ⟩ = 2 S↓ ⟩ ⎨ ⎪ also as a consequence of E.Q. (4.4.9)⎩⎪ S↓ ⟩ = S↑ ⟩ 1 1 H 0 i 1 H 0 iHH 1 2 2 1 √ S ⟩ ± S ⟩ = √ S ⟩i± S ⟩ 2 1 = √ 0 S+⟩ ±1 S−i⟩ 0 i 1 2 1 = S1⟩ +i S 0⟩ ± S 1⟩ ∓i S1⟩ 2 1 i 1 i = ( ±0)S ⟩ + ( 1∓ )S ⟩ 2 1 i 1±i ∓ = S0⟩ +i 1 S. ⟩ 2 ± ± = S ⟩ ∓ S ⟩ 4.4.3 CNOT gate Def 4.4.2 (CNOT). The Controlled-Not gate, or CNOT gate for short, is deﬁned as

CNOT a b a, b a, a b , a b a b mod 2, (4.4.13) where we call a as∶ theS controlled⟩ ⊗ S ⟩ = S qubit,⟩ ↦ Sb as⊕ the⟩ target⊕ = qubit.( + The) CNOT gate is usually represented with the diagram S ⟩ S ⟩ a a CNOT ab Sb⟩ ● Sb⟩ a . S ⟩ = Remarks: S ⟩ S ⊕ ⟩ (1) With a 0, then CNOT 0 b 0 b . (4.4.14) = With a 1, then S ⟩ ⊗ S ⟩ = S ⟩ ⊗ S ⟩ CNOT 1 b 1 ¯b . (4.4.15) = It is the reason why the CNOT gateS is⟩ ⊗ calledS ⟩ = S controlled⟩ ⊗ S ⟩ not gate, namely if target bit is 1, we do the NOT operation.

28 (2) The controlled-U gate is deﬁned as

CU c t c U c t . (4.4.16)

S ⟩S ⟩ ∶= S ⟩ S ⟩ CU U● See more details and properties of the= controlled-U gate in following section 8.1.2. (3) From the deﬁnition of the CNOT gate, we can verify that

2 CNOT I 4, (4.4.17)

since ( ) = CNOT 2 a, b CNOT a, b a a, b a a ( ) S ⟩ = S ⊕ ⟩ a, b , = S ( ⊕ )⊕ ⟩ with a, b 0, 1. = S ⟩

(4) The CNOT= gate can be expressed as

CNOT 0 0 I 2 1 1 XX. (4.4.18)

by which we can get the matrix formalism,= S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 1 0 1 0 0 0 0 1 CNOT 0 0 0 1 0 1 1 0 namely = ⊗ + ⊗ 1 1 CNOT . (4.4.19) ⎛ 1 ⎞ ⎜ ⎟ = ⎜ 1 ⎟ ⎜ ⎟ (5) CNOT gate is the quantum analogy of⎝ the classical⎠ gate XOR (the Exclusive OR gate): XOR a, b a b , irreversible,

2 bits 1 bit ⎪⎧ ∶ ( ) → ⊕ ⎪ ± ⎪ ² ⎪ CNOT a , b a , a b , reversible. ⎨ ⎪ 2 qubits 2 qubits ⎪ ∶ (S ⟩ S ⟩) → (S ⟩ S ⊕ ⟩) ⎪ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ (6) CNOT SU 4 .⎩⎪ Since we have derived the matrix formalism, we can see ∈ ( ) det CNOT 1, (4.4.20)

and = CNOT † CNOT. (4.4.21) On the other hand, we know that CNOT 2 I , thus ( ) = 4 † † 2 CNOT CNOT CNOT( CNOT) = CNOT I 4.

Therefore, we can infer( that CNOT) = SU(4 . ) = ( ) =

∈29 ( ) Lemma 4.4.3.1. The CNOT gate has the following properties

X X CNOT CNOT X I 2 , (4.4.22a) ⎧ X I 2 CNOT CNOT X X , (4.4.22b) ⎪ ( ⊗ ) = ( ⊗ ) ⎪ I 2 X CNOT CNOT I 2 X , (4.4.22c) ⎪( ⊗ ) = ( ⊗ ) ⎪ Z Z CNOT CNOT I 2 Z , (4.4.22d) ⎪( ⊗ ) = ( ⊗ ) ⎨ I 2 Z CNOT CNOT Z Z , (4.4.22e) ⎪ ( ⊗ ) = ( ⊗ ) ⎪ Z I 2 CNOT CNOT Z I 2 . (4.4.22f) ⎪ ( ⊗ ) = ( ⊗ ) ⎪ Proof. Now, we are going⎪ to verify E.Q.(4.4.22a) E.Q.(4.4.22f), one by one, from the ⎩⎪ ( ⊗ ) = ( ⊗ ) expression (4.4.18) of CNOT gate. ∼ (a) X X CNOT CNOT X I 2

( ⊗ ) =X X CNOT( ⊗ ) X X 0 0 I 2 1 1 X X 0 0 X X 1 1 X 2 ( ⊗ ) = ( ⊗ ) S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 1 1 X XXII 2 0 0 X I 2I 2 = S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 1 1 X 0 0 I X I = S ⟩⟨ S ⊗ + S ⟩⟨ S 2 ⊗ 2 = CNOT S ⟩⟨ S ⊗X +IS2 ⟩⟨. S ⊗ ⊗ (4.4.23) = ⊗ (b) X I 2 CNOT CNOT X X

( ⊗ ) =X I 2 CNOT( ⊗ ) X I 2 0 0 I 2 1 1 X

X 0 0 I 2 X 1 1 X ( ⊗ ) = ( ⊗ ) S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 1 1 X XX 0 0 X I 2X = S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 1 1 X 0 0 I X X = S ⟩⟨ S ⊗ + S ⟩⟨ S 2 ⊗ CNOT X X . (4.4.24) = S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ ( ⊗ ) = ( ⊗ ) (c) I 2 X CNOT CNOT I 2 X

( ⊗ ) =I 2 X CNOT( ⊗ ) I 2 X 0 0 I 2 1 1 X 0 0 X 1 1 X 2 ( ⊗ ) = ( ⊗ ) S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 0 0 I 2X 1 1 XX = S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 0 0 I 1 1 X I X = S ⟩⟨ S ⊗ 2 + S ⟩⟨ S ⊗ 2 CNOT I X . (4.4.25) = S ⟩⟨ S ⊗ 2 + S ⟩⟨ S ⊗ ( ⊗ ) = ( ⊗ ) (d) Z Z CNOT CNOT I 2 Z

( ⊗ ) =Z Z CNOT( ⊗ ) Z Z 0 0 I 2 1 1 X Z 0 0 Z Z 1 1 ZX ( ⊗ ) = ( ⊗ ) S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 0 0 Z 1 1 XZ = S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 0 0 I 2Z 1 1 XZ = S ⟩⟨ S ⊗ − S ⟩⟨ S ⊗(− ) 0 0 I 1 1 X I Z = S ⟩⟨ S ⊗ 2 + S ⟩⟨ S ⊗ 2 CNOT I Z . (4.4.26) = S ⟩⟨ S ⊗ 2 + S ⟩⟨ S ⊗ ( ⊗ ) = ( ⊗ ) 30 (e) I 2 Z CNOT CNOT Z Z

( ⊗ ) =I 2 Z CNOT( ⊗ ) I 2 Z 0 0 I 2 1 1 X 0 0 Z 1 1 ZX ( ⊗ ) = ( ⊗ ) S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 0 0 Z I 2Z 1 1 Z XZ = S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 0 0 I 1 1 X Z Z = S ⟩⟨ S ⊗ 2 + S ⟩⟨ S ⊗ CNOT Z Z . (4.4.27) = S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ ( ⊗ )

(f) Z I 2 CNOT CNOT Z I 2 = ( ⊗ )

( ⊗ ) =Z I 2 CNOT( ⊗ ) Z I 2 0 0 I 2 1 1 X Z 0 0 I 2 Z 1 1 X ( ⊗ ) = ( ⊗ ) S 2⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 0 0 Z I 1 1 Z XXII 2 = S ⟩⟨ S ⊗ 2 + S ⟩⟨ S ⊗ 0 0 I 1 1 X Z I = S ⟩⟨ S ⊗ 2 + S ⟩⟨ S ⊗ 2 CNOT Z I . (4.4.28) = S ⟩⟨ S ⊗ + S2 ⟩⟨ S ⊗ ( ⊗ ) There we get all the six relations (4.4.22a)= (4.4.22f)( ⊗ veriﬁed.)

On the other hand, we can present these properties∼ of the CNOT gate, namely the six relations (4.4.22a) (4.4.22f), in the form of quantum circuit diagram, which is shown in Figure 4.4. ∼ X X X X

X ● ● ● ● X = = (a) X X CNOT CNOT X I 2 (b) X I 2 CNOT CNOT X X

( ⊗ ) = ( ⊗ ) ( ⊗ ) = ( ⊗ ) Z

● ● X ● ● X Z Z = = (c) I 2 X CNOT CNOT I 2 X (d) Z Z CNOT CNOT I 2 Z

( ⊗ ) = ( ⊗ ) ( ⊗ ) = ( ⊗ ) Z Z Z Z

Z ● ● Z ● ● = = (e) I 2 Z CNOT CNOT Z Z (f) Z I 2 CNOT CNOT Z I 2

( ⊗ )Figure 4.4:= Properties( ⊗ ) of the CNOT(gate⊗ ) (4.4.22a)= (4.4.22f).( ⊗ )

∼ Lemma 4.4.3.2. The Hadmard gate and CNOT gate have the following connection

H H CNOT12 H H CNOT21, (4.4.29) with CNOT12 and CNOT( 21⊗deﬁned) as ( ⊗ ) = CNOT a, b a, a b , 12 (4.4.30) CNOT21 a, b a b, b . ∶ S ⟩ ↦ S ⊕ ⟩ ∶ S ⟩ ↦ S ⊕ ⟩ 31 Proof. The prove is very simple. We can utilize the properties of the Hadmard gate as expressed in E.Q.(4.4.2) (4.4.6) in the following way

H∼ H CNOT12 H H H H 0 0 I 1 1 X H H ( ⊗ ) (2 ⊗ ) 2 = H( 0⊗ 0)H S ⟩⟨H S ⊗ H+1S ⟩⟨1 HS ⊗ HXH( ⊗ ) I 2 Z = S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ 0 0 1 1 0 0 1 1 = S+⟩⟨+S ⊗ + S−⟩⟨−S ⊗ = S+⟩⟨+S ⊗ S ⟩⟨ S + S ⟩⟨0 S0 + S−⟩⟨−S ⊗ S ⟩⟨ S − S ⟩⟨1 S 1 I 0 0 X 1 1 = 2 S+⟩⟨+S + S−⟩⟨−S ⊗ S ⟩⟨ S + S+⟩⟨+S − S−⟩⟨−S ⊗ S ⟩⟨ S CNOT21. (4.4.31) = ⊗ S ⟩⟨ S + ⊗ S ⟩⟨ S In this derivation,= we have also used the following facts

CNOT12 0 0 I 2 1 1 XX,

⎧ CNOT12 I 2 0 0 X 1 1 , ⎪ = S ⟩⟨ S ⊗ + S ⟩⟨ S ⊗ ⎨ ⎪ and ⎩⎪ = ⊗ S ⟩⟨ S + ⊗ S ⟩⟨ S I 2 0 0 1 1 , ⎧ I 2 , ⎪ = S ⟩⟨ S + S ⟩⟨ S ⎪ Z 0 0 1 1 , ⎪ = S+⟩⟨+S + S−⟩⟨−S ⎨ X . ⎪ = S ⟩⟨ S − S ⟩⟨ S ⎪ ⎪ And we can also represent the⎩⎪ relation= (4.4.29)S+⟩⟨+S − betweenS−⟩⟨−S the Hadmard gate and CNOT gate in the diagram formalism, see Figure 4.5:

H H

● H H = Figure 4.5: H H CNOT12 H H ● CNOT21

( ⊗ ) ( ⊗ ) = 4.4.4 Quantum circuit model of Bell states Thm 4.4.4.1. With the help of the Hadmard gate and CNOT gate, we can construct the Bell state ψ i, j from the product state j, i

S ( )⟩ ψ i, j CNOT HS I 2⟩ j, i , with i, j 0, 1, (4.4.32) which can also be expressedS ( )⟩ in= the Quantum( ⊗ circuit)S ⟩ model =

phase-bit j H ψ i, j . S ⟩ ● parity-bit i = S ( )⟩ Here we present two typesS of⟩ proofs.

32 Proof. The ﬁrst type of proof. From the index formulation of the Hadamard gate E.Q. (4.4.11),

1 j H I 2 j i 1 j i ¯j i . (4.4.33) 2 Following the CNOT gate,( ⊗ we)S get⟩S ⟩ = √ (− ) S ⟩S ⟩ + S ⟩S ⟩

1 j CNOT H I 2 j i 1 j i j ¯j i ¯j . (4.4.34) 2 And considering the case(j ⊗0 and)S ⟩Sj ⟩ =1,√ (− ) S ⟩S ⊕ ⟩ + S ⟩S ⊕ ⟩

j = 0 √1 =0 i 1 ¯i ψ i, 0 2 (4.4.35) ⎪⎧ j 1 √1 0 i 1 ¯i ψ i, 1 ⎪ = ∶ 2 S ⟩S ⟩ + S ⟩S ⟩ = S ( )⟩ ⎨ Conclusively, we have ⎪ = ∶ S ⟩S ⟩ − S ⟩S ⟩ = S ( )⟩ ⎩⎪ CNOT H I 2 j i ψ i, j . (4.4.36)

( ⊗ )S ⟩S ⟩ = S ( )⟩ Proof. The second type of proof. To prove E.Q.(4.4.32), we may make use of the deﬁnition (4.3.2) of the Bell states and E.Q. (4.3.3). We are going to reach our destination by two main steps.

Step 1: We will prove that ψ 0, 0 CNOT H I 2 00 . As we can show that

CNOT HS (I 2 00)⟩ = CNOT( ⊗ H)S0 ⟩ 0 1 ⊗ S ⟩ = CNOT S ⟩ ⊗ S ⟩0 2 1 = √ CNOT S00+⟩ ⊗ S 10⟩ 2 1 = √ 00 11 S ⟩ + S ⟩ 2 = √ψ 0, S0 ⟩.+ S ⟩ (4.4.37)

i j Step 2: With the equation ψ i, j I=2 XS (Z )⟩ψ 0, 0 , we can get i j ψ i, j S (I 2 )⟩X=Z( ⊗ψ 0, 0)S ( )⟩ i j S ( )⟩ = I 2⊗X Z SCNOT( )⟩ H I 2 00 i j = I 2⊗X I2 Z CNOT ⊗ H S I 2⟩ 00 i j = I 2⊗X CNOT ⊗ Z Z H ⊗I 2 S00 ⟩ i j = CNOT ⊗ I 2 X Z⊗Z H⊗I 2 S 00⟩ ,

in the last two jumps, we= have used the ⊗ relations ⊗ (4.4.22e) ⊗ and S (4.4.22c),⟩ which we can veriﬁed in the following context:

i i−1 I 2 X CNOT I 2 X CNOT I2 X

( ⊗ ) = ( ⊗ ) i ( ⊗ ) CNOT I 2 X , ⋮ =33 ( ⊗ ) and

j i−1 I 2 Z CNOT I 2 Z CNOT Z Z

( ⊗ ) = ( ⊗ ) ( ⊗ ) CNOT Z Z j. ⋮ And we should notice that = ( ⊗ )

i j j j i j I 2 X Z Z Z X Z 1 jZ j X i−1Z jX ⊗ ⊗ = ⊗ = (− ) ⊗ 1 ijZ j Z jX i ⋮ ij j j i 1 Z Z I 2 X . (4.4.38) = (− ) ⊗ Thus, we can infer that = (− ) ( ⊗ )( ⊗ )

ij j j i ψ i, j 1 CNOT Z Z I 2 X H I 2 00 1 ijCNOT Z j Z j H X i 00 S ( )⟩ = (− ) ( ⊗ )( ⊗ ) ⊗ S ⟩ ij j j i 1 CNOT Z Z H I 2 I 2 X 00 = (− )ij ( j⊗ j)( ⊗ )S ⟩ 1 CNOT Z Z H I 2 0i . = (− ) ( ⊗ )( ⊗ )( ⊗ )S ⟩ On the other hand, if we’ve= (− noticed) that( ⊗ )( ⊗ )S ⟩

HZH X 2 ZH HXHX, (4.4.39) H I 2 = ¡ ⇒ = then =

Z jH Z j−1HX

= HX j. (4.4.40) ⋮ Hence, =

ψ i, j 1 ijCNOT Z jH Z j 0i 1 ijCNOT HX j Z j 0i S ( )⟩ = (− )ij ( ⊗ )Sj ⟩ j 1 CNOT H I 2 X Z 0i = (− )ij ( ⊗ )Sj ⟩ j 1 CNOT H I 2 X 0 Z i = (− )ij ( ⊗ )( ⊗ )Sij ⟩ 1 CNOT H I 2 j 1 i = (− ) ( ⊗ )( S ⟩ ⊗ S ⟩) CNOT H I 2 ji , = (− ) ( ⊗ )(S ⟩ ⊗(− ) S ⟩) which is equivalent to E.Q.= (4.4.32). ( ⊗ )S ⟩

34 Chapter 5

No-Cloning, Dense Coding, Teleportation and Cryptography

Reference:

[Preskill] Chapter 4: Quantum entanglement.

● [Nielsen & Chuang] Chapter 2: Introduction to quantum mechanics.

● [Nielsen & Chuang] Chapter 12: quantum information theory.

● 5.1 No-cloning theorem

Def 5.1.1 (Cloning Machine). The cloning machine is a unitary transformation U, which satisﬁes U φ 0 φ φ (5.1.1) for arbitrary state φ . It can be represented S ⟩ ⊗ S ⟩ in = theS ⟩ ⊗ diagramS ⟩ shown in Figure 5.1. S ⟩ Target object φ φ U Blank object S0⟩ Sφ⟩

S ⟩ S ⟩ Figure 5.1: Copy machine in quantum mechanics

Thm 5.1.0.2 (No-cloning theorem 1). The cloning machine doesn’t exist (in Quantum Mechanics).

Proof. For simplicity, we deal with the two-dimensional Hilbert space. As the deﬁnition (5.1.1) shows, we choose the blank object to be 0 and the target object to be be 0 or 1 , and the copy machine satisﬁes the following equations: S ⟩ S ⟩ S ⟩ U 0 0 0 0 , (5.1.2) ⎧ U 1 0 1 1 . ⎪ S ⟩ ⊗ S ⟩ = S ⟩ ⊗ S ⟩ ⎨ ⎪ ⎩⎪ S ⟩ ⊗ S ⟩ = S ⟩ ⊗ S ⟩

35 For example, the most popular two-qubit quantum gate in quantum computation is the CNOT gate, which has the property,

CNOT 0 0 0 0 , (5.1.3) ⎧ CNOT 1 0 1 1 . ⎪ S ⟩ ⊗ S ⟩ = S ⟩ ⊗ S ⟩ ⎨ ⎪ For φ H2, which has the⎩⎪ form of S ⟩ ⊗ S ⟩ = S ⟩ ⊗ S ⟩ 2 2 ∀ S ⟩ ∈ φ a 0 b 1 , with a b 1, (5.1.4) we obtain S ⟩ = S ⟩ + S ⟩ S S + S S = U φ 0 U a 0 b 1 0 aU 0 0 bU 1 0 S ⟩ ⊗ S ⟩ = S ⟩ + S ⟩ ⊗ S ⟩ a 0 0 b 1 1 , = S ⟩ ⊗ S ⟩ + S ⟩ ⊗ S ⟩ i.e., = S ⟩ ⊗ S ⟩ + S ⟩ ⊗ S ⟩ U φ 0 a 0 0 b 1 1 . (5.1.5) However, the copy machine (5.1.1) tells us another thing: S ⟩ ⊗ S ⟩ = S ⟩ ⊗ S ⟩ + S ⟩ ⊗ S ⟩ U φ 0 φ φ a 0 b 1 a 0 b 1 S ⟩ ⊗ S ⟩ = S ⟩ ⊗ S ⟩ a2 0 0 b2 1 1 ab 0 1 ab 1 0 . (5.1.6) = S ⟩ + S ⟩ ⊗ S ⟩ + S ⟩ In general, E.Q. (5.1.5) and= E.Q.S ⟩ (5.1.6)⊗ S ⟩ + areS ⟩ not⊗ S consistent⟩ + S ⟩ ⊗ withS ⟩ + oneS ⟩ another.⊗ S ⟩ Therefore, the cloning machine is not available for φ H2.

Remarks: ∀ S ⟩ ∈ • The copy machine is a non-linear process, but Quantum Mechanics respects linear super-position principle.

• The no-cloning theorem is compatible with the Heisenberg uncertainty relation. If a state can be exactly copied, then it can be exactly measured, which violates the Heisenberg uncertainty relation. Thm 5.1.0.3 (No-cloning theorem 2). There is no unitary transformation (cloning machine) which can make copies on distinct non-orthogonal states. Proof. We ﬁrstly consider that case that if such a unitary transformation U exists, what we can obtain. And choose two unital state vectors φ and ψ : being distinct means that

φ ψ 1. S ⟩ S ⟩ (5.1.7)

While, being non-orthogonal means that⟨ S ⟩ ≠ φ ψ 0. (5.1.8)

Because the copy machine U satisﬁes E.Q.⟨ S (5.1.1),⟩ ≠ we ﬁnd out φ 0 ψ 0 φ 0 U †U ψ 0 φ φ ψ ψ (⟨ S⟨ S) (S ⟩S ⟩) = ⟨ S⟨ S S ⟩S ⟩ φ ψ 2 , = (⟨ S⟨ S) (S ⟩S ⟩) =36 ⟨ S ⟩ thus φ ψ φ ψ 2 . (5.1.9) It’s clear that the E.Q.(5.1.9) will either violate the relation (5.1.7) or (5.1.8). Therefore, ⟨ S ⟩ = ⟨ S ⟩ the assumption is invalid, i.e., such a unitary transformation U does not exist. Remark: Two orthogonal states can be exactly copied:

U 0 0 00 , (5.1.10) ⎧ U 1 0 11 ⎪ (S ⟩ ⊗ S ⟩) = S ⟩ ⎨ where U can be the CNOT gate⎪ in quantum computation. ⎩⎪ (S ⟩ ⊗ S ⟩) = S ⟩ Thm 5.1.0.4 (No-cloning theorem 3). There is no unitary transformation to distinguish two non-orthogonal states without disturbing them. This is actually the third version of the no-cloning theorem. Proof. We denote two arbitrary distinct non-orthogonal normalized states with φ and ψ , S ⟩ φ ψ 1, S ⟩ (5.1.11) ⎧ φ ψ 0. ⎪ ⟨ S ⟩ ≠ ⎨ Assume that the unitary transformation U can distinguish φ and ψ , without disturbing ⎪ ⟨ S ⟩ ≠ them, namely ⎩ S ⟩ S ⟩ U φ 0 φ e , (5.1.12) ⎧ U ψ 0 ψ f ⎪ S ⟩ ⊗ S ⟩ = S ⟩ ⊗ S ⟩ ⎨ where e f 1, means φ can⎪ be distinguished from ψ . Because ⎩⎪ S ⟩ ⊗ S ⟩ = S ⟩ ⊗ S ⟩ φ 0 ψ 0 φ 0 U †U ψ 0 ⟨ S ⟩ ≠ S ⟩ S ⟩ φ e ψ f ⟨ S ⊗ ⟨ S S ⟩ ⊗ S ⟩ = ⟨ S ⊗ ⟨ S S ⟩ ⊗ S ⟩ φ ψ e f , = ⟨ S ⊗ ⟨ S S ⟩ ⊗ S ⟩ which is = ⟨ S ⟩⟨ S ⟩ φ ψ φ ψ e f . (5.1.13) From E.Q.(5.1.13) we know that either ⟨ S ⟩ = ⟨ S ⟩⟨ S ⟩ φ ψ 0 (5.1.14) or ⟨ S ⟩ = e f 1 (5.1.15) must be true.Therefore we are able to distinguish two orthogonal states without disturbing ⟨ S ⟩ = them, or we cannot distinguish two non-orthogonal states without disturbing them. Remarks: • The no-cloning theorem means that no quantum cloning machine exists, which may be bad news to quantum mechanics, but is good to quantum information security or quantum cryptography.

• The no-cloning theorem denies the possibility for a third party to extract information from the communication between the other two parties, without being noticed (without disturbance on the communication), if they make use of the resource of non-orthogonal states. Therefore, the security of information can be ensured.

37 5.2 Dense coding

Alice sends a qubit

(1st qubit) Alice φ+ Bob (2nd qubit)

S ⟩ Bob gets two-bits of information

Dense coding is that, with the entangled resource, Alice sends Bob two classical bits of information by transmitting a qubit to Bob1. Dense coding can be executed in the following manner step by step:

Step 1: Experiment setup. Alice and Bob share a maximally entangled state, e.g.

+ Alice φ AB Bob

S ⟩ which is the cup representation for the Bell state φ+ deﬁned in (4.3.7).

Step 2: Local unitary transformation. S ⟩ Alice chooses one of the four unitary transformation I2, X, Z, ZX and performs it on her qubit. { } 1 ψ 0, 0 00 11 (5.2.1) 2 S ( )⟩ = √ S ⟩ + S ⟩ ψ 00 AB I2 I2

S ( )⟩ =Alice Bob = Alice Bob

1 ψ 0, 1 00 11 2 S ( )⟩ = I√2 ZS ψ⟩ −0,S0 ⟩ Bob Z I2 ψ 0, 0 Alice (5.2.2) = ⊗ S ( )⟩ = ⊗ S ( )⟩ ψ 01 AB Z Z

S ( )⟩ =Alice Bob = Alice Bob

1 ψ 1, 0 01 10 2 S ( )⟩ = I√2 XS ψ⟩ +0,S 0 ⟩ Bob X I2 ψ 0, 0 Alice (5.2.3) = ⊗ S ( )⟩ 1 Note: The Holevo bound (Old Chapter 5.4.1,= page 36,⊗ JohnS ( Preskill’s)⟩ lecture notes) says that without entanglement at most a classical bit information can be transmitted via sending a qubit.

38 ψ 10 AB X X

S ( )⟩ =Alice Bob = Alice Bob

1 ψ 1, 1 01 10 2 S ( )⟩ = I√2 XZS ⟩ ψ− S0, 0⟩ Bob ZX I2 ψ 0, 0 Alice (5.2.4) = ⊗ S ( )⟩ = ⊗ S ( )⟩ ψ 11 AB ZX XZ

S ( )⟩ =Alice Bob = Alice Bob

The texts “Bob” and “Alice” appearing in the above equations mean that the corresponding systems, that are “Bob” and “Alice”, on which the associated nontrivial local unitary transformations are to be performed to obtain the rightful target states.

Step 3: Qubit transmission. Alice sends her qubit to Bob.

XiZj XiZj Alice Bob Bob

Step 4: Bell measurements. Bob performs the Bell measurement:

X X ψ ij 1 j ψ ij , (5.2.5)

which gives the parity bit i,⊗ and S ( )⟩ = (− ) S ( )⟩

Z Z ψ ij 1 i ψ ij (5.2.6)

from which gives the phase bit⊗ j. S With( )⟩ =i,( j−,) BobS ( get)⟩ two bits of information.

As we see that, for each two-bit i, j that Alice( wants) to send to Bob, she just modiﬁes the qubit in her system with the corresponding local unitary transformation by utilizing the protocol as illustrated in Table( 5.1,) and then transmits such the qubit to Bob. Remarks:

• On the one hand, the word “dense” in dense coding means that sending one qubit is to transmit two classical bits.

• On the other hand, we still have that sending two qubits is to transmit two classical bits, if we think about it the following way: Alice prepares the entangled state φ+ and then sends one qubit to Bob, so Alice sends two qubits to Bob in the entire procedure. S ⟩

39 Table 5.1: Dense coding

Local unitary transf. Final state Two bits Alice Bob Bob + I2 φ ψ 0, 0 0, 0 X ψ+ ψ 1, 0 1, 0 S ⟩ = S ( )⟩ ( ) Z φ− ψ 0, 1 0, 1 S ⟩ = S ( )⟩ ( ) ZX ψ− ψ 1, 1 1, 1 S ⟩ = S ( )⟩ ( ) S ⟩ = S ( )⟩ ( ) 5.3 Quantum teleportation

• Y.Z., “Braid Group, Temperley–Lieb Algebra, and Quantum Information and Com- putation”, arXiv: quant-ph/0601050.

• Y.Z., Jinglong Pang, “Space-Time Topology in Teleportation-Based Quantum Com- putation”, arXiv:1309.0955.

• Y.Z., Kun Zhang, “Bell Transform, Teleportation Operator and Teleportation-Based Quantum Computation”, arXiv:1401.7009.

In some sense, Quantum Teleportation is a kind of inverse process of dense coding (see Table 5.2).

Alice sends two classical bits to Bob

(1st qubit) Alice φ+ Bob (2nd qubit)

S ⟩ Bob gets one qubit from Alice

Table 5.2: Dense coding vs. Quantum teleportation

Resource Send transmit Dense coding 1 qubit 2 bits φ+ Quantum Teleportation 2 bits 1 qubit S ⟩ Task: Alice wants to send an unknown qubit to Bob, who is far away from her.

Lemma 5.3.0.1. 1 ψ φ+ φ+ ψ X I φ+ X ψ Z I φ+ Z ψ 2 2 2

S ⟩ ⊗ S ⟩ = S ⟩ ⊗ S ⟩ + +⊗ S ⟩ ⊗ S ⟩ + ⊗ S ⟩ ⊗ S ⟩ ZX I2 φ XZ ψ , (5.3.1)

+ ⊗ S ⟩ ⊗ S ⟩

40 which can be represented in the diagram formulism also,

1 2 X X Z Z ZX XZ . ⎧ ⎫ ⎪ ⎪ ψ φ+ = ⎨ + ● ● + ● ● + ● ● ⎬ ⎪ ⎪ (5.3.2) ⎩⎪ ⎭⎪ S ⟩ S ⟩ Proof. We can give an expression to unknown state ψ

2 2 ψ a 0 b 1 , with a, b CS and⟩ a b 1. (5.3.3)

From the deﬁnition ofS ⟩ the∶= fourS ⟩ + BellS ⟩ states (4.3.2),∈ we can get+ =

00 √1 φ+ φ− , 2 ⎧ ⎪ S01⟩ = √1 Sψ+⟩ + Sψ−⟩ , ⎪ 2 ⎪ (5.3.4) ⎪ ⎪ S10⟩ = √1 Sψ+⟩ + Sψ−⟩ , ⎪ 2 ⎨ ⎪ 1 + − ⎪ S11⟩ = √ Sφ ⟩ − Sφ ⟩ . ⎪ 2 ⎪ ⎪ S ⟩ = S ⟩ − S ⟩ With these materials we can make⎩⎪ the following derivation ψ φ+ 1 a 0 b 1 00 11 S ⟩S2 ⟩ = √1 S ⟩ + S ⟩ S ⟩ + S ⟩ a 0 00 11 b 1 00 11 2 = √1 S ⟩ S ⟩ + S ⟩ + S ⟩ S ⟩ + S ⟩ a 00 0 01 1 b 10 0 11 1 2 = 1√ S ⟩ + S ⟩ + S ⟩ + S ⟩ a φ+¯ φ− ¯0 b ψ+¯ ψ− ¯0 2 = S ⟩ + S ⟩ S ⟩ + S ⟩ − S ⟩ S ⟩ a ψ+ ψ− 1 b φ+ φ− 1

1+ S ⟩ + S ⟩ S ⟩ + S ⟩ − S ⟩ S ⟩ φ+ a 0 b 1 φ− a 0 b 1 2 = S ⟩ S ⟩ + S ⟩ + S ⟩ S ⟩ − S ⟩ ψ+ a 1 b 0 ψ− a 1 b 0

1+ S ⟩ S ⟩ + S ⟩ + S ⟩ S ⟩ − S ⟩ φ+ ψ Z I φ+ Z ψ X I φ+ X ψ 2 2 2

= S ⟩⊗S ⟩ + + ⊗ S ⟩⊗ S ⟩ + ⊗ S ⟩⊗ S ⟩ ZX I 2 φ XZ ψ ,

+ ⊗ S ⟩ S ⟩ which is equivalent to E.Q. (5.3.1). Therefore, Lemma 5.3.0.1 has been proved.

Remarks: Magic of QM.

• With the superposition principle in QM, for one state, it can be realized by the superposition of many other states, namely Lemma 5.3.0.1. (1 N)

→ 41 • In QM, wave function collapse due to the measurement process, i.e., measurement can extract one state from the superposition of many other states. (N 1)

The Quantum Telecportation can be accomplished with the following steps: → Step 1: State preparation. Alice and Bob share the entangled state φ+ . And Alice has the unknown qubit

ϕ A in hand. This can also be represented in the form of diagram: S ⟩ S ⟩ Alice & Bob:

+ ϕ A A φ B.

S ⟩ S ⟩ Step 2: Bell measurement by Alice. Alice makes joint measurement for the observables X X and Z Z, on the composite of the subsystem A and the unknown particle that Alice wants to send to Bob. The measurement results and the two-bit⊗ information⊗ associated with the measurement datum, along with post measurement states, are listed in Table 5.3. Table 5.3: Alice’s measurement result and the two-bit information

post-measurement state Z Z X X two-bit φ+ 1 1 0, 0 ⊗ ⊗ φ− 1 1 0, 1 S ⟩ ( ) ψ+ 1 1 1, 0 S ⟩ − ( ) ψ− 1 1 1, 1 S ⟩ − ( ) S ⟩ − − ( ) It is transparent to that the measurement of the observables X X and Z Z are equivalent to the projection measurements for the Bell measurement, deﬁned as ⊗ ⊗ + + E00 φ φ , ⎧ − − ⎪ E01 ∶= Sφ ⟩⟨φ S , ⎪ (5.3.5) ⎪ E ψ+ ψ+ , ⎪ 10 ∶= S ⟩⟨ S ⎨ − − ⎪ E11 ∶= Sψ ⟩⟨ψ S , ⎪ which can be represented in⎪ a diagrammatic formalism shown as ⎩⎪ ∶= S ⟩⟨ S X , , E 00 E 01 X ● = = Z ZX ● , E 10 E 11 . ●Z XZ ● = = ● ●

42 We now utilize Lemma 5.3.0.1, which actually means 1 ψ φ+ φ+ ψ φ− Z ψ ψ+ X ψ 2 S ⟩ ⊗ S ⟩ = S ⟩⊗S ⟩ + S ⟩⊗ S ⟩ + S ⟩⊗ S ⟩ ψ− XZ ψ . (5.3.6)

Therefore, after the Bell measurement+ S ⟩ carriedS ⟩ out on the qubit of Alice and the unknown state, we obtain

+ + + 1 + φ φ I2 ψ φ 2 φ ψ ,

⎧ − − + 1 − ⎪ Sφ ⟩⟨φ S ⊗I2 Sψ⟩ ⊗ Sφ ⟩ = Sφ ⟩ ⊗SZ ⟩ψ , ⎪ 2 ⎪ (5.3.7) ⎪ Sψ+ ⟩⟨ψ+S ⊗I Sψ⟩ ⊗ Sφ+⟩ = 1 Sψ+⟩ ⊗XS ψ⟩ , ⎪ 2 2 ⎨ ⎪ − − + 1 − ⎪ Sψ ⟩⟨ψ S ⊗I2 Sψ⟩ ⊗ Sφ ⟩ = 2 Sψ ⟩ ⊗XZS ⟩ψ . ⎪ ⎪ E.Q.(5.3.7) can⎪ also S be⟩⟨ rewrittenS ⊗ Sin⟩ the⊗ S diagram⟩ = languageS ⟩ ⊗ shownS ⟩ in the following context: ⎩ + + + 1 + • φ φ I2 ψ φ 2 φ ψ S ⟩⟨ S ⊗ S ⟩ ⊗ S ⟩ = S ⟩ ⊗S ⟩ Measurement 1 Preparation 2 A A B = A A B (5.3.8)

− − + 1 − • φ φ I2 ψ φ 2 φ Z ψ

S ⟩⟨ S ⊗ S ⟩ ⊗ S ⟩ = ZS ⟩ ⊗ S ⟩ Z Measurement Z Z Z ● 1 ● Z 2 Preparation ● ● A A B = A A B (5.3.9)

+ + + 1 + • ψ ψ I2 ψ φ 2 ψ X ψ

S ⟩⟨ S ⊗ S ⟩ ⊗ S ⟩X = S ⟩ ⊗ S ⟩X Measurement X X X ● 1 ● X 2 Preparation ● ● A A B = A A B (5.3.10)

− − + 1 − • ψ ψ I2 ψ φ 2 ψ XZ ψ

S ⟩⟨ S ⊗ S ⟩ ⊗ S ZX⟩ = S ⟩ ⊗ S ⟩ZX Measurement XZ XZ XZ ● 1 ● XZ 2 Preparation ● ● A A B = A A B (5.3.11)

43 Step 3: Classical communication between Alice and Bob. Alice informs her results i, j (two-bit of information as shown in Table 5.3) to Bob. ( ) Step 4: Unitary correction by Bob. Bob performs unitary corrections on his particle to obtain ψ . The protocol between Alice’s measurement results and Bob’s unitary corrections shows in Table 5.4. S ⟩ Table 5.4: Quantum Teleportation Protocol

Bell measurement Classical communication Unitary correction Alice two-bit Bob + φ 0, 0 I2 φ− 0, 1 Z S ⟩ ( ) ψ+ 1, 0 X S ⟩ ( ) ψ− 1, 1 ZX S ⟩ ( ) S ⟩ ( ) Remarks:

• The state ψ , which Alice means to transmit to Bob, is unknown to Alice.

• No-cloningS theorem⟩ 5.1.0.2 is consistent with teleportation process, since once Bob gets the state ψ B, Alice’s state ψ A has been destroyed by measurement. Quantum copy machine: S ⟩ U Sφ⟩ 0 φ φ ; (5.3.12) Quantum teleportation: (S ⟩ ⊗ S ⟩) = S ⟩ ⊗ S ⟩ T φ 0 X φ . (5.3.13)

• Quantum teleportation has the(S interpretation⟩ ⊗ S ⟩) = S of⟩ ⊗ space-timeS ⟩ topology.

5.4 The quantum teleportation using continuous variables

Problem description2:

One complete orthonormal basis for the Hilbert space of two particles on the real line is the (separable) position eigenstate basis q1 q2 . Another is the entangled basis Q, P , where {S ⟩ ⊗ S ⟩} 1 {S ⟩} Q, P dq eiP ⋅q q q Q ; (5.4.1) 2π these are the simultaneousS eigenstates⟩ = √ ofS the relativeS ⟩ position⊗ S + ⟩ operator Q q2 q1 and the total momentum operator P p1 p2. ≡ − (a) Verify that ≡ + Q′,P ′ Q, P δ Q′ Q δ P ′ P ; (5.4.2)

2 Originated from the exercise 4.3,⟨ revisedS Chapter⟩ = 4( of John− ) Preskill’s( − online) lecture notes.

44 (b) Since the states Q, p are a basis, we can expand a position eigenstate as

S q⟩1 q2 dQdP Q, P Q, P q1 q2 . (5.4.3)

Evaluate the coeﬃcientsS ⟩ ⊗ SQ,⟩ P= Sq1 q2S . ⟩⟨ S(S ⟩ ⊗ S ⟩)

(c) Alice and Bob have prepared⟨ theS(S entangled⟩ ⊗ S ⟩) state of two particles A and B; Alice has kept particle A ad Bob has particle B; Now Alice has received an unknown single particle wavepacket ψ C dq q CC q ψ C that she intends to teleport to Bob. Design a protocol that they can execute to achieve the teleportation. What should Alice measure? WhatS ⟩ information= ∫ S ⟩ should⟨ S ⟩ she send to Bob? What should Bob do when he receives this information, so that particle B will be prepared in the state ψ B?

1○ BackgroundS ⟩ : Quantum teleportation is a quantum information protocol in which Alice and Bob are space-like separated, but Alice can send Bob a qubit based on the ) application of both quantum entanglement and quantum measurement. This protocol usually consists of the four steps:

i) State preparation ii) Bell measurement iii) Classical communication iv) Unitary correction

2○ Notation: Ω dq qq Q 0,P 0 ) (5.4.4) Deﬁne the U 1 phase operatorS ⟩ = ∫ asS ⟩ = S = = ⟩ =

−iP ⋅qˆ ′ −iP ⋅q′ ′ ′ ( ) U P e dq e q q , (5.4.5)

= = iP ⋅q S ⟩⟨ S U −P qS e q . (5.4.6)

The unitary formalism shows as S ⟩ = S ⟩ † † iP ⋅q U P U −P , q U P q U −P q e . (5.4.7) Deﬁne the translation operator= as ⟨ S = ⟨ S = ⟨ S

−ipˆ⋅Q ′ ′ ′ T Q e dq q Q q , (5.4.8)

= T Q q= S q SQ +. ⟩⟨ S (5.4.9)

The unitary formalism shows as S ⟩ = S + ⟩ † † T Q T −Q, q T Q q T −Q q Q , (5.4.10)

′= ′ ⟨′ S = ⟨ ′S = ⟨ ′ + S q dq q Q q dq δq,q′−Q q q Q . (5.4.11)

Note we have the relations⟨ S S S − ⟩⟨ S = S ⟨ S = ⟨ + S −iP ⋅(q+Q) U P T Q q U P q Q e q Q , (5.4.12)

S ⟩ = S +45 ⟩ = S + ⟩ −iP ⋅q −iP ⋅q T QU P q T Qe q e q Q , (5.4.13) therefore S ⟩ = −iPS ⋅Q⟩ = S + ⟩ U P T Q e T QU P . (5.4.14)

The entangled basis is deﬁned as =

iP ⋅q Q, P U −P T Q Ω dqe q, q Q (5.4.15)

with the cup conﬁgurationS ⟩ = ( ⊗ )S ⟩ = S S + ⟩ Q, P −P ● ● Q (5.4.16) And the cap conﬁguration expressesS as⟩ =

Q, P P ● ● −Q (5.4.17) ⟨ S = The normalization relation in diagrammatical language:

P ′ ● ● −Q′ Q′,P ′ Q, P δ P P ′ δ Q Q′ . − ● ● Q P (5.4.18) ⟨ S ⟩ = = ( − ) ( − ) Other diagrammatical rules:

−P ● ● −P Q ● ● −Q (5.4.19) = =

● Q −P ● ● Q ● −P (5.4.20) = The product state has the diagrammatical representation:

q1, q2 ∇ ∇ S ⟩ = Sq1⟩ Sq2⟩

The inner product of product state with the entangled state Q, P has the expression

S ⟩ P ● ● −Q √1 e−iP ⋅q1 δ Q q q . Q, P q1, q2 2π 2 1 ∇ ∇ ⟨ S ⟩ = Sq1⟩ Sq2⟩ = ( − ( − ))

Note that Q stands for the relative position of two particles, namely Q q2 q1.

○ 3 Continuous teleportation: = −

)

46 i) State preparation

−P ● ● Q ∇ Sψ⟩C SQ, P ⟩AB (5.4.21)

where ψ C is the unknown state, expressed as

S ⟩ ψ C dq q C q ψ C dqψ q q . (5.4.22)

ii) Bell measurement S ⟩ = S S ⟩ ⟨ S ⟩ = S ( )S ⟩

−P ′ ● ● Q′

P ′ ● ● −Q′

C A B (5.4.23) iii) Topological operation

−P ′ ● ● Q′ −P ′ ● ● Q′ ● Q ● = −P ′ P ′ ● ● −Q′ ● Q ● P ′ −P ● ● Q ∇ ∇

Sψ⟩C SQ, P ⟩AB Sψ⟩B iv) Unitary correction After Bell measurement, Bob will obtain the state

ψ B T QU −P T Q′U P ′ ψ B. (5.4.24)

Therefore, he is required toS perform⟩ = the unitaryS correction⟩ † −iP ⋅Q′ U U −P ′T −Q′U P T −Q e U P −P ′T −Q−Q′ , (5.4.25)

where the commutative= relation (5.4.14)= has been applied.

5.5 Quantum cryptography (information security)

5.5.1 Classical cryptography Alice and Bob want to communicate (transmit information) with each other. 1. To ensure information security, they choose to get an encryption key, e.g.

K 1 1 1 1 1 .

2. Now, Alice wants to send Bob some∶= ( information, e.g.)

A 0 1 0 0 0 .

For information security consideration,∶= ( she would) encrypt her information before sending out, i.e., A K 1 0 1 1 1 .

+ = ( ) 47 3. Bob then receives the message, namely

B A K.

To read the message, Bob will have to∶= decrypt+ it, i.e., B K A K K A.

Remark: The encryption key is the+ most= important+ + thing= for information security. In classical physics, the key can be copied without Alice and Bob’s notice, by a third party diﬀerent from Alice and Bob, who we can name “Eve”.

5.5.2 Quantum key distribution (QKD) In Quantum physics, if the encryption keys are encoded in non-orthogonal states, for example, x and z , it cannot be copied without disturbing Alice and Bob, which is ensured by the non-cloning theorem. The process to conduct a Quantum key distribution is shown stepS↑ ⟩ by stepS↑ ⟩ in the following: Step 1: State preparation. Alice and Bob share the Bell state.

+ 1 φ z z z z 2 S ⟩ = √1 S↑ ⟩S↑ ⟩ + S↓ ⟩S↓ ⟩ x x x x 2 2 = √ S↑ ⟩ + S↓ ⟩ ⊗ S↑ ⟩ + S↓ ⟩ x x x x

1 + S↑ ⟩ − S↓ ⟩ ⊗ S↑ ⟩ − S↓ ⟩ x x x x , 2 i.e., = √ S↑ ⟩ ⊗ S↑ ⟩ + S↓ ⟩ ⊗ S↓ ⟩ 1 1 φ+ 00 11 . (5.5.1) 2 2 Step 2: Random local measurementsS ⟩ = √ S ⟩ by+ S Alice.⟩ = √ S++⟩ + S−−⟩ Alice makes purely random (Prob 1 2) choices from σz, σx to conduct measurements on her qubit with the chosen observables (see Table 5.5). = ~ { } Table 5.5: Alice’s purely random local measurements

Alice 1 2 3 4 5 6 Z Z X X Z X random prob 1 2 A A A A A A σz σz σx σx σz σx = ~ eigenstate z z x x z x eigenvalue 1 1 1 1 1 1 S↑ ⟩ S↑ ⟩ S↑ ⟩ S↓ ⟩ S↑ ⟩ S↓ ⟩ + + + − + − Step 3: Random local measurements by Bob. At the same time, Bob does the same thing on his particle as Alice does to hers system (see Table 5.6).

48 Table 5.6: Bob’s purely random local measurements

Alice 1 2 3 4 5 6 X Z Z X X X prob 1 2 B B B B B B σx σz σz σx σx σx = ~ eigenstate uncertain z uncertain x uncertain x eigenvalue 1 1 1 1 1 1 S↑ ⟩ S↓ ⟩ S↓ ⟩ + + + − + − Step 4: Inform each other the experiments. Alice and Bob inform each other which observable they have chosen in every experiment, but do not mention the measurement results. In this circumstance, the third party, i.e., Eve, can get the observables without notice, but cannot get the measurement results.

Step 5: Get the Key. Alice and Bob choose the same results, which are shown in Table 5.7.

Table 5.7: The same observables that Alice and Bob share

2 4 6

measurement σz σx σx eigenvalue 1 1 1

And then encode them as + − −

x , z 0

+1 ⎪⎧ S↑ ⟩ S↑ ⟩ Ð→ (5.5.2) ⎪ x , z 1. ⎪ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ ⎨ −1 ⎪ S↓ ⟩ S↓ ⟩ Ð→ ⎪ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ Finally, we can get the encryption⎩⎪ key, i.e., K 0 1 1 (5.5.3)

Step 6: For practical consideration, we may= ( have to) develop from this row key to a practical key.

Remark: If Eve wants to know the key K, then he (she) must detect the states, but disturb Alice’s and Bob’s systems at the same time.

5.5.3 BB84 quantum key distribution The BB84 Quantum key distribution was ﬁrstly presented in 1984.

Principle 5.5.3.1. Non-orthogonal state (e.g. x , z ), cannot be distinguished without any changes, so Quantum infromation can be securely encoded in non-orthogonal states. S↑ ⟩ S↑ ⟩ To obtain the BB84 Quantum key distribution, we have the following steps:

49 Step1: Alice randomly prepares one of the four states

z , z , x , x ,

and labels S↑ ⟩ S↓ ⟩ S↑ ⟩ S↓ ⟩ z , z

with its observable Z, while S↑ ⟩ S↓ ⟩ x , x

with X. S↑ ⟩ S↓ ⟩ Step2: Alice sends her state to Bob. Bob then makes a random measurement on observable X or Z.

Step3: Alice and Bob inform each other their observables, not including measurement outcomes.

Step4: Alice and Bob keep states in which the same observable are exploited, and keep the remaining outcomes as the raw key.

, 0 x z . x , z 1 S↑ ⟩ S↑ ⟩ Ð→ Step5: Develop the row key to a practicalS↓ ⟩ key.S↓ ⟩ Ð→

50 Chapter 6

Bell Inequalities

what is proved by impossibility proofs is lack of imagination. —John Bell ⋯ The true logic of this world is in the calculus of probabilities. —James Clerk Maxwell

Reference:

[Preskill] Chapter 4: Quantum entanglement;

● Lorenzo Maccone, “A simple proof of Bell’s inequality”, arXiv: 1212.5214v2.

● 6.1 Einstein’s quantum mechanics: local hidden variable theory (LHV)

6.1.1 What hidden variable (HV) theory? In Einstein’s opinion, quantum theory is not complete and the reason is that a complete theory should be deterministic. He explained further that quantum randomness is a result of our ignorance of local hidden variables, and the local hidden variable theory is complete. In other words, quoting the famous statement by Einstein, “God does not play dice.” We may show it in the table 6.1. But, Bohr didn’t agree with him, and argued that the

Table 6.1: QM and LHV

Quantum theory Hidden variable theory

ψ α 0 β 1 ψ , λ , λ: hidden variable

S ⟩ = S ⟩ + S ⟩ (S ⟩ ) quantum theory is complete, and the measurement output has to be probabilistic, which is the intrinsic character of quantum mechanics. They were the two greatest minds in the 20th century, but held opposite opinions concerning quantum mechanics. Whose opinion is right? It should be answered by the experiment, and there is no other way. Up to now, experiments always agree with Bohr’s viewpoint.

51 6.1.2 What local theory? As we know that one of the two axioms of Special Relativity is that there is no faster-than- light communication, which ensures the causality. We call this the “relativistic locality”. From that, we can infer that two events at space-like separated regions can not have any causal connection. Einstein thought that if two subsystems A and B are space-like separated, measurements on subsystem A cannot modify subsystem B, neither measurements on subsystem B can modify subsystem A. We call this principle as Einstein’s locality. Einstein’s locality can be violated in quantum mechanics, under a given circumstance. For example, when the two subsystems A and B share the Bell state φ+ (4.3.2), Alice measures her subsystem A along the z-axis, and Bob measures his subsystem B soon after Alice’s measurements, and Bob’s results will be the same as Alice’s results.S ⟩ Afterwards, Alice measures her subsystem A along the x-axis, and Bob measures his subsystem B soon after Alice’s measurements, and Bob’s results will be still the same as Alice’s results. Hence, Bob’s measurement results can be modiﬁed by Alice’s measurements. With the GHJW theorem 14.4.2.1, we know that if two subsystems A and B share an entangled state in the composite system HA HB, local measurements on subsystem (e.g., B) can lead to diﬀerent state ensembles for another subsystem (e.g., A), even if the two subsystems are space-like separated, but,⊗ which does not cause the faster-than-light information communication. Note that “information is physical”. If there are no classical communication between Alice and Bob, then Alice’s measurements do not modify the ensemble description for subsystem B. Therefore, the relativistic causality survives quantum mechanics but Einstein’s locality does not.

6.1.3 The rule to justify the rightful theory The local hidden variable theory (LHV) contradicts with quantum mechanics, but which of these two theories is right? Of course, it should be eventually answered by experiments. How can an experiment tell us which one is right, or what kind of experiments can distinguish these two theories? Bell’s inequality is derived for this purpose, and we can directly check the Bell’s inequality in our experiment, from the result we will know which theory is telling the truth.

6.2 Bell’s inequality in the local hidden variable theory

Let’s consider the following experiment:

(1) Alice has three coins, each with head or tail face. We can label the three coins, which Alice holds, with 1A, 2A and 3A respectively. For each coin we assign a speciﬁc variable to describe the state of the coin, i.e., x for coin 1A, y for coin 2A and z for coin 3A, and

x, y, z H, T , with “H” for Head, and “T” for Tail.

(2) The local hidden theory∈ allows us to assign the probability distribution of the three coins faces, denoted as Prob x, y, z , with x, y, z H, T . And we shall know that the total probability should be ( ) ∈ Prob x, y, z 1. x,y,z∈ H,T Q ( ) = 52 The probability that the i-th coin and the j-th coin have the same value can be denoted as Psame i, j , e.g.

Psame 1, 2 Prob( H), H, H Prob H, H, T Prob T, T, H Prob T, T, T . (6.2.1)

Now,we( can) ∶= deﬁne ( )+ ( )+ ( )+ ( )

BI Psame 1, 2 Psame 1, 3 Psame 2, 3 , (6.2.2)

and evaluate it in the following∶= ( way) + ( ) + ( )

BI Prob H, H, H Prob H, H, T Prob T, T, H Prob T, T, T Prob H, H, H Prob H, T, H Prob T, H, T Prob T, T, T = ( ) + ( ) + ( ) + ( ) Prob H, H, H Prob T, H, H Prob H, T, T Prob T, T, T + ( ) + ( ) + ( ) + ( ) 1 2 Prob H, H, H Prob T, T, T (6.2.3) + ( ) + ( ) + ( ) + ( ) i.e., = + ( ) + ( ) BI Psame 1, 2 Psame 1, 3 Psame 2, 3 1, (6.2.4)

which is the so-called= Bell’s inequality.( ) + ( ) + ( ) ≥ In the logical of probabilities distribution, we have the following Venn diagram to describe Prob x, y, z , shown in Figure 6.1. For another method of calculation BI, with the help of the diagram, we obtain ( ) BI Psame 1, 2 Psame 1, 3 Psame 2, 3 A C B C C D ∶= ( ) + ( ) + ( ) 1 2C = ( + ) + ( + ) + ( + ) 1 2 Prob H, H, H Prob T, T, T = + 1, (6.2.5) = + ( ) + ( ) where, through derivation,≥ we have applied the relation A B C D 1, due to every time at leats two coins share the same face value. Note: The Bell’s inequality actually says one simple thing,+ + i.e.,+ for= three coins placed on the table, there would be at least two coins shown the same face up or down, in any circumstances.

(3) Mutually Exclusive Experiments (MEE) or Complementary experiment: Each time Alice is allowed to measure just one coin, namely, the other two coins are forbidden to measure. Then x, y and z are called complementary variables, which means you cannot have two values of them simultaneously. Remark: LHV agrees with MEE.

(4) Perfect correlation.

• Alice and Bob are space-like separated.

• Bob also has three coins, which is numbered with 1B, 2B, and 3B. • Alice and Bob obtain the same results when measuring their coins with the same label, i.e., when Alice measures the coin iA and Bob measures the coin iB, with i 1, 2, 3, then their results are always coincide.

= 53 A 1 2 C B D

Figure 6.1: Venn diagram of probabilities distribution Prob x, y, z . The circles labeled by 1, 2 and 3 represent the corresponding probabilities distribution of coin 1, 2 and 3. The overlap regions labeled by A, B, C and D stand for the( probability) of the coin i and j have the same face value, e.g. region A tells us that coin 1 and 2 share the same face value, whereas coin 1 and 3, coin 2 and 3, have the diﬀerent values.

• This correlation can be proved by classical communications between Alice and Bob.

Remark: LHV agrees with the perfection correlation because Alice and Bob are space-like separated.

(5) With the perfect correlation, for a set of complementary variables, the Bell’s inequalities can be checked, and so the local hidden variable theory.

6.3 Bell’s inequality in quantum mechanics

(1) Experiment setup: Alice and Bob share the spin singlet state ψ− , 1 ψ− 01 10 . S ⟩ (6.3.1) 2 Here, we replaced a coin withS a⟩ spin= √ polarization S ⟩ − S ⟩ direction. For example, we can choose the three orientations ai, i 1, 2, 3, for Alice’s system, and bi, i 1, 2, 3, for Bob’s system, with ⃗ = ⃗ = ai bi 1, and ai aj, bi bj, if i j.

A spin pointing alongY⃗ theY = positiveY⃗ Y = (negative)⃗ ≠⃗ direction⃗ ≠⃗ of an≠ orientation corresponds to a coin with “Head” (“Tail”) face.

(2) Mutually Exclusive Experiment (MEE). Because σ ai,σ aj 0, if i j,

⎧ σ bi,σ bj 0, if i j, ⎪ [⃗⋅⃗ ⃗⋅⃗ ] ≠ ≠ ⎨ which means that σ ai and⎪σ aj ⃗have⃗ no simultaneous eigenstates, neither σ ai and ⎩⎪ [⃗⋅ ⃗⋅ ] ≠ ≠ σ bi. As an result, there comes some diﬀerences between the quantum mechanics and the local hidden⃗⋅⃗ variable⃗⋅⃗ theory: ⃗ ⋅ ⃗ ⃗⋅⃗

54 • QM: Complementary variables cannot be assigned values simultaneously, because of uncertainty relation. • LHV: Complementary variables can be assigned values simultaneously, and then the classical logic can be applied, because LHV supports pre-existing variables before measurement.

(3) Perfect correlation.

• QM: Perfect correlation exists and agrees with the Special Relativity causality. • LHV: No information transformation in space-like regions, namely no classical communication in space-like regions.

(4) Calculation on successive measurements

1 E A a a a I2 1 a σA , 2 with a b 1, and a, b 3. 1 R ⎧ E B b I2 b b 1 b σB , ⎪ (⃗) ∶= S⃗⟩⟨⃗S ⊗ ∶= 2 + ⃗⋅⃗ ⎨ Y⃗Y = Y⃗Y = ⃗ ⃗(6.3.2)∈ ⎪ (⃗) ∶= ⊗ S⃗⟩⟨⃗S ∶= + ⃗⋅⃗ The⎩⎪ density matrix of the composite system is expressed as

− − ρAB ψ AB ψ . (6.3.3)

The probability, that Alice’s spin is∶= pointingS ⟩ ⟨ alongS the direction a and Bob’s spin along the orientation of b when we do the measurements, is calculated by ⃗ Prob⃗a, b ψ− E a E b ψ− AB (⃗ ⃗) A B AB 1 = ⟨ ψS− 1 (⃗a)σ (⃗)1 S b σ⟩ ψ− 4 AB A B AB 1 = ⟨ψ−S 1+ ⃗a⋅⃗σ b+σ⃗⋅⃗ a Sσ ⟩ b σ ψ− . (6.3.4) 4 AB A B A B AB part I = ⟨ S + ⃗⋅⃗ + ⃗⋅⃗ + ⃗⋅⃗part⃗ II⋅⃗ S ⟩ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ And we can calculate the two parts of expression´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ separately. Firstly, for part I,

− − − − part I 1 AB ψ a σA ψ AB AB ψ b σB ψ AB ,

where = + ⟨ S ⃗⋅⃗ S ⟩ + ⟨ S⃗⋅⃗ S ⟩

− − AB ψ a σA ψ AB 1 01 10 a σ 01 10 2 AB⟨ S ⃗⋅⃗ ABS ⟩ A AB AB 1 = 0⟨ a Sσ− 0 ⟨ S ⃗⋅1⃗ aσS 1⟩ − S ⟩ 2 A A A A A A 1 = tr ⟨ aS ⃗σ⋅⃗ S ⟩ + ⟨ S ⃗⋅⃗ S ⟩ 2 A A = 0, ⃗⋅⃗ (6.3.5) =

55 and − − AB ψ b σB ψ AB 1 01⃗ 10 b σ 01 10 2 AB⟨ S ⋅⃗ ABS ⟩ B AB AB 1 = 1⟨ b σS − 1 ⟨ S ⃗0⋅⃗b σ S 0⟩ − S ⟩ 2 B B B B B B 1 = tr ⟨ bS⃗σ⋅⃗ S ⟩ + ⟨ S⃗⋅⃗ S ⟩ 2 B B = 0, ⃗⋅⃗ (6.3.6) because Pauli matrices= are traceless. Therefore, part I 1. (6.3.7)

Next, we come to part II, = − − part II AB ψ a σA b σB ψ AB 1 = ⟨ 01S ⃗⋅⃗ ⃗10⋅⃗ a S σ ⟩ b σ 01 10 2 AB AB A B AB AB 1 = 0⟨ a σS − 0 ⟨ S 1⃗b⋅σ⃗ 1⃗⋅⃗ S0 a⟩σ −1S ⟩ 1 b σ 0 2 A A A B B B A A A B B B ⃗ ⃗ = A⟨ 1S ⃗a⋅⃗σAS 0⟩ A B⟨ 0S b⋅⃗σBS 1⟩ B− A⟨ 1S ⃗a⋅⃗σAS 1⟩ A B⟨ 0S b⋅⃗σBS 0⟩ B 1 − a⟨ Sb⃗⋅⃗ aS ⟩ ia⟨ S⃗b⋅⃗ ibS ⟩ + a⟨ S ⃗ia⋅⃗ S b⟩ ib⟨ S⃗⋅⃗a bS ⟩ 2 3 3 1 2 1 2 1 2 1 2 3 3 1 = −a b −2(a b − 2a)(b +a b) − ( + )( − ) − 2 3 3 1 1 2 2 3 3 = a(−b. − − − ) (6.3.8)

In another method= −⃗ on⋅⃗ the calculation of part II, from the physical point that Bell − state ψ AB is a singlet state, namely spin-less state, we get

− S ⟩ σA σB ψ AB 0, (6.3.9) therefore the part II can be rewritten⃗ + ⃗ as S ⟩ = − − part II AB ψ a σA b σB ψ AB − ⃗ − = AB⟨ ψS ⃗⋅a⃗σA ⋅b⃗σA S ψ⟩ AB − A A − = −AB ⟨ψ S a⃗iσ⋅⃗i bjσ⃗j⋅⃗ψ SAB ⟩ ψ− a b δ ψ− ψ− iε a b σA ψ− = −AB ⟨ S i j ij S AB⟩ AB ijk i j k AB a b, (6.3.10) = − ⟨ S S ⟩ − ⟨ S S ⟩ where the repeated= indexes−⃗⋅⃗ imply sum and the relation of the Pauli matrices is applied, σiσj δijI2 iεijkσk. (6.3.11) Hence, integrating the calculation of part I and part II, we get = + 1 1 Prob a, b a b 4 4 1 (⃗ ⃗) = −1 ⃗cos⋅⃗ a, b . (6.3.12) 4 = − (⃗ ⃗) 56 Following, we obtain the probability that Alice’s spin points along ai and Bob’s spin points along bj or Alice’s spin points along ai and Bob’s spin bj, namely measurement results are same, ⃗ ⃗ −⃗ −⃗ ij Psame Prob a, b Prob a, b 1 1 cos⃗ a , b . ⃗ (6.3.13) ∶= 2 (⃗ ) +i j (−⃗ − ) = − (⃗ ⃗ ) There, we can see that with bj ai, E.Q.(6.3.13) should be ⃗ ij = −⃗ Psame 1,

i.e., physical systems Alice and Bob are perfectly= anticorrelated. Now, we can calculate the quantity BI in Quantum Mechanics view point,

12 13 23 BI Psame Psame Psame, with bi ai, i 1, 2, 3. (6.3.14)

For the explanation= of E.Q.(6.3.14),+ + we may have⃗ some= −⃗ discussions.=

• E.Q.(6.3.14) is derived from the Mutually Exclusive Experiments, due to the fact, in Quantum Mechanics, that we cannot get any two components of the angular momentum in different directions, simultaneously. • With two subsystems, we still cannot find out the values of two complementary variables out of three in Mutually Exclusive Experiments, but with the help from subsystem Bob, we can obtain the inequality (6.2.4) and relation (6.3.14) from the view point of LHV and QM respectively. But, of course, when we consider all the matters in the frame of LHV or Quantum Mechanics, the results are different. • The correlation between Alice and Bob is specified by the maximal entangled bipartite state, namely ψ− . The reason why we have to set the constraint bi ai, i 1, 2, 3 is to ensure that, when Alice and Bob measure the same axis, or same coin, theirS results⟩ are perfect correlated. ⃗ = −⃗ = Now, we can calculate out some special cases by specifically setting the orientations of ai, i 1, 2, 3.

Case 1:⃗ ai =with diﬀerent i 1, 2, 3 are mutually orthogonal, which is shown in Fig- ure 6.2a. Therefore, we can see ⃗ ∈ { } cos a1, b2 cos a1, b3 cos a2, b3 0. (6.3.15)

Following, for BI, (⃗ ⃗ ) = (⃗ ⃗ ) = (⃗ ⃗ ) = 1 1 1 BI 2 2 2 3 = ,+ + 2 i.e., = BI 1. (6.3.16)

In this case, the Bell inequality (6.2.4)≥ is correct, and it means QM and LHV are consistent with each other.

57 Case 2: ai, i 1, 2, 3 are placed in a counterclockwise way on a plane, as is shown in Figure 6.2b. {⃗ = } 1 cos a , b cos a , b cos a , b , (6.3.17) 1 2 1 3 2 3 2 ⃗ ⃗ ⃗ (⃗ ) = (⃗ 1 ) 1= 1(⃗ ) = BI 4 4 4 3 = ,+ + 4 namely = BI 1, (6.3.18) which violates the inequality (6.2.4), which implies QM and LHV disagrees with ≤ each other at this time. The Bell’s inequality is violated in QM.

a3 a2 b3 ⃗ 120○ b1 b2 ⃗ ⃗ b1 a1 ⃗ a 120○ 120○ ⃗ a 2 1 ⃗ ⃗ b3 ⃗ a3 b2 ⃗ (a) Mutually⃗ orthogonal setting (b)⃗ Planar setting⃗

Figure 6.2: Orientation settings for ai and bi, i 1, 2, 3, where ai bi

⃗ ⃗ = ⃗ = −⃗ 6.4 The CHSH inequality

The CHSH inequality, instead of the Bell’s inequality, is often tested in experiment. In experiments, it often uses light polarization instead of spin polarization to stand for qubit. (1) Experimental setup, see Table 6.2.

Table 6.2: Experiment setup

Alice ψ− Bob observables observables A, BBCS ⟩ C, D A 1, B 1 C 1, D 1

= ± = ± = ± = ± (2) Local Hidden Variable theory

• Observables AA,BBB,CCC,DD can be assigned values simultaneously, because they are pre-determined or pre-existing. A B 0 B A A B 2A A B 2. A B 0 B A A B 2A A B 2. + = = − − = − = ± ¡ ⇒ ¡ ⇒ ¡ ⇒ Deﬁne− the new= observable,=M + = + = ± M A B C A B D 2. (6.4.1)

∶= ( + ) + ( − ) = ± 58 • Since HV is unknown, we only have probability distribution.

M 2 M M 2, (6.4.2)

i.e., = ± ⇒ S⟨ ⟩S≤ ⟨S S⟩ = M AC BC AD BD 2, (6.4.3)

which is called CHSHS⟨ ⟩S inequality.= T ⟨ ⟩ + ⟨ ⟩ + ⟨ ⟩ − ⟨ ⟩ T ≤ (3) Quantum Mechanics We can set these four observable AA,BBB,CCC,DD as

A a σA, B a′ σ , ⎧ A (6.4.4) ⎪ C = b⃗⋅σ⃗ , ⎪ B ⎪ = ⃗′ ⋅⃗ ⎨ D b σB. = ⃗⋅⃗ ⎪ ′ ⎪ ⃗ ′ The relative positions of a to a and⎩⎪ that= of b⋅⃗to b are shown in Figure 6.3. ⃗ ⃗ a ′ ⃗ ⃗ b a b′ ⃗ θ ⃗ θ′ ⃗ ⃗

(a) Alice (b) Bob

Figure 6.3: Choice of the operators

With the help of E.Q. (6.3.8), we can evaluate the left-hand-side of the CHSH inequality, M cos a, b cos a′, b cos a, b′ cos a′, b′ . (6.4.5)

Now consideringS⟨ the⟩S = specialS − ( case⃗ ⃗) depicted− (⃗ in⃗) Figure− (⃗ 6.4,⃗ ) + We can(⃗ derive⃗ )S A D π 4 π 4 C π 4 ~ ~ B ~ Figure 6.4: Special setting for AA,BBB,CCC,DD

2 AC AD BC BD . (6.4.6) √2 Therefore, ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = − ⟨ ⟩ = − 2 M 4 2 2 2. (6.4.7) √2 √ It violates the CHSH inequaltiyS⟨ (6.4.3).⟩S = × = ≥ Remarks:

• The CHSH inequality is violated for all entangled pure states, see Chapter 4.3.4, John Preskill’s lecture notes.

59 1 • What is involved in our calculation is spin- 2 system, but usually it is the photon- polarization represented for qubit utilized in experiments.

• There are disagreements on experiment tests of the CHSH inequality, see Chapter 4.3.6, John Preskill’s lecture notes.

6.5 Hints for the violation of the Bell inequality

The violation of the Bell inequality denies the local hidden variable theory (LHV) for Quantum Mechanics, but does favor the following two theories:

Copenhagen’s quantum mechanics: local non-hidden variable theory. (Standard quantum mechanics in textbook) ● De Broglie-Bohm’s quantum mechanics: non-local hidden variable theory. Parti- cle’s trajectories are inﬂuenced with each other non-locally. ●

True randomness

Quantum non-locality

Classical locality

No LHV theory

v Randomness is intrinsic c QM is complete ≪ Perfect correlation

6.6 Hardy’s theorem

Problem description1:

Bob (in Boston) and Claire (in Chicago) share many identically prepared copies of the two-qubit state ψ 1 2x 00 x 01 x 10 . (6.6.1) » √ √ where x is a real numberS between⟩ = ( − 0 and)S 1⟩2.+ TheyS ⟩ conduct+ S ⟩ many trials in which each measures his/her qubit in the basis 0 , 1 , and they learn that if Bob’s outcomes is 1 then Claire’s is always 0, and if Claire’s outcome~ is 1 then Bob’s is always 0. Bob and Claire conduct further{S experiments⟩ S ⟩} in which Bob measures in the basis 0 , 1 and they learn that if Bob’s outcome is 1 then Claire’s is always 0, and if Claire’s outcomes is 1 then Bob’s is always 0. {S ⟩BobS ⟩} and Claire conduct further experiments in which Bob measures in the basis 0 , 1 and Claire measures in the orthonormal basis ϕ , ϕ . They discover that if Bob’s outcome is 0, then Claire’s outcome is always ϕ and never ϕ. Similarly, if Claire {S ⟩ S ⟩} {S ⟩ S ⟩} 1Originated from the exercise 4.1, revised Chapter 4 of John Preskill’s online lecture notes.

60 measures in the basis 0 , 1 and Bob measures in the basis ϕ , ϕ , then if Claire’s outcome is 0, Bob’s outcome is always ϕ and never ϕ. Bob and Claire now{S ⟩ wonderS ⟩} what will happen if they both{S ⟩ measureS ⟩} in the basis ϕ , ϕ . Their friend Albert, a ﬁrm believer in local realism predicts that it is impossible for both to obtain the outcome ϕ(a prediction knows as Hardy’s theorem). Albert {Sargues⟩ S as⟩} follows:

When both Bob and Claire measures in the basis ϕ , ϕ , it is reasonable to consider what might have happened if one or the other had measured in the basis 0 , 1 instead. {S ⟩ S ⟩}

So{S ⟩ supposeS ⟩} that Bob and Claire both measure in the basis ϕ , ϕ , and that they both obtain the outcome ϕ, Now if Bob had measured in the basis 0 , 1 instead, we can be certain that his outcome would have been 1, since{S experiment⟩ S ⟩} has shown that if Bob had obtain 0 then Claire could not have obtained ϕ. Similarly,{S ⟩ S ⟩} if Claire had measured in the basis 0 , 1 , then she certainly would have obtained the outcome 1. We conclude that if Bob and Claire both measured in the basis 0 , 1 , both would have obtained the{S ⟩ outcomeS ⟩} 1. Bit this a contradiction, for experiment has shown that it is not possible for both Bob and Claire to obtain the outcome{S ⟩ S 1⟩} if they both measure in the basis 0 , 1 .

We are therefore forced to conclude{S ⟩ thatS ⟩} if Bob and Claire both measure in the basis ϕ , ϕ , it is impossible for both to obtain the outcome ϕ.

Though{S ⟩ S impressed⟩} by Albert’s reasoning, Bob and Claire decide to investigate what prediction can be inferred from quantum mechanics.

(a) Express the basis ϕ , ϕ in terms of the basis 0 , 1 .

(b) If Bob and Claire{S both⟩ S measure⟩} in the basis {Sϕ ,⟩ϕS ⟩}, what is the quantum- mechanical prediction for the probability P x that both obtain the outcome ϕ? {S ⟩ S ⟩} (c) Find the ”maximal violation” of Hardy’s theorem:( ) show that the maximal value of P x is P 3 5 2 5 5 11 2 .0902. √ √ (d) Bob( ) and Claire[( − conduct)~ ] = ( an experiment− )~ ≈ that conﬁrms the prediction of quantum mechanics. What was wrong with Albert’s reasoning?

1○ Motivation: Hardy’s theorem is able to make a suitable judgement on the conﬂict between the local hidden variable model and standard quantum mechanics, as well as ) the Bell inequalities do. 2○ In local hidden variable model, Bob and Charlie are space-like separated and perfect correlated. ) Bob Charlie

The Box represents for the hidden variable that correlates Bob and Charlie.

• The observables ub and uc satisfy the experimental fact:

ubuc 0, (6.6.2)

where ub, uc 0, 1. =

= 61 • The observables wb and wc satisfy the experimental facts:

ub 0 wc 0; (6.6.3)

uc 0 wb 0. (6.6.4) = Ô⇒ = Conclusion: = Ô⇒ = ubuc 0 ub 0 or uc 0 wb 0 or wc 0. (6.6.5) Therefore, = Ô⇒ = = Ô⇒ = = Prob wb 0 & wc 0 ubuc 0 LHV 0, (6.6.6) namely ( ≠ ≠ S = ) = Prob wb 1 & wc 1 ubuc 0 LHV 0. (6.6.7) In following, we will calculate it in quantum mechanics, and verify ( = = S = ) =

Prob wb 1 & wc 1 ubuc 0 QM 0. (6.6.8)

○ 3 In quantum mechanics, Bob( and= Charlie are= S correlated= ) by≠ entangled state ψ BC .

) Bob ψ BC Charlie S ⟩

S ⟩

ψ BC ϕ B 0 C x 0 B 1 C

0 B ϕ C √x 1 B 0 C S ⟩ = S ⟩ S ⟩ + S ⟩ S ⟩ 1 2x 00√BC x 01 BC x 10 BC (6.6.9) = S ⟩ S ⟩ + S ⟩ S ⟩ 1 √ √ √ where 0 x 2 . = − S ⟩ + S ⟩ + S ⟩

• The< observables< ub and uc (ub, uc 0, 1) are corresponding to the measurements U b and U c deﬁned as = ′ U b 0 B 0 ,UU b 1 B 1 , (6.6.10) U 0 0 ,UU ′ 1 1 . (6.6.11) c = S ⟩C ⟨ S c = S ⟩C ⟨ S Note that = S′ ⟩ ⟨ S = S ⟩′ ⟨ S U b U b 112,UU c U c 112. (6.6.12)

The value ub,c 0 or ub,c 1 is associated with the measurement result U b,c or ′ + = + = U b,c respectively. We have ubuc 0, because state ψ BC does not have the term 11 BC . = = = S ⟩ • The observables wb and wc (wb, wc 0, 1) are corresponding to the measurements S ⟩ W b and W c deﬁned as = ′ W b ϕ˜ B ϕ˜ ,WW b ϕ˜ B ϕ˜ , (6.6.13) ′ W c = Sϕ˜⟩C ⟨ϕ˜S,WW c = Sϕ˜ ⟩C ⟨ϕ˜ S, (6.6.14) where state ϕ˜ is the normalized state ϕ deﬁned in (6.6.9), and state ϕ˜ is = S ⟩ ⟨ S = S ⟩ ⟨ S orthogonal to the state ϕ˜ . Note that S ⟩ S ⟩ S ⟩ ′ ′ SW⟩b W b 112,WW c W c 112. (6.6.15)

The value wb,c 0 or wb,c 1+ is associated= with+ the= measurement result W b,c or ′ W b,c respectively. = = 62 Bob ψ BC Charlie

U b S ⟩ W c

0 B ϕ C

S ⟩ S ⟩

ub 0 wc 0

= = After Bob’s measurement U b, he obtains the state 0 B, and with certainty, Charlie ﬁnd his state in ϕ C . Namely, observable value ub 0 leads to wc 0. S ⟩ S ⟩ = = Bob ψ BC Charlie

W b S ⟩ U c

ϕ B 0 C

S ⟩ S ⟩

wb 0 uc 0

= = After Charlie’s measurement U c, he obtains the state 0 C , and with certainty, Bob ﬁnd his state in ϕ B. Namely, observable value uc 0 leads to wb 0. S ⟩ ○ 4 The task is to calculateS ⟩ the probability = =

2 ) Prob wb 1 & wc 1 ubuc 0 QM BC ϕ˜ ϕ˜ ψ BC . (6.6.16)

From ( = = S = ) = T ⟨ S ⟩ T 0 B 0 ψ AB 0 B ϕ C , (6.6.17)

we know the state ϕ can be expressedS ⟩ ⟨ S ⟩ as = S ⟩ S ⟩ S ⟩ ϕ 1 2x 0 x 1 , (6.6.18) √ √ and the normalized form S ⟩ = − S ⟩ + S ⟩ 1 ϕ˜ 1 2x 0 x 1 . (6.6.19) 1 x √ √ S ⟩ = √ ( − S ⟩ + S ⟩ Find the orthogonal state of ϕ˜ , denoted− as ϕ˜ ,

S ⟩ ϕ˜ A 0S ⟩B 1 , (6.6.20) ϕ˜ B 0 A 1 , S ⟩ = S ⟩ + S ⟩ where S ⟩ = S ⟩ − S ⟩ 1 2x x A ,B . (6.6.21) √ 1 x √1 x − Rewrite it into matrix formalism= √ = √ − − ϕ˜ AB 0 , (6.6.22) ϕ˜ B A 1 S ⟩ S ⟩ = S ⟩ 63 − S ⟩ where the transformation matrix is a real orthogonal matrix, namely

AB AB A2 B2 0 1 0 . (6.6.23) B A B A 0 B2 A2 0 1 + = = Therefore, − − + 0 AB ϕ˜ , (6.6.24) 1 B A ϕ˜ S ⟩ S ⟩ 0 = A ϕ˜ B ϕ˜ , S ⟩ − S ⟩ (6.6.25) 1 B ϕ˜ A ϕ˜ . S ⟩ = S ⟩ + S ⟩ Consider the term ϕ˜ A ϕ˜ B solelyS ⟩ in= stateS ⟩ −ψ ABS : ⟩ 2 S ⟩ S ⟩00 AB B ϕ˜S A⟩ ϕ˜ B; (6.6.26) 01 AB AB ϕ˜ A ϕ˜ B; (6.6.27) S ⟩ ∝ S ⟩ S ⟩ 10 AB AB ϕ˜ A ϕ˜ B. (6.6.28) S ⟩ ∝ − S ⟩ S ⟩ Hence S ⟩ ∝ − S ⟩ S ⟩ x ψ 1 2xB2 2 xAB ϕ˜ϕ˜ 1 2x ϕ˜ϕ˜ . (6.6.29) AB BC 1 x BC √ √ √ S ⟩ ∝ ( − − )S ⟩ = − − S ⟩ − 2 Prob wb 1 & wc 1 ubuc 0 QM BC ϕ˜ ϕ˜ ψ BC x2 ( = = S = ) = T 1 ⟨2x S ⟩ T 1 x 2 = p( x− . ) (6.6.30) ( − ) Find the maximal value of function p x : = ( ) 2x p′ x ( ) x2 3x 1 0. (6.6.31) 1 x 3 ( ) = ( − + ) = And ( − ) 3 5 x± , x0 0. (6.6.32) 2√ 1 ± Due to 0 x 2 , we have = =

< < 5 5 11 pmax x− 0.0902 0. (6.6.33) √ 2 − The result is conﬂicted with local( ) hidden= variable≈ theory.≠

5○ Reason for conﬂict between quantum mechanics and local hidden variable theory:

) U b,WW b 0, U c,WW c 0. (6.6.34)

Due to uncertainty relation, [ub, wb and] ≠ uc, w[ c can not] ≠ be determined simultaneously. Note: XX,ZZ 0 X XX,ZZ Z 0. (6.6.35)

[ ] ≠ ⇏ [ ⊗ ⊗ ] =

64 Table 6.3: A comparison between Bell’s inequalities, Hardy’s theorem and GHZ theorem

Parties Formalism Correlation Statement Bell inequalities 2 Inequality Statistical Hardy’s theorem 2 Equality Statistical LHV hypothesis is denied GHZ theorem 3 Equality Perfect

6.7 The GHZ theorem

1○ A comparison between Bell’s inequalities, Hardy’s theorem and GHZ theorem is shown in Table 6.3. ) 2○ GHZ theorem (the 3-qubit system)

) Step I: Alice, Bob and Charlie are space-like from each other and share the GHZ state, i.e., 1 GHZ 000 111 , (6.7.1) 3 2 as shown in the diagramS formalism⟩ = √ inS Figure⟩ + S 6.5.⟩

Charlie(III)

GHZ

Alice(I) S ⟩ Bob(II)

Figure 6.5: 3-qubit GHZ state.

Step II: Chose four observables with the form of

(1) (2) (3) A σx σy σy ; (6.7.2a) B σ(1)σ(2)σ(3); (6.7.2b) ⎪⎧ ∶= y x y ⎪ (1) (2) (3) ⎪ C σy σy σx ; (6.7.2c) ⎪ ∶= ⎨D σ(1)σ(2)σ(3). (6.7.2d) ⎪ ∶= x x x ⎪ ⎪ For these four observables,⎩⎪ we∶ may= notice the following things: 2 2 2 2 • A B C D I 8; • AA,BBB,CCC,DD are mutually commutative; = = = = • ABCD I 8; • D X 1X 2X 3, i.e. phase-bit operator, = − 2 AD σx σyσx σyσx I 2 Z Z, i.e. the second parity- bit= operator, 2 −CD = −(σyσ) x⊗( σyσ)x⊗( σx ) = Z ⊗Z ⊗I 2, i.e. the ﬁrst parity-bit operator; − = −( )⊗( )⊗( ) = ⊗ ⊗

65 •

A GHZ 3 GHZ 3 , (6.7.3a)

⎧B GHZ 3 GHZ 3 , (6.7.3b) ⎪ S ⟩ =− S ⟩ ⎪ C GHZ 3 GHZ 3 , (6.7.3c) ⎪ S ⟩ =− S ⟩ ⎨D GHZ 3 GHZ 3 , (6.7.3d) ⎪ S ⟩ =− S ⟩ ⎪ since ⎪ ⎩⎪ S ⟩ = S ⟩

A X 1 iZZ2X 2 iZZ3X 3 X 1 Z 2X 2 Z 3X 3 , ⎧B iZZ1X 1 X 2 iZZ3X 3 Z 1X 1 X 2 Z 3X 3 , ⎪ = ⊗(− )⊗(− )=− ⊗( )⊗( ) ⎪ C iZZ1X 1 iZZ2X 2 X 3 Z 1X 1 Z 2X 2 X 3, ⎪ =(− )⊗ ⊗(− )=−( )⊗ ⊗( ) ⎨D X 1 X 2 X 3. ⎪ =(− )⊗(− )⊗ =−( )⊗( )⊗ ⎪ Step III: For LHV hypothesis,⎪ A, B, C, D can be exactly determined, i.e. ⎩⎪ = ⊗ ⊗ (1) (2) (3) A mx my my ; (6.7.4a) (1) (2) (3) ⎧ B my mx my ; (6.7.4b) ⎪ = ⎪ C m(1)m(2)m(3); (6.7.4c) ⎪ y y x ⎪ = (1) (2) (3) ⎨D mx mx mx ; (6.7.4d) ⎪ = ⎪ with ⎪ = (i)⎩ (j) mx , my 1, i, j 1, 2, 3 . (6.7.5)

Thus, = ± = 2 (1) (1) (2) (2) (2) (2) ABCD mx my mx my my mx 1. (6.7.6) = Step IV: While for QM, we can get= from E.Q.(6.7.3a) (6.7.3d) that

ABCD GHZ GHZ∼ , (6.7.7)

i.e. S ⟩ = − S ⟩ ABCD 1. (6.7.8)

Step V: Experiment support E.Q.(6.7.8), which= − means LHV is denied. (i) (i) (i) (i) Remark: In QM, σx ,σσy 0, with i 1, 2, 3, thus mx and my cannot be determined simultaneously. [ ] ≠ =

66 Part II

Quantum Computing and Quantum Algorithm

67 Chapter 7

Classical Circuit Model

Computers are physical objects, and computations are physical processes. What computers can or can not compute is determined by the laws of physics alone, and not by pure mathematics. —David Deutsch

By raising these issues we wish to introduce the question of the completeness of the quantum circuit model, and reemphasize the fundamental point that information is physical. —Nielsen & Chuang

Whether physically reasonable models of computation exist, which beyond the quantum circuit model is a fascinating question which we leave open for you. —Nielsen & Chuang

A detailed examination and attempted justiﬁcation of the physics underlying the quantum circuit model is outside of the scope of the present discussions, and indeed outside the scope of the present knowledge. —Nielsen & Chuang

In our attempts to formulate the models of information processing, we should always attempt to go back to fundamental physical laws. —Nielsen & Chuang

Reference:

[Preskill] New Chapter 5: Classical circuit and quantum circuit;

● [Nielsen & Chuang] Chapter 3: Introduction to computer science;

● [Nielsen & Chuang] Chapter 4: Quantum circuits.

● 7.1 Classical circuit

Def 7.1.1. Classical circuit, a circuit model of classical computation, is a ﬁnite sequence of elementary gates applied to a ﬁnite string of input bits.

68 7.1.1 Elementary logical gates (1) NOT gate: NOT x x¯ 1 x, (7.1.1)

from which we shall see that ∶ ↦ = − NOT NOT Id, (7.1.2)

i.e. “NOT” gate is a reversible gate. The○ truth= table of the NOT gate is shown in Table 7.1. Table 7.1: Truth table of NOT gate.

input output 0 1 1 0

(2) AND gate: AND x, y x y mod 2 x y. (7.1.3)

The “AND” gate is irreversible.∶ ( Its) truth↦ ( table⋅ ) is Table= ∧ 7.2. Table 7.2: Truth table of AND gate.

input output 0 0 0 0 1 0 1 0 0 1 1 1

(3) OR gate: OR x, y x y xy x y. (7.1.4)

As we can see that “OR” gate∶ is( irreversible.) ↦ + The− corresponding= ∨ truth table is Table 7.3. Table 7.3: Truth table of OR gate.

input output 0 0 0 0 1 1 1 0 1 1 1 1

69 7.1.2 Universal gate set Def 7.1.2. An universal gate is a finite set of elementary gates which suffice to evaluate any function of a finite number of input bits.

Note: For a general function:

n n general function F 0, 1 0, 1 n F f1, f2, , fn , typical function fi 0, 1 0, 1 ⎫ ∶ ↦ ⎪ ⎬ ⇒ = ( ⋯ ) ⎪ i.e., the general function can be represented∶ ↦ by typical ⎭⎪ functions. Examples:

• Thm1: NOT, AND is an universal gate set;

• Thm2: NOT, OR is an universal gate set;

• Thm3: NAND, COPY is an universal gate set; Remark: NAND NOT AND, (7.1.5)

and ∶= ○ COPY x x, x . (7.1.6)

Since ∶ ↦ ( )

NOT x 1 x 1 x2 1 AND x NOT AND x, x , (7.1.7)

thus ( ) = − = − = − ( ) = ○ ( ) NOT x NAND x, x NAND COPY x . (7.1.8)

• Thm4: NOR, COPY is( an) = universal( gate) = set. ○ ( )

NOR NOT OR. (7.1.9)

∶= ○ 7.2 Reversible classical computation

7.2.1 Irreversible computation Classical computation in terms of irreversible gates, such as AND, OR, NAND, NOR gates, is irreversible.

Thm 7.2.1.1 (Landauer’s principle). Irreversible logic gates erase information with irreducible expenditure of power.

This means that if we want to erase information, we have to pay for it with work (energy), i.e., energy loss. On the other hand, the reversible gates would ensure that there is no expenditure of power, i.e., there is no energy loss. Quantum Computer is reversible, which is one of the reasons people prefer Quantum Computer.

70 Table 7.4: Irreversible and reversible computation

Computer Gates Erasure of information Irreversible Irreversible gates Irreducible expenditure Reversible Reversible gates No expenditure

7.2.2 Classical reversible gate One-bit reversible gates:

NOT x 1 x x,¯ NOT gate, Id x x, Identity gate. ( ) = − = Two-bit reversible gate: XOR,( ) = i.e., Exclusive OR gate. The quantum analogy of XOR is the CNOT gate. XOR x, y x, x y . (7.2.1)

With three XOR gate, we can form a∶ swap( gate,) ↦ ( which⊕ is) deﬁned as SWAP x, y y, x . (7.2.2)

And the construction should be ∶ ( ) ↦ ( )

SWAP XOR12 XOR21 XOR12. (7.2.3)

And, we can verify that, = ○ ○

XOR12 XOR21 XOR12 x, y

XOR12 XOR21 x, x y ○ ○ ( ) XOR12 x x y, x y = ○ ( ⊕ ) x x y, x y x x y = ( ⊕ ⊕ ⊕ ) y, x . = ( ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ) Similarly, the quantum analogy= should( ) be

SWAP CNOT12CNOT21CNOT12, (7.2.4) with SWAP deﬁned as = SWAP ψ φ φ ψ . (7.2.5)

And in quantum circuit model, it has∶ theS ⟩ diagrammatical⊗ S ⟩ ↦ S ⟩ ⊗ S expression⟩ ψ φ

Sφ⟩ × Sψ⟩ Thm 7.2.2.1. One-bit gates and two-bit gates cannot suﬃce universal classical reversible × computation. S ⟩ S ⟩

71 7.2.3 Three-bit Toffoli gate Def 7.2.1 (Three-bit Toffoli gate). Toffoli gate θ(3) x, y, z x, y, z xy . (7.2.6) which can also be shown in the diagram∶ formalism,( ) ↦ ( ⊕ ) x x y y z ● z xy. ● As we can see from the definition of the 3-bit Toffoli gate, it works as the controlled- ⊕ controlled NOT gate with two control-bit and one target bit, besides it is a nonlinear gate. Thm 7.2.3.1. The 3-bit Toffoli gate with constant bits is universal for classical reversible computation. Proof. The strategy is by setting constant bits for the Toffli gate, we can obtain an universal gate set made up of two-bit gates and one-bit gates. 1○ Toffli gate with z 1:

(3) ) = θ x, y, 1 x, y, 1 xy ( ) x, y, NAND x, y , (7.2.7) = ( − ) here we get the NAND gate; = ( ( )) 2○ Toﬄi gate with x 1, z 0:

(3) ) = = θ 1, y, 0 1, y, 0 y ( ) 1, y, y = ( + ) 1, COPY y , (7.2.8) = ( ) the COPY gate for this time; = ( ( )) 3○ Toﬄi gate with x 1:

(3) ) = θ 1, y, z 1, y, z y ( ) 1, XOR y, z , (7.2.9) = ( ⊕ ) and we get the XOR gate. = ( ( )) Actually, the NAND gate and COPY gate together can make an universal gate set, hence they can construct a universal reversible computer, i.e., the 3-bit Toﬀoli gate with constant bits is universal for classical reversible computation, since the 3-bit Toﬄi gate is reversible: 2 θ(3) x, y, z (3) θ x,( y, z xy) x, y, z xy xy = ( ⊕ ) x, y, z , = ( ⊕ ⊕ ) = ( 72) namely 2 θ(3) Id. (7.2.10)

= Remark: To construct given classical gate, one may need the exponential number of θ(3) gates, which is terrible in practice.

7.2.4 Three-bit Fredkin gate Def 7.2.2 (Three-bit Fredkin gate). The three bit Fredkin gate is deﬁned as

Fredkin gate x, y, z x, xz xy,¯ xy xz¯ (7.2.11)

This deﬁnition shows that the three-bit∶ ( Fredkin) ↦ ( gate is+ nonlinear,+ too.)

Thm 7.2.4.1. The three-bit Fredkin gate with constant bits is universal for classical reversible computation.

Proof. The strategy is the same as the proof that three-bit Toﬀoli gate is universal for the classical reversible computation.

1○ Fredkin gate with z 0:

) = Fredkin x, y, 0 x, xy,¯ xy ( ) x, xy,¯ AND x, y , (7.2.12) = ( ) here we get the AND gate; = ( ( ))

2○ Fredkin gate with y 0, z 1:

) = = Fredkin x, 0, 1 x, x, x¯ ( ) COPY x , NOT x , (7.2.13) = ( ) and we obtain here the COPY= gate( and NOT( ) gate;( ))

3○ Fredkin gate with x 1:

) = Fredkin 1, y, z 1, z, y ( ) 1, SWAP y, z , (7.2.14) = ( ) and we get the SWAP gate. = ( ( ))

Because the NAND gate can be constructed by using the NOT gate and the AND gate, and the NAND gate and COPY gate can make an universal gate set. Therefore, the AND

73 gate, the NOT gate and the COPY gate together make an universal gate set. On the other hand, the Fredkin gate is reversible too, since

Fredkin2 x, y, z Fredkin x, xz xy,¯ xy xz¯ ( ) x, x xy xz¯ x¯ xz xy¯ , x xz xy¯ x¯ xy xz¯ = ( + + ) x, x2y xxz¯ xxz¯ x¯2y, x2z xxy¯ xxy¯ x¯2z = ( + ) + ( + ) ( + ) + ( + ) x, xy 0 0 xy,¯ xz 0 0 xz¯ = + + + + + + x, x x¯ y, x x¯ z = ( + + + + + + ) x, y, z , = ( + ) ( + ) i.e., = ( ) Fredkin2 Id. (7.2.15) Hence, the three-bit Fredkin gate with constant bits is universal for classical reversible = computation.

7.3 The construction of an n-bit Toﬀoli gate using the 3-bit Toﬀoli gate

Def 7.3.1 (n-bit Toﬀoli gate).

(n) θ x1, x2, , xn−1, y x1, x2, , xn−1, y x1x2 xn−1 (7.3.1) which can also be shown( in the⋯ diagram) = formalism,( ⋯ ⊕ ⋯ )

x1 x1 x2 x2 ● ● xn−1 xn−1 ⋮ ⋮ xn y x1x2 xn−1 ● which has n 1 control bits and one target bit. ⊕ ⋯ (n) Thm 7.3.0.2.− With only one bit of scratch space, performing θ gate needs at least number of 2n−3 2n−2 2 of θ(3) gates.

(3) (n) Remark: With+ more− bit of scratch space, the number of θ needed to perform θ can be reduced in polynomial scale. Thm 7.3.0.3. With n 3 scratch bits returning the initial value 0 at the end of computation, the number of θ(3) gates needed to perform θ(n) gate is 2n 5. − Proof. The proof is done by induction. When n 4, we need 1 scratch− bit and 2n 5 4 3 number of θ(3) gates. = ( − )S = x1 x1 x2 θ(3) θ(3) x2 0 x1x2 x1x2 0 x3 θ(3) x3 ⊕ = y y x1x2x3

74 ⊕ To prevent possible notation confusion and to maintain the notation consistence in following, in above quantum circuit diagram, we apply the notation of θ 3 gate as a three-qubit gate. And the new constructed gate functional as the θ 4 gate, we will denote it as θ′ 4 . Note gate θ′ 4 is a 5-bit gate with scratch bit. ( ) ( ) ( ) Assume we can( ) construct θ n gate with n 3 scratch bits and 2n 5 number of θ(3) gates. The gate we have constructed to perform n-bit Toﬀoli gate is denoted as θ′(n). In the manner of recursive construction,( ) to construct− θ(n+1) gate, we( have− )

x1 x1 x2 θ(3) θ(3) x2 0 0 x3 x3 0 0 x4 x4 0 θ′(n) 0 x5 x5

xn xn ⋮ ⋮ y y x1x2x3 xn

Therefore, the number of scratch space needed to construct⊕θ(n+1) gate⋯ is n 3 1 n 1 3. And the number of θ(3) gates is 2n 5 2 2 n 1 5. ( − ) + = (Thm+ ) 7.3.0.4.− With n 3 scratch bits returning the− + initial= ( value+ ) at− the end of computation, the number of θ(3) gates needed to perform θ(n) gate is 4n 12. − (3) Proof. When n 4, we need 1 scratch bit and 4n 12 4 − 4 number of θ gates.

= x1 ( − )Sx1 = x2 x2 s1 θ′(4) s1 x3 θ(3) x3 y y x1x2x3

After the ﬁrst θ′(4) gate, we have ⊕

′ y y x1x2 s1 x3 y x1x2x3 s1x3. (7.3.2)

After the second θ(3) gate,= we⊕ obtain( ⊕ ) = ⊕ ⊕

′′ y y x1x2x3. (7.3.3)

In the case of construction θ(n) gate, we= can⊕ ﬁnd we need one θ′(n) gate and one θ′(n−1)

75 gate, shown as x1 x1 x2 x2 s1 s1 x3 x3 s2 s2 ′( ) x4 θ n x4 ′( − ) s3 θ n 1 s3 x5 x5

xn−1 xn−1 ⋮ ⋮ y y x1x2x3 xn ′(n) After the ﬁrst θ gate, we have ⊕ ⋯ ′ y y x1x2 s1 x3 s2 x4 s3 x5 xn−1

y x1x2x3 xn−1 s1x3 s2 x4 s3 x5 xn−1 = ⊕ (((( ⊕ ) ⊕ ) ⊕ ) ⊕ ⋯) y x1x2x3 xn−1 z. (7.3.4) = ⊕ ⋯ ⊕ ((( ⊕ ) ⊕ ) ⊕ ⋯) Note the redundant= term⊕ z can be⋯ rewritten⊕ as

′ ′ ′ ′ ′ ′ ′ z x1x2 s1 x3 s2 x4 xn−2. (7.3.5)

Therefore, one more θ′(n−1) =gate((( can wipe⊕ ) it out,⊕ ) ⊕ ⋯)

′′ ′ y y z y x1x2x3 xn−1. (7.3.6)

We used totally n 3 scratch bits= and⊕ =2n ⊕5 2 ⋯n 1 5 4n 12 number of θ(3) gates here. − ( − ) + ( ( − ) + ) = −

76 Chapter 8

Quantum Circuit Model

8.1 Deﬁnition of quantum circuit

Def 8.1.1 (Quantum Circuit). Quantum Circuit, the circuit model of quantum computation, is a sequence of a ﬁnite number of Quantum gates acting on a ﬁnite number of qubits.

As we shall see that this definition is similar to the definition of Classical Circuit, but the following results can be very different. A comparison of Classical Circuit and Quantum Circuit is shown in Table 8.1. Table 8.1: Classical Circuit and Quantum Circuit ` `` Note ``` Object Operation Reversibility Complexity Computation ```` ⎪⎧ n-bit logic gate: ⎪ irreversible (most cases) Classical Computation ⎨ P vs. NP n ⎪ 0, 1 AND, OR, NOT,etc. ⎩⎪ reversibe n-qubit quantum gate: Quantum Computation reversible BPP vs. BQP n H2n U(2 ) group

8.1.1 One-qubit gates The one qubit gates are U 2 or SU 2 operators. And we shall see that U 2 operator is equivalent to SU 2 operator up to a phase coeﬃcient. There are some examples of the one-qubit gate: ( ) ( ) ( ) ( ) • Hadamard gate H, 1 1 1 H ; (8.1.1) 2 1 1 = • Phase gate S, − 1 0 S ; (8.1.2) 0 i

π = • 8 gate T : 1 0 e−iπ~8 0 T eiπ~8 ; (8.1.3) 0 eiπ~4 0 eiπ~8 = =

77 • Pauli X gate: 0 1 X σx ; (8.1.4) 1 0 = = • Pauli Z gate: 1 0 Z σz ; (8.1.5) 0 1 = = • Phase-shift gate Rθ: − 1 0 R ; (8.1.6) θ 0 eiθ = Thm 8.1.1.1. Any U 2 group element can be expressed as

iα ( ) U e D z β D y γ D z δ , (8.1.7) where D z β ,D y γ , and D z δ are= rotation( ) gates( ) deﬁned( ) as

−iβ~2 ( ) ( ) ( ) −iσσzβ~2 e 0 D z β e , (8.1.8a) 0 eiβ~2 ⎪⎧ ⎪ ( ) ∶= = γ γ ⎪ −iσσyγ~2 cos 2 sin 2 ⎪ D y γ e γ γ , (8.1.8b) ⎪ sin 2 cos 2 ⎪ − ⎨ ( ) ∶= = −iδ~2 ⎪ −iσσzδ~2 e 0 ⎪ D z δ e . (8.1.8c) ⎪ 0 eiδ~2 ⎪ ⎪ ( ) ∶= = Proof. Now, we are going⎩⎪ to prove the theorem 8.1.1.1 step by step. Step 1. Any U 2 group element U should satisfy the unitary condition

† † ( ) UUUU U U I 2. (8.1.9)

Let’s assume that = = a′ b′ U , with a′, b′, c′, d′ . (8.1.10) c′ d′ C From E.Q. (8.1.9) we can∶= get ∈

detU 2 det UUUU † 1,

namely S S = = 2iα detU e , with 0 α 2π and α R. (8.1.11)

Step 2. With the E.Q. (8.1.11)= in hand, we≤ can< denote U as∈

U eiαU ′, (8.1.12)

with = ′ ′ ′† ′† ′ detU 1, and U U U U I 2, (8.1.13)

and = = = a b U ′ SU 2 , with a, b, c, d . (8.1.14) c d C ∶= ∈ ( ) ∈ 78 Thus a∗ a b aa∗ bb∗ 1, (8.1.15a) b∗ ⎧ ⎪ ∗= + = ⎪ c ∗ ∗ ⎪ c d ∗ cc dd 1, (8.1.15b) ⎪ d ⎪ ⎪ = + = ⎪ c∗ ⎪ a b ac∗ bd∗ 0, (8.1.15c) ⎨ d∗ ⎪ ⎪ = + = ⎪ a b ⎪ det ad bc 1. (8.1.15d) ⎪ c d ⎪ ⎪ Thus, we can conclude that ⎪ = − = ⎩⎪ • E.Q.(8.1.15a) and E.Q.(8.1.15b) both can set a constraint on U ′, respectively. As we shall see that the imaginary parts of the both two equations are absolutely 0. • E.Q.(8.1.15c) can spell two constraint on U ′. • It seems that E.Q.(8.1.15d) can also put two constraint on U ′. But, we can ﬁnd out that the constraint (8.1.15d) can be derived from the former three equations (8.1.15a) (8.1.15c).

Therefore, the E.Q.(8.1.15a)∼ (8.1.15d) would set totally ﬁve constraints on the parameters of U ′. Consequently, there are only 8 5 3 degrees of freedom left for the SU 2 group element U ′. ∼ − = Step( 3). From the constraint E.Q.(8.1.15c) we can derive ac∗ bd∗ 0 ac∗ bd∗, ∗ ∗ ∗ ∗ a c b d 0 ⎫ ⎧ a c b d, + = ⎪ ⎪ =− ⎬ ⇒ ⎨ namely ⎪ ⎪ + = ⎭⎪ ⎩⎪ =− a 2 c 2 b 2 d 2. (8.1.16) Combine E.Q.(8.1.15a) and E.Q.(8.1.15b) with E.Q(8.1.16), and we obtain S S S S = S S S S a 2 1 d 2 1 a 2 d 2,

i.e., S S − S S = − S S S S a d . Therefore a S S =dS ,S b c , (8.1.17) ⎧ S S = S S2 2 ⎪ 1 a b . ⎨ S S = S S Thus, ⎪ ⎩⎪ = S S + S S a d , (8.1.18a) b c , (8.1.18b) ⎧S S=S S2 2 ⎪ 1 a b , (8.1.18c) ⎪ S S=S S∗ ∗ ⎪ 0 ac bd , (8.1.18d) ⎨ =S S + S S ⎪ 1 ad bc. (8.1.18e) ⎪ = + ⎪ ⎩⎪ 79= − Hence, we can denote a, b, c, d in the form of γ a f cos , (8.1.19a) a 2 ⎧ γ ⎪ b= fb sin , (8.1.19b) ⎪ 2 ⎪ γ ⎪ c=−fc sin , (8.1.19c) ⎪ 2 ⎨ ⎪ γ ⎪d=fd cos . (8.1.19d) ⎪ 2 ⎪ ⎪ = Step 4. Then, we have ⎩⎪

2 2 2 2 fa fb fc fd 1, (8.1.20a) ∗ ∗ ⎧ fa fc fb fd, (8.1.20b) ⎪ S S = S S = S S = S S = ⎪ fafd 1 fbfc, (8.1.20c) ⎨ = ⎪ namely ⎩⎪ = =

fa fb fc fd 1, (8.1.21a) f f ∗, (8.1.21b) ⎧ c b ⎪ S S = S∗ S = S S = S S = ⎪ fd f . (8.1.21c) ⎨ = a ⎪ Therefore, we can assume that⎪ ⎩⎪ = −i(β+δ)~2 fa e , (8.1.22a) f e−i(β−δ)~2, (8.1.22b) ⎪⎧ b ⎪ = i(β−δ)~2 ⎪ fc e , (8.1.22c) ⎪ = i(β+δ)~2 ⎨fd e , (8.1.22d) ⎪ = ⎪ i.e., ⎪ ⎩⎪ = −i(β+δ)~2 γ −i(β−δ)~2 γ iα e cos 2 e sin 2 U e i(β−δ)~2 γ i(β+δ)~2 γ e sin 2 e cos 2 − = −iβ −iδ~2 γ iδ~2 γ iα e 0 e cos 2 e sin 2 e iβ~2 −iδ~2 γ iδ~2 γ 0 e e sin 2 e cos 2 − = −iβ γ γ −iδ~2 iα e 0 cos 2 sin 2 e 0 e iβ~2 γ γ δ~2 0 e sin 2 cos 2 0 e − iα = e D z β D y γ D z δ . (8.1.23)

= ( ) ( ) ( )

Notes:

• Relations between rotations: π π D y D x α D y D z α (8.1.24) 2 2 π π ⎧ − ( ) = ( ) ⎪D z D x α D z D y α (8.1.25) ⎨ 2 2 ⎪ ⎩⎪ ( ) − = ( ) 80 • Examples: π π H D x π D y Ph , Hadamard gate, (8.1.26) 2 2 π ⎧ = ( ) ⎪NOT D x π Ph , NOT gate, (8.1.27) ⎨ 2 ⎪ with ⎩⎪ = ( ) iθ Ph θ e I 2. (8.1.28)

These can be veriﬁed easily. ( ) ∶= – Hadamard gate: π π D x π cos i sin σx iσσx, 2 2 ⎧ ⎪ (π )= π − π =−1 ⎪D y cos i sin σy I 2 iσσy , ⎨ 2 4 4 2 ⎪ ⎪ = − =√ ( − ) from which we⎩⎪ can get the left-hand-side of E.Q. (8.1.26) π π 1 D x π D y Ph σx I 2 iσσy 2 2 2 ( ) = √1 ( − ) σx iσσxσy 2 = √1 ( − ) σx σz 2 = HH.√ ( + )

– NOT gate X: = π π π D x π Ph cos i sin σx i 2 2 2 σ ( ) = x − X. = Notes: For an arbitrary single-qubit gate, we= exploit the symbol D to denote it, since such the notation is often used to denote the spinor representation of the SU 2 group. In the literature, we may use another the symbol R instead of D, since this notation means the an arbitrary single-qubit gate may represent a rotation in three dimensional( ) space.

8.1.2 Controlled two-qubit gates and controlled three-qubit gates Def 8.1.2 (Controlled two-qubit gates). Controlled operation is deﬁned as

if A is true, then do B; ⎧ if A is false, then do C. ⎪ ⎨ ⎪ Controlled operation is essential⎩⎪ in computer science. Def 8.1.3 (Controlled U gate). Let’s denote the Controlled U gate as CU , then

CU c t c U c t , (8.1.29)

∶ S ⟩S ⟩81↦ S ⟩ S ⟩ which means if c 0, 0 t 0 t , if c 1, 1 t 1 U t , = S ⟩S ⟩ ↦ S ⟩S ⟩ where c is the control qubit, and =t isS the⟩S ⟩ target↦ S qubit.⟩ S ⟩ This can also be represented in the diagrammatical formalism: S ⟩ S ⟩ CU U● = One example of the Controlled U gate is the CNOT gate, i.e.,

● X● . = 8.1.2.1 Quantum Toﬀoli gate and Fredkin gate Remarks: Quantum Toﬀoli gate and Fredkin gate can be regarded as controlled-controlled- NOT gate and controlled-swap gate respectively.

Def 8.1.4. Analogy to the classical reversible gate, the Quantum Toﬀoli gate is deﬁned as

θ(3) x, y, z x, y, z xy , (8.1.30) where x, y, z 0, 1 and it has the diagramS formalism,⟩ = S ⊕ ⟩

= x x y y Sz⟩ ● Sz⟩ xy . S ⟩ ● S ⟩ Remark: Quantum Toﬀoli gate withS ⟩ the HadamardS ⊕ gate⟩ and phase gate is the universal quantum gate set, namely these three gates can perform universal quantum computation.

Def 8.1.5. The Quantum Fredkin gate is deﬁned as

θ(3) x, y, z x, xz xy,¯ xy xz¯ , (8.1.31) where x, y, z 0, 1 and it has theS diagram⟩ = S formalism,⊕ ⊕ ⟩

= x x y xz xy¯ Sz⟩ ● Sxy⟩ xz¯ . S ⟩ × S ⊕ ⟩ S ⟩ × S ⊕ ⟩ 8.1.3 Quantum circuit model of GHZ states Most important and popular state ¨¨* Bell states - Two-qubit maximally entangled HHj Bell inequality Most important and popular state ¨¨* GHZ states - Multi-qubit maximally entangled HHj GHZ theorem

82 Thm 8.1.3.1.

1 x1 GHZ 0x2x3 xn 1 1¯x2x¯3 x¯n n 2 S ⟩ ∶= CNOT√ S 1nCNOT⋯ ⟩ +1((n−−1) CNOTS ⋯12H⟩1 x1x2 xn , (8.1.32) where CNOTij is the CNOT= gate with the i-th qubit⋯ as the controlS qubit⋯ ⟩ and j-th qubit as the target qubit. And this can be represented in the diagram formulism, shown in Figure 8.1.

x1 H

x2 S ⟩ ● ● ● GHZ x3 S ⟩ S ⟩ Sxn⟩ ⋮ ⋮ ⋮ ⋮ S ⟩ Figure 8.1: GHZ state generated by the CNOT and H gates.

S ⟩ e.g.1. Three qubit GHZ state. 1 GHZ 000 111 . (8.1.33) 3 2 S ⟩ = √ S ⟩ + S ⟩

0 H

S ⟩ 0 GHZ 3

S0⟩ S ⟩

S ⟩ t0 t1 t2 t3

Figure 8.2: GHZ 3 state generated by the CNOT and H gates.

S ⟩ • t0 000 ; 1 • t1 000 100 ; ∶ S 2 ⟩ ∶ √1 S ⟩ + S ⟩ • t2 000 110 ; 2 ∶ √1 S ⟩ + S ⟩ • t3 000 111 . 2 e.g.2. Four qubit∶ √ GHZS state.⟩ + S ⟩ 1 GHZ 0011 1100 . (8.1.34) 4 2 S ⟩ = √ S ⟩ − S ⟩

83 1 H

S0⟩ GHZ 4 S1⟩ S ⟩ S1⟩

S ⟩ t0 t1 t2 t3 t4

Figure 8.3: GHZ 4 state generated by the CNOT and H gates.

S ⟩ • t0 1011 ; 1 • t1 0011 1011 ; ∶ S 2 ⟩ ∶ √1 S ⟩ − S ⟩ • t2 0011 1111 ; 2 ∶ √1 S ⟩ + S ⟩ • t3 0011 1101 ; 2 ∶ √1 S ⟩ + S ⟩ • t4 0011 1100 . 2 Let’s now consider∶ √ S the complete⟩ − S ⟩ set of observables deﬁning the GHZ state.

• The phase-bit operator X 1 X 2 X n:

X 1 X 2 X n CNOT⊗ ⊗⋯⊗1nCNOT1(n−1) CNOT12H 1 x1x2 xn CNOT1n X 1 X 2 X n−1 I 2 CNOT ( − ) CNOT12H 1 x1x2 xn ( ⊗ ⊗⋯⊗ ) ⋯1 n 1 S ⋯ ⟩ CNOT1nCNOT ( − ) = ( ⊗ 1 ⊗⋯⊗n 1 ⊗ ) ⋯ S ⋯ ⟩ X 1 X n−2 I 2 I 2 CNOT ( − ) CNOT12H 1 x1x2 xn = 1 n 2 ( ⊗⋯⊗ ⊗ ⊗ ) ⋯ S ⋯ ⟩ CNOT1nCNOT ( − ) CNOT12 X 1 I 2 I 2 H 1 x1x2 xn ⋮ 1 n 1 n−1 = ⋯ ⊗ ⊗⋯⊗ S ⋯ ⟩ CNOT1nCNOT1(n−1) CNOT12H 1 Z 1´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶I 2 I 2 x1x2 xn , n−1 = ⋯ ⊗ ⊗⋯⊗ S ⋯ ⟩ i.e., ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

X 1 X 2 X n CNOT1nCNOT1(n−1) CNOT12H 1 x1x2 xn x1 1 CNOT1nCNOT ( − ) CNOT12H 1 x1x2 xn , (8.1.35) ⊗ ⊗⋯⊗ 1 n 1 ⋯ S ⋯ ⟩ where we= have(− ) utilized the relation (4.4.22a),⋯ namely S ⋯ ⟩

X 1 X 2 CNOT12 CNOT12 X 1 I 2 ,

and the property ( ⊗ ) = ( ⊗ ) HXH Z 2 XH HZHZ. H I 2 = ¡ ⇒ = = 84 • The ﬁrst parity-bit operator Z 1 Z 2 I 2 I 2: n−2 ⊗ ⊗ ⊗⋯⊗ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ Z 1 Z 2 I 2 I 2 CNOT1nCNOT1(n−1) CNOT12H 1 x1x2 xn n−2 ⊗ ⊗ ⊗⋯⊗ ⋯ S ⋯ ⟩ CNOT1nCNOT´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶1(n−1) CNOT13 Z Z I I CNOT H x x x = 1 2 2 2 ⋯ 12 1 1 2 n n−2 ⊗ ⊗ ⊗⋯⊗ S ⋯ ⟩ CNOT1nCNOT´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶1(n−1) CNOT12 I 2 Z 2 I 2 I 2 H 1 x1x2 xn n−2 = ⋯ ⊗ ⊗ ⊗⋯⊗ S ⋯ ⟩ CNOT1nCNOT1(n−1) CNOT12H 1 I 2 Z´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶2 I 2 I 2 x1x2 xn n−2 = x2 ⋯ ⊗ ⊗ ⊗⋯⊗ S ⋯ ⟩ 1 CNOT1nCNOT1(n−1) CNOT12H 1 x1´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶x2 xn , (8.1.36) where= we( have− ) used the relations (4.4.22d)⋯ and (4.4.22f),S namely⋯ ⟩

Z Z CNOT12 CNOT12 I 2 Z ,

⎧ Z I 2 CNOT12 CNOT12 Z I 2 . ⎪ ⊗ = ⊗ ⎨ ⎪ ⊗ = ⊗ • The second parity-bit I⎩2 Z 2 Z 3 I 2 I 2: n−3 ⊗ ⊗ ⊗ ⊗⋯⊗ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ I 2 Z 2 Z 3 I 2 I 2 CNOT1nCNOT1(n−1) CNOT12H 1 x1x2 xn n−3 ⊗ ⊗ ⊗ ⊗ ⋯ S ⋯ ⟩ CNOT1n CNOT´¹¹¹¹¸¹¹¹¹¶14 I 2 Z 2 Z 3 I 2 I 2 CNOT13CNOT12H 1 x1x2 xn n−3 = ⋯ ⊗ ⊗ ⊗ ⊗ S ⋯ ⟩ CNOT1nCNOT1(n−1) CNOT13 ´¹¹¹¹¸¹¹¹¹¶ Z Z Z I I CNOT H x x x = 1 2 3 2 2 ⋯ 12 1 1 2 n n−3 ⊗ ⊗ ⊗ ⊗ S ⋯ ⟩ CNOT1nCNOT´¹¹¹¹¸¹¹¹¹¶1(n−1) CNOT12 I 2 Z 2 Z 3 I 2 I 2 H 1 x1x2 xn n−3 = ⋯ ⊗ ⊗ ⊗ ⊗ S ⋯ ⟩ CNOT1nCNOT1(n−1) CNOT12H 1 I 2 Z 2 Z´¹¹¹¹¸¹¹¹¹¶3 I 2 I 2 x1x2 xn n−3 = x2+x3 ⋯ ⊗ ⊗ ⊗ ⊗ S ⋯ ⟩ 1 CNOT1nCNOT1(n−1) CNOT12H 1 x1x´¹¹¹¹¸¹¹¹¹¶2 xn , (8.1.37) to derive= (− this) result we have to employ⋯ the relations (4.4.22d)S ⋯ and⟩ (4.4.22f), i.e.,

Z Z CNOT12 CNOT12 I 2 Z ,

⎧ I 2 Z CNOT12 CNOT12 Z Z . ⎪ ⊗ = ⊗ ⎨ ⎪ ⎩⎪ ⊗ = ⊗

85 • The third parity-bit I 2 I 2 Z 3 Z 4 I 2 I 2: n−4 ⊗ ⊗ ⊗ ⊗ ⊗ ´¹¹¹¹¸¹¹¹¹¶ I 2 I 2 Z 3 Z 4 I 2 I 2 CNOT1nCNOT1(n−1) CNOT12H 1 x1x2 xn n−4 ⊗ ⊗ ⊗ ⊗ ⊗⋯⊗ ⋯ S ⋯ ⟩ CNOT1n CNOT´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶15 I 2 I 2 Z 3 Z 4 I 2 I 2 CNOT14CNOT13CNOT12H 1 x1x2 xn = ⋯ n−4 ⊗ ⊗ ⊗ ⊗ ⊗⋯⊗ S ⋯ ⟩ CNOT1n CNOT´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶14 Z 1 I 2 Z 3 Z 4 I 2 I 2 CNOT13CNOT12H 1 x1x2 xn = ⋯ n−4 ⊗ ⊗ ⊗ ⊗ ⊗⋯⊗ S ⋯ ⟩ CNOT1n CNOT´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶13 I 2 I 2 Z 3 Z 4 I 2 I 2 CNOT12H 1 x1x2 xn n−4 = ⋯ ⊗ ⊗ ⊗ ⊗ ⊗⋯⊗ S ⋯ ⟩ CNOT1n CNOT12H 1 I 2 I 2 Z 3 Z´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶4 I 2 I 2 x1x2 xn n−4 = x3+x4 ⋯ ⊗ ⊗ ⊗ ⊗ ⊗⋯⊗ S ⋯ ⟩ 1 CNOT1n CNOT12H1 x1x2 x´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶n . (8.1.38)

• The= n(− 1) -th parity-bit ⋯I 2 I 2 Z n−S 1 Z⋯n: ⟩ n−2 ( − ) ⊗⋯⊗ ⊗ ⊗ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ I 2 I 2 Z n−1 Z n CNOT1nCNOT1(n−1) CNOT12H 1 x1x2 xn n−2 ⊗⋯⊗ ⊗ ⊗ ⋯ S ⋯ ⟩ CNOT´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶1n Z 1 I 2 I 2 Z n−1 Z n CNOT1(n−1) CNOT12H 1 x1x2 xn n−3 = ⊗ ⊗⋯⊗ ⊗ ⊗ ⋯ S ⋯ ⟩ CNOT1nCNOT´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶1(n−1) I I Z Z CNOT CNOT H x x x = 2 2 n−1 n 1(n−2) 12 1 1 2 n n−2 ⊗⋯⊗ ⊗ ⊗ ⋯ S ⋯ ⟩ CNOT´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶1n CNOT12H 1 I 2 I 2 Z n−1 Z n x1x2 xn n−2 = xn−1+x⋯n ⊗⋯⊗ ⊗ ⊗ S ⋯ ⟩ 1 CNOT1n CNOT´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶12H 1 x1x2 xn . (8.1.39) = (− ) ⋯ S ⋯ ⟩ 8.2 Universal quantum computation

8.2.1 Quantum universal gate set The deﬁnition of the Universal Quantum Gate set is analog to the Universal Classical gate set. Def 8.2.1 (Universal Quantum Gate set). A set of quantum gates is universal if any unitary operator U SU 2 (or U U 2 , up to a global phase), can be expressed as a product of the elementary gates from this very set. ∈ ( ) ∈ ( ) One example of the Universal Quantum Gate set is

CNOT, one-qubit gate SU 2 .

This is a Universal Quantum{ Gate set, but the number∈ of( the)} elementary gates in the set is inﬁnity. Thus it is not practical to construct a Quantum Computer with this gate set.

86 Def 8.2.2 (Finite Approximately Universal Quantum Gate set). A ﬁnite gate set is approximately universal, if any unitary operator in SU 2n can be approximately expressed as a product of elementary gates in this set to arbitrary precisions. ( ) Examples: 1) HH,SSS,θθ(3) , H being the Hadamard gate, S representing the Phase gate, and θ(3) for the three-bit quantum Toﬀoli gate, i.e., 1 1 1 1 0 H ,SS , θ(3) x, y, z x, y, z xy . 2 1 1 0 i = √ = S ⟩ = S ⊕ ⟩ For this case, there are− only three elementary gates in this set, we can construct a very small Quantum Computer. And we can get arbitrary precisions, though this gate set is not a “real” Universal gate set.

π 2) HH,TTT, CNOT with T as the 8 gate and CNOT denoted the CNOT gate,namely,

−i π { } i π e 8 0 T e 8 i π , CNOT x, y x, x y . 0 e 8 = S ⟩ = S ⊕ ⟩ Remark: S T 2.

NOTE: Designing= of a unitary operator as product of elementary gates may require exponential number of gates, therefore, this sort of design is not suﬃcient and Quantum Computer may run very slowly in certain conditions.

8.2.2 Universal quantum gate set of two-qubit gates

Def 8.2.3 (Generic two-qubit gates). The two-qubit gate with eigenvalues eiθ1 , eiθ2 , eiθ3 and eiθ4 , where θi and θi are irrational numbers, is deﬁned as Generic two-qubit gate. π θj To completely understand this, we have to learn the Number Theory in mathematics. Thm 8.2.2.1. Any Generic two-qubit gate is an universal quantum gate set.

Thm 8.2.2.2 (Barenco’s{ gate). Barenco’s gate} (1995) is deﬁned as π θ Controlled Ph D x , (8.2.1) 4 2

θ − − where π is an irrational number. It can be represented in the diagram formalism as

U ● with π θ U Ph D x 4 2

π θ θ ∶= −i − cos 4 i sin 4 e 4 θ θ . (8.2.2) i sin 4 cos 4 − = Barenco’s gate is not a generic gate. One− thing to keep in mind is that we want to avoid the irrational number. The set Barenco’s gate, SWAP is a universal quantum gate set.

{ 87 } Thm 8.2.2.3. CNOT, arbitrary single-qubit gates is a universal quantum gate set.

Proof. Since Barenco’s gate together with the SWAP gate can perform universal quantum computation, we describe the SWAP gate as

SWAP CNOT12CNOT23CNOT12, (8.2.3) and with the following Lemmas,= reformulate Barenco’s gate as an expression in terms of the CNOT gate and single-qubit gates.

Lemma 8.2.2.1. Any single-qubit gate U , can be described in terms of Euler angles, namely iα U e D z β D y γ D z δ , with α, β, γ, δ R. (8.2.4)

Lemma 8.2.2.2. Any single-qubit= ( ) gate( )U can( ) be formulated as ∈

U Ph α AXBXCAXBXC, with ABC I 2,AAA,BBB,CC SU 2 and α R. (8.2.5)

Lemma= 8.2.2.3.( ) Any controlled two-qubit= gate can be decomposed∈ ( ) in the∈ following way

Rα (8.2.6) U ● C ● B ● A = with 1 0 R . α 0 eiα

Proof. We are now going to prove the three lemmas= 8.2.2.1, 8.2.2.2 and 8.2.2.3.

Lemma 1: The lemma 8.2.2.1 is proved in subsection 8.1.1, section 8, chapter 7.

Lemma 2: The lemma 8.2.2.2 can be proved through lemma 8.2.2.1. Because AA,BBB,CC are all elements of the SU 2 group, therefore, from lemma 8.2.2.1, we can construct AA,BBB,CC in the form of: ( ) A D z φ D y ϕ D z ψ , (8.2.7a) B φ′ ϕ′ ψ′ , (8.2.7b) ⎧ D z D y D z ⎪ = ( ′′) ( )′′ ( ) ′′ ⎪C D z φ D y ϕ D z ψ , (8.2.7c) ⎨ = ( ) ( ) ( ) ⎪ where we have introduced 9⎩⎪ parameters.= ( ) The( ) expressions( ) (8.2.7a) (8.2.7c) ensures that A, B and C all belong to SU 2 , therefore ABC SU 2 . Since we can pa- rameterize an arbitrary SU 2 operator with only two independent∼ real variables, the expressions (8.2.7a) (8.2.7c) have( ) two constraints. And∈ the( ) relation ( ) ∼ ABC I 2 (8.2.8)

gives rise to another constraint on the parameters= that we have introduced in the expressions (8.2.7a) (8.2.7c). Therefore, we actually have 6 independent variables in the expressions (8.2.7a) (8.2.7c). We do not need that much of independent ∼ ∼

88 variables, since U U 2 can be determined with three independent real variables. We may set ∈ ( ) γ A D z β D y , (8.2.9a) 2 γ β δ ⎧ = ( ) ⎪B D y D z , (8.2.9b) ⎪ 2 2 ⎪ β δ + ⎨C=D z − . − (8.2.9c) ⎪ 2 ⎪ ⎪ − + As we shall see that ⎩⎪ = γ γ β δ β δ ABC D z β D y D y D z D z 2 2 2 2 + − + = D z(β)D y 0 D z −β − I 2, = ( ) ( ) (− ) and = γ γ β δ β δ AXBXC D z β D y XD y D z XD z 2 2 2 2 γ γ β δ + β δ − + = D z(β)D y D y −Dz − D z 2 2 2 2 + − + = D z(β)D y γ D z δ , (8.2.10) from which we can= see that( ) theorem( ) 8.2.2.2( ) can be derived from theorem 8.2.2.1. Lemma 3: The lemma 8.2.2.3 can be proved by utilizing the lemma 8.2.2.2. The controlled-U gate can be expressed as

I CU 0 0 I 1 1 U 2 (8.2.11) 2 U

By inserting E.Q.(8.2.5) into= S E.Q.(8.2.11),⟩⟨ S ⊗ + S ⟩⟨ andS ⊗ we= can get

CU ABC eiαAXBXC = I 2 A I 2 B I 2 C iα e I 2 A X B X C

= A B C C Ph α CNOT CNOT , (8.2.12) A B C

where C Ph= α is− the( controlled-phase) gate,

iα − ( ) C Ph α 0 0 I 2 1 1 e I 2

− ( ) = S I⟩⟨2 S ⊗ + S ⟩⟨ S ⊗ iα e I 2

= 1 0 I 0 eiα 2

= R α I 2, ⊗ (8.2.13)

= 89 ⊗ namely

Rα (8.2.14) Ph α ● . On the other hand, we can also obtain = ( ) A I A (8.2.15a) A 2 ⎧ A ; ⎪ ⎪ = ⊗ = ⎪ B ⎪ I 2 B (8.2.15b) ⎪ B B ; ⎪ ⎨ = ⊗ = ⎪ C ⎪ I 2 C (8.2.15c) ⎪ C C . ⎪ ⎪ = ⊗ = On the other hand we know⎪ that ⎩⎪ CU (8.2.16) U . ● = By substituting E.Q. (8.2.13), (8.2.15a) (8.2.15c) and (8.2.16) in E.Q. (8.2.12), we can acquire E.Q. (8.2.6). ∼ 8.2.3 Deutsch’s gate is a universal quantum gate Thm 8.2.3.1 (Deutsch’s gate). Deutsch gate (1989) is a universal quantum gate,

Deutch’s gate x y z x y Rxy z , (8.2.17) where ∶ S ⟩S ⟩S ⟩ ↦ S ⟩S ⟩ S ⟩ π R Ph D x θ 2 −iσσ θ ∶= ie −x 2, ( ) (8.2.18)

θ with π being an irrational real number.= And− it has the diagrammatical formalism,

Deutch’s gate ● = R● This is the ﬁrst universal quantum gate, and is a three-qubit gate. Note that 2 π π θ 2 R Ph D x θ Ph D x U , 2 4 2 Deutsch Barenco = − ( ) = − = ( ) and we show that® Deutsch’s gate can be expressed in terms of the® two-qubit gates, i.e., CNOT and Barenco’s gate,

● ● ● ● (8.2.19) † R● = U● U● U t0 t1 t2 t3 t4 t5

90 If we let the quantum circuit act on x y z state, then we can get the tripartite states in diﬀerent steps as labeled in the right-hand-side of E.Q. (8.2.19): S ⟩ ⊗ S ⟩ ⊗ S ⟩ (1) ψ t0 x y z ;

y (2) Sψ(t1)⟩ = Sx⟩Sy⟩SU⟩ z ; y (3) Sψ(t2)⟩ = Sx⟩Sx⟩ y US ⟩ z ;

† x⊕y y (4) Sψ(t3)⟩ = Sx⟩Sx⊕y⟩ U S ⟩ U z x x y U −x⊕yU y z ; S ( )⟩ = S ⟩S ⊕ ⟩( ) S ⟩ −x⊕y y (5) ψ t4 = Sx⟩Sx⊕y⟩ x U US ⟩ z x y U −x⊕yU y z ; S ( )⟩ = S ⟩S ⊕ ⊕ ⟩ S ⟩ x −x⊕y y (6) ψ t5 = Sx⟩Sy⟩U U US ⟩ z x y U x+y−x⊕y z S ( )⟩ = Sx⟩Sy⟩U x+y−(x+y−2xyS )⟩ z = Sx⟩Sy⟩U 2xy z S ⟩ = Sx⟩Sy⟩Rxy z . S ⟩ = S ⟩S ⟩ S ⟩ Note: the calculation= S ⟩S ⟩ hereS ⟩ is binary, namely

x y x y mode 2 x y 2xy, (8.2.20) ⊕ ∶= ( + ) where x, y 0, 1 . = + −

∈ { }

91 Chapter 9

Physical Realization of Quantum Computers

92 Chapter 10

Quantum Algorithms

Quantum mechanics: Real Black Magic Calculus. —Albert Einstein

Computer programming is an art of form, like the creation of poetry or music. —Donald Knuth

Deutsch’s algorithm combines quantum parallelism with a property of quantum mechanics known as interference. —Nielsen & Chuang

The quantum search algorithm is essentially optimal, is both exciting and disappointing. It is exciting because it tells us that for this problem, at least we have fully plumbed the depths of quantum mechanics, no further improvement is possible. The disappointment arises because we might have hoped to do much better than the square root speedup oﬀered by the quantum search algorithm. —Nielsen & Chuang

The essence of the design of many quantum algorithms is that a clever choice of function and ﬁnal transformation allows eﬃcient determination of useful global information about the function–information which can not be attained quickly on a classical computer. —Nielsen & Chuang

Reference: [Preskill] Chapter 6: Quantum computation;

● [Nielsen & Chuang] Chapter 6: Quantum search algorithms

● 10.1 Classical and quantum algorithm

Def 10.1.1 (Algorithm). An algorithm is 1. a well-deﬁned procedure;

2. with ﬁnite descriptions;

93 3. designed for realising an information processing task;

4. designed to solve a computational problem.

This definition is proper for both classical algorithm and quantum algorithm. The only thing we have to notice is that, the classical algorithm is designed to process classical information, while the quantum algorithm is designed to perform information processing task using the fundamental principle’s of Quantum Mechanics. But, what advantage can we get from the Quantum Algorithm? One of the benefit is that Quantum Algorithm is more efficient than the Classical Algorithm in some specific problems. And there are some examples, as shown in Table 10.1:

Table 10.1: Examples of Quantum Algorithm

Speed up Examples Non-exponential speed up Grover’s search algorithm (Oracle model) Relativized exponential speed up Simon’s algorithm (Oracle model) Exponential speed up Shor’s factoring algorithm

10.2 Oracle model

Oracle model is the black-box model.

Def 10.2.1 (Oracle (black box)). A subroutine evaluating a function, but we have no idea about how this subroutine is performed. We are only concerned about how this subroutine is performed, or we only make a query, on the Oracle (Black-box).

Remark: For the oracle (black box), usually we don’t care about how the black box works, and we are only concerned about how to use it.

• We can make a query on U f .

• We are only concerned about the input and output.

• No idea about how U f is performed. Examples:

(1) Classical black box: x f f x .

which may be not a reversible operation. And we( ) can deﬁne the classical black box which is a reversible gate to compute the function f x shown as

x x ( ) Uf y y f x .

⊕ ( )

94 (2) Quantum black box is a quantum gate Uf computing f x :

x x ( ) U f y y f x , S ⟩ S ⟩ where x is called the register qubit, and y is called the ancilla qubit. Note that S ⟩ S ⊕ ( )⟩ quantum gate U f satisﬁes the unitary (reversible) condition given by S ⟩ S ⟩ † † U fU f U fU f I d. (10.2.1) The Query Complexity: = = • In the oracle model, we only count the minimal number of the black box to char- acterize the circuit complexity. Because the time of performing U f gate takes far greater than the time of performing other gates.

• Complexity is very abstract, but very important in computer science.

Note: Time of performing U f Time of perform other gates. For example

≫Time U f 1 year; Time (other gates) 1 sec. (10.2.2) ( ) = = 10.3 Deutsch’s algorithm

The Deutsch’s algorithm is the ﬁrst Quantum Algorithm, presented in 1989.

10.3.1 Deﬁnitions Def 10.3.1 (Constant function and Balanced function). Let’s deﬁne a function f: f x 0, 1 y f x 0, 1 . (10.3.1)

Then, the function f is ∶ ∀ ∈ { } ↦ = ( ) ∈ { } Constant function, if f 0 f 1 ; (10.3.2) Balanced function, if f 0 f 1 . (10.3.3) ( )= ( ) • In the case that f is constant, i.e., f 0 f 1 (, we)≠ can( ) acquire f 0 f 1 0(, ) =or ( f) 0 f 1 1. (10.3.4) Therefore, ( ) = ( ) = ( ) = ( ) = f 0 f 1 0 (10.3.5) with the binary addition. ( ) ⊕ ( ) = • In the case that f is balanced, i.e., f 0 f 1 , we can know

f 0 0, f 1 1(, ) ≠or ( f) 0 1, f 1 0. (10.3.6) Similarly, ( ) = ( ) = ( ) = ( ) = f 0 f 1 1 (10.3.7) which suggests that the number of x which satisﬁes f x 0 and x that f x 1 is ( ) ⊕ ( ) = same, namely ( ) = ( ) = 1 # x f x 0, x 0, 1 # x f x 1, x 0, 1 . (10.3.8)

= T ( ) = ∀ ∈ { } = T ( ) = ∀ ∈ { } 95 Question: Assume that the Uf is a black box of computing f x , what is the minimum number of performing Uf to judge whether f x is a constant or balanced function? ( ) ( ) 10.3.2 Classical algorithm

The minimum number of performing Uf to know about whether f x is constant or balanced is two. For x x ( ) Uf y y f x , we can construct the following classical circuit to verify whether f x is constant or bal- ⊕ ( ) anced, 0 0 ( ) Uf 0 f 0 (10.3.9) 1 1 Uf ● ( ) 0 f 0 f 1 , where obviously the f 0 and f 1 have to be calculated indepently. ( )⊕ ( ) 10.3.3 Deutsch’s( problem) ( ) Deutsch’s problem is ﬁrstly presented in 1989, and it is of conceptual importance. This problem can be stated in the following way.

• Input: A black-box for computing an unknown function f x , i.e.,

x x ( ) U f (10.3.10) y y f x . S ⟩ S ⟩ • Promise: f x is constant or balanced.S ⟩ S ⊕ ( )⟩

• Question: Determine( ) f x constant or balanced.

• Answer: 1 time of using( U)f . Note: In quantum mechanics, f 0 and f 1 can be computed simultaneously, because of quantum superposition principle. ( ) ( ) 10.3.4 Phase kick-back

Lemma 10.3.4.1. With the quantum black box U f deﬁned as

U f x y x y f x , (10.3.11) with the diagrammatical representation∶ S ⟩S ⟩ ↦ S ⟩S ⊕ ( )⟩ x x U f (10.3.12) y y f x , S ⟩ S ⟩ we can prove the following algebraicS formula,⟩ S ⊕ ( )⟩

0 1 f(x) 0 1 U f x 1 x , (10.3.13) 2 2 S ⟩ − S ⟩ S ⟩ − S ⟩ S ⟩ √ = (− ) S ⟩ √ 96 i.e., the quantum circuit model given by

x x (10.3.14) U f S ⟩−S ⟩ S ⟩−S ⟩ 0√S 1⟩ S ⟩1 f(x) 0√ 1 . 2 2 or equivalently (− ) x 1 f(x) x (10.3.15) U f S ⟩−S ⟩ S ⟩−S ⟩ 0√S 1⟩ (0−√ )1 . S ⟩ 2 2 Note: Why is the above quantum circuit model called the phase kick-back? The global phase 1 f(x) can be kicked back from the target qubit to the control quibt, so that S ⟩−S ⟩ the target qubit 0√ 1 seems unchanged from the input to the output. (− ) 2 This lemma can be proved in the following manner.

Proof.

0 1 U f x 0 U f x 1 U f x 2 2 S ⟩ − S ⟩ S ⟩S ⟩ − S ⟩S ⟩ S ⟩ √ = x f x √ x 1 f x 2 S ⟩S ( )⟩ − S ⟩S ⊕ ( )⟩ = x f x √x f¯ x 2 S ⟩S ( )⟩ − S ⟩S ( )⟩ = f x √ f¯ x x 2 S ( )⟩ − S ( )⟩ = S ⟩if f x √ 0, x √1 0 1 ; 2 1 1 ⎧ if f x 1, x √ 1 0 1 x √ 0 1 ; ⎪ ( ) = S ⟩ 2 (S ⟩ − S ⟩) 2 = ⎨ 0 1 ⎪ 1 f((x))x= S .⟩ (S ⟩ − S ⟩) = (− )S ⟩ (S ⟩ − S ⟩)(10.3.16) ⎩ 2 S ⟩ − S ⟩ There we get E.Q. (10.3.13)= (− veriﬁed.) S ⟩ √

Example: If f x x, then U x CNOT, since

( ) = =U x x y x y f x x y x S ⟩S ⟩ = S ⟩S ⊕ ( )⟩ CNOT x y . (10.3.17) = S ⟩S ⊕ ⟩ And we can see that = S ⟩S ⟩ 0 1 x x x x¯ CNOT x 2 2 S ⟩ − S ⟩ S ⟩S ⟩ − S 0⟩S ⟩ 1 S ⟩ √ = 1 x√x , (10.3.18) 2 S ⟩ − S ⟩ so that in such the circumstance we have =CNOT(− ) xS ⟩ √1 x x .

Note: The phase kick-back is the key technique in theS ⟩ = performance(− ) S ⟩ of such quantum algorithms as Deutsch’s algoithm, Deutsch-Jozsa’a algorithm, Grover’s search algorithm, and so on.

97 10.3.5 Deutsch’s algorithm This is the ﬁrst quantum algorithm in quantum information science. Although this algorithm may be nonsense in practice, but is of conceptual importance. The associated quantum circuit is of the form,

0 H H U f S ⟩−S ⟩ (10.3.19) 0√ 1 2 t1 t2 t3 t4 where we study it in the following ﬁve steps:

1. Input: state preparation. S ⟩−S ⟩ ψ t 0 0√ 1 , 1 2

2. Sψ(t2)⟩ = S H⟩ ⊗I 2 ψ t1 S ⟩+S ⟩ S ⟩−S ⟩ 0√ 1 0√ 1 , S ( )⟩ = ( ⊗2 )S 2( )⟩ We have= utilized⊗ the relation that 0 1 H 0 1 1 2 H i 1 i⋅j j (10.3.20) S0⟩ + S1⟩ ⎪⎫ H S ⟩= √ ⎪ 2 j=0 H 1 ⎪ 2 ⎬ ⇐⇒ S ⟩ = √ Q(− ) S ⟩ S ⟩ − S ⟩ ⎪ S ⟩= √ ⎪ where i j is AND operation. ⎭⎪

3. ψ t3 ⋅ U f ψ t2 S ⟩−S ⟩ √1 1 U i 0√ 1 S ( )⟩ = 2 S i(=0 )⟩f 2 S ⟩−S ⟩ √1 1 1 f(i) i 0√ 1 , = 2 ∑i=0 S ⟩ ⊗ 2

4. ψ t4 = H ∑I 2 (−ψ )t3 S ⟩ ⊗ S ⟩−S ⟩ √1 1 1 f(i) H I i 0√ 1 S ( )⟩ = 2 ⊗ i=0 S ( )⟩ 2 2 S ⟩−S ⟩ 1 1 1 f(i)+i⋅j j 0√ 1 . = 2 ∑i,j=0 (− ) ⊗ S ⟩2⊗ 5. Output:= quantum∑ (− measurement.) S ⟩ ⊗ ______L And the icon ______stands for the measurement with respect to the standard basis 0 , 1 . If we perform the measurement 0 0 I, there we can get the state after measurement for the bipartite system as, {S ⟩ S ⟩} S ⟩⟨ S ⊗ 1 1 f(i) 0 1 j 0 ψ˜ t4 1 0 . (10.3.21) 2 i=0 2 S ⟩ − S ⟩ = ∶ S ( )⟩ = Q(− ) S ⟩ ⊗ √ And we can see that,

98 (1) if f is a constant function, i.e., f 0 f 1 , the probability that the ﬁrst qubit is in the state 0 is 1, since ( ) = ( ) S ⟩ f(0) 0 1 0 0 I 2 ψ t4 1 0 , (10.3.22) 2 1 S ⟩ − S ⟩ (S ⟩⟨ S ⊗ )S ( )⟩ = (1− ) Sf(⟩i⊗)+i √ 0 1 1 1 I 2 ψ t4 1 1 2 i=0 2 S ⟩ − S ⟩ (S ⟩⟨ S ⊗ )S ( )⟩ = Q(− )1 1 S ⟩ ⊗ √0 1 1 f(0) 1 i 1 2 i=0 2 S ⟩ − S ⟩ = (0;− ) Q(− ) S ⟩ ⊗ √ (10.3.23)

(2) if f is a balanced function, i.e., f= 0 f 1 , the probability that the ﬁrst qubit is in the state 0 is 0, due to ( )≠ ( ) 1 S ⟩ 1 f(i) 0 1 0 0 I 2 ψ t4 1 0 2 i=0 2 S ⟩ − S ⟩ (S ⟩⟨ S ⊗ )S ( )⟩ = 0.Q(− ) S ⟩ ⊗ √ (10.3.24)

Note: What is underlying Deutsch’s algorithm?= The f 0 and f 1 are calculated simultaneously so that f 0 f 1 is calculated using the black box Uf once. This is called quantum parallelism due to quantum interference or quantum( superposition) ( ) principle. ( )⊕ ( ) 10.4 Deutsch-Jozsa’s algorithm

As we have seen that the Deutsch’s algorithm uses the one-qubit register, we are going to consider the Deutsch-Jozsa’s Algorithm which uses an n-qubit register.

10.4.1 Constant and balanced function in n-qubit Def 10.4.1 (Constant and Balanced function (n-qubit)). Let’s deﬁne the function f to be f x 0, 1 n f x 0, 1 . (10.4.1)

Then function f is ∶ ∀ ∈ { } ↦ ( ) ∈ { } n • Constant function, if f x c, x 0, 1 , where c is a constant in 0, 1 ;

n • Balanced function, if the( ) number= ∀ of∈ elements{ } in the set x f x 0, x{ 0}, 1 is n n−1 the same as that in x f x 1, x 0, 1 , i.e., the numbers T ( ) both= ∀ equal∈ { to }2 . T ( ) = ∀ ∈ { } # x f x 0, x 0, 1 n # x f x 1, x 0, 1 n 2n−1. (10.4.2)

T ( ) = ∀ ∈ { } = T ( ) = ∀ ∈ { } = 10.4.2 Constant or balanced function? Question: If we know that f is either a Constant function or a Balanced function, how can we determine it? And, how many queries would have to be made?

There are some optional ways to solve this problem:

99 • If we choose to use the classical deterministic algorithm, there we should make at least 2n−1 1 queries in the worst case.

• If we use the+ classical random algorithm, there are at least order 1 queries.

• If we choose Deutsh-Jozsa’s algorithm, one query will suﬃce, which( ) is due to the application of the quantum superposition principle.

10.4.3 Notation and lemma • State:

x x1x2 xn , (10.4.3a)

y y1y2 yn . (10.4.3b) S ⟩ ∶= S ⋯ ⟩ • “Product”: S ⟩ ∶= S ⋯ ⟩ x y x1y1 x2y2 xnyn, (10.4.4) with ⋅ ∶= ⊕ ⊕⋯⊕ xiyi xi AND yi. (10.4.5)

• Summation: ∶= 1 1 1 . (10.4.6) y∈{0,1}n y1=0 y2=0 yn=0 Q ∶= Q Q ⋯ Q Lemma 10.4.3.1.

(n) H x H H H x1x2 xn n S ⟩ ∶= 1 ⊗ ⊗⋯⊗ S ⋯ ⟩ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶1 x⋅y y . (10.4.7) n~2 2 y∈{0,1}n = Q (− ) S ⟩ Proof.

(n) H x H H H x1x2 xn n S ⟩ = ⊗ ⊗⋯⊗ S ⋯ ⟩ H´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶x1 H x2 H xn n 1 1 1 1 x1⋅y1 x2⋅y2 xn⋅yn = S ⟩ ⊗ S ⟩1⊗ ⋯⊗ y1S ⟩ 1 y2 1 yn 2 y1=0 y2=0 yn=0 = √ n Q1 (−1 ) 1S ⟩ Q (− ) S ⟩ ⋯ Q (− ) S ⟩ 1 x1⋅y1+x2⋅y2+⋯+xn⋅yn 1 y1y2 yn 2 y1=0 y2=0 yn=0 = √ n Q1 Q1 ⋯ Q1 (− ) S ⋯ ⟩ 1 x1⋅y1⊕x2⋅y2⊕⋯⊕xn⋅yn 1 y1y2 yn 2 y1=0 y2=0 yn=0 = √1 Q Q ⋯ Q (− ) S ⋯ ⟩ 1 x⋅y y , (10.4.8) n~2 2 y∈{0,1}n = Q (− ) S ⟩ where we have utilized E.Q. (10.3.20). Therefore, we have veriﬁed E.Q. (10.4.7).

Note: 1 H (n) 0 x . (10.4.9) n~2 2 x∈{0,1}n S ⟩ = Q S ⟩ 100 With the action of the n-fold Hadamard gates, we have the superposition of all the computational basis vectors, namely, all product basis vectors. In other words, all computational basis vectors can be encoded in the one quantum state due to the quantum superposition principle.

Lemma 10.4.3.2 (Phase kick-back for n-qubit). Let’s deﬁne U f as

U f a y a y f a (10.4.10) with ∶ S ⟩S ⟩ ↦ S ⟩S ⊕ ( )⟩ f a 0, 1 n f a 0, 1 , and ∶ ∀ ∈ { } ↦ ( )∈ { } y 0, 1 . And we call a as the ﬁrst register, which is an n-qubit state, and y as the second register, ∈ { } which is a single qubit. Therefore, S ⟩ S ⟩ 0 1 f(x) 0 1 U f x 1 x , (10.4.11) 2 2 S ⟩ − S ⟩ S ⟩ − S ⟩ with x 0, 1 n. This can alsoS ⟩ be presented√ = (− in) the diagramS ⟩ √ formalism:

∈ { } x1 x1 x x S 2⟩ S 2⟩ U f (10.4.12) Sxn⟩ Sxn⟩ S0⟩−⋮S1⟩ ⋮ S0⟩−S1⟩ √ 1 f(x) √ . S 2 ⟩ S ⟩ 2 Proof. (− ) 0 1 U f x 2 f xS ⟩ − S1⟩ f x x S ⟩ √ 2 S ( )⟩ − S + ( )⟩ = S ⟩ f x √f¯ x x 2 S ( )⟩ − T ( )f 0 1 = S ⟩1 f(x) √x . (10.4.13) 2 S ⟩ − S ⟩ We have made use of the same technique= (− ) exploitedS ⟩ √ to prove lemma 10.3.4.1, here.

10.4.4 Deutsch-Jozsa’s algorithm Deutsch-Jozsa’s algorithm can be expressed in quantum circuit model:

0 1 H H

S0⟩2 H H U f (10.4.14) S ⟩ 0 n H H S ⟩−⋮S ⟩ ⋮ ⋮ ⋮ S ⟩−S ⟩ 0√ 1 1 f(x) 0√ 1 , S 2⟩ 2 t1 t2 t3 t4 (− ) where

101 (1) Input: state preparation. ⊗ S ⟩−S ⟩ ψ t 0 n 0√ 1 , 1 2 ⊗n (2) Sψ(t2)⟩ = S ⟩H ⊗ I 2 ψ t1 1 S0⟩−S1⟩ n x √ , S ( )⟩ = 2n~2 x∈⊗{0,1} S ( )⟩2 (3) The phase= kick-back∑ technique.S ⟩

ψ t3 U f ψ t2 1 S0⟩−S1⟩ U n x √ S ( )⟩ = 2n~2S (f )⟩x∈{0,1} 2 1 f(x) S0⟩−S1⟩ n 1 x √ , = 2n~2 x∈∑{0,1} S ⟩ 2 ⊗n (4) ψ t4 = H ∑ I 2 (ψ−t3) S ⟩ 1 ⊗n f(x) S0⟩−S1⟩ H I n 1 x √ S ( )⟩ = 2n~2 ⊗ S 2( )⟩x∈{0,1} 2 1 f(x)+x⋅y S0⟩−S1⟩ n 1 y √ . = 2n x,y∈{0,1}⊗ ∑ (− ) 2 S ⟩ (5) Output:= quantum∑ measurement.(− ) S ⟩ If the post-measurement state of the ﬁrst n-qubit is 0 ⊗n, then the corresponding state of the n 1-qubit system should be S ⟩ 1 0 1 + ψ˜ t 1 f(x) 0 ⊗n . (10.4.15) 4 n 2 ∈{ }n 2 x 0,1 S ⟩ − S ⟩ T ( )f = Q (− ) S ⟩ √ Then, we can infer that

• if f is a constant function, i.e., f x f x′ , x, x′ 0, 1 n ,

⊗n then the probability to ﬁnd( out) = that( ) the∀ ﬁrst∈ {n-qubit} is in the state 0 is 1, since S ⟩ 1 0 1 0 ⊗n 0 I ψ t 1 f(0) 0 ⊗n 2 4 n 2 ∈{ }n 2 x 0,1 S ⟩ − S ⟩ S ⟩ ⟨ S ⊗ S ( )⟩ = Q (− )0 S 1⟩ √ 1 f(0) 0 ⊗n , (10.4.16) 2 S ⟩ − S ⟩ and = (− ) S ⟩ √ 1 0 1 1 ⊗n 1 I ψ t 1 f(x)+x 1 ⊗n 2 4 n 2 ∈{ }n 2 x 0,1 S ⟩ − S ⟩ S ⟩ ⟨ S ⊗ S ( )⟩ = 1 Q (− ) S ⟩ 0 √ 1 1 f(0)+x 1 ⊗n n 2 ∈{ }n 2 x 0,1 S ⟩ − S ⟩ = (− ) S ⟩ √ Q 1 0 1 1 f(0) 1 x 1 ⊗n n 2 ⎡ n ⎤ 2 ⎢x∈{0,1} ⎥ ⎢ ⎥ S ⟩ − S ⟩ = 0(−, ) ⎢ Q (− ) ⎥ S ⟩ √ (10.4.17) ⎢ ⎥ ⎣ ⎦ with = 1 x 1 x1+⋯+xn .

(− ) ∶= (− ) 102 • if f is a balanced function, i.e.,

# x f x 0, x 0, 1 n # x f x 1, x 0, 1 n ,

then the probability T ( ) to= ﬁnd∀ out∈ { that} the= ﬁrst T n(-qubit) = is∀ in∈ { the state} n ⊗n is 0, since S ⟩ 1 0 1 0 ⊗n 0 I ψ t 1 f(x) 0 ⊗n 2 4 n 2 ∈{ }n 2 x 0,1 S ⟩ − S ⟩ S ⟩ ⟨ S ⊗ S ( )⟩ = (− ) S ⟩ √ 1 Q 0 1 1 f(x) 0 ⊗n n 2 ⎡ n ⎤ 2 ⎢x∈{0,1} ⎥ ⎢ ⎥ S ⟩ − S ⟩ = 0. ⎢ Q (− ) ⎥ S ⟩ √ (10.4.18) ⎢ ⎥ ⎣ ⎦ Remark: Quantum superposition principle= makes it possible that computing f x for all x 0, 1 n simultaneously, which is impossible in classical computation. ( ) ∈{ } 10.5 Bernstein-Vazirani’s algorithm

Firstly, let’s introduce a lemma.

Lemma 10.5.0.1.

(a+y)⋅x n n 1 2 δa,y, a, y 0, 1 . (10.5.1) x∈{0,1}n Q (− ) = ∀ ∈ { } This lemma can proved in the following way.

Proof.

1 1 1 1 (a+y)⋅x 1 (a1+y1)⋅x1 1 (a2+y2)⋅x2 1 (an+yn)⋅xn x∈{0,1}n x1=0 x2=0 xn=0 (− ) = n (− ) (− ) ⋯ (− ) Q 2Qδa1,y1 δa2,y2 δan,yQn Q n 2 δa,y, (10.5.2) = ⋯ which is equivalent to E.Q.= (10.5.1).

Let’s assume that there is a function fa deﬁned as

n fa x 0, 1 fa x 0, 1 . (10.5.3)

And we also know some facts∶ about ∀ ∈ the{ function} ↦ (black( ) ∈ box){ } that

fa x a x

a1x1 a2x2 anxn. (10.5.4) ( ) = ⋅ Question: How can we determine= a, which⊕ ⊕⋯⊕ is an n-bit string?

y Solution 1 With classical algorithm, we need totally n queries, i.e., n equations, to get a.

103 y Solution 2 With Quantum Algorithm, only one query is suﬃcient to obtain a. For the Quantum Algorithm, the corresponding Quantum Circuit model should be

______L 0 1 H H ______

S0⟩2 H H

U fa S ⟩ (10.5.5) 0 n H H S ⟩−⋮S ⟩ ⋮ ⋮ ⋮ S ⟩−S ⟩ 0√ 1 1 f(x) 0√ 1 , S 2⟩ 2 t1 t2 t3 t4 (− ) where

⊗ S ⟩−S ⟩ (1) ψ t 0 n 0√ 1 , 1 2 ⊗n (2) Sψ(t2)⟩ = S ⟩H ⊗ I 2 ψ t1 1 S0⟩−S1⟩ n x √ , S ( )⟩ = [(2n~2 x)∈⊗{0,1}]S ( )⟩2 (3) ψ t U ψ t 3 = fa ∑ 2 S ⟩ 1 S0⟩−S1⟩ U n x √ S ( )⟩ = 2n~2 S f(a )⟩x∈{0,1} 2 1 a⋅x S0⟩−S1⟩ n 1 x √ , = 2n~2 x∈{∑0,1} S ⟩ 2 (4) ψ t H ⊗n I ψ t 4 = ∑ 2 (− )3 S ⟩ 1 ⊗n a⋅x S0⟩−S1⟩ H I n 1 x √ S ( )⟩ = [(2n~2 ) ⊗ ]S 2( )⟩x∈{0,1} 2 1 x⋅(a+y) S0⟩−S1⟩ n 1 y √ = 2n [(x,y∈{0,1)}⊗ ] ∑ (− )2 S ⟩ S0⟩−S1⟩ n δ y √ = y∈∑{0,1} a,y (− ) 2 S ⟩ S ⟩−S ⟩ a 0√ 1 . = ∑ 2 S ⟩ With the quantum= S ⟩ ⊗ measurement on the n register qubits, we can get a , and equivalently we obtain a a1, a2, , an . S ⟩ = ( ⋯ ) 10.6 Simon’s algorithm

Simon’s Algorithm

• has no phase kickback;

• is the 1-st algorithm that shows Quantum Algorithm can have exponential speed up for apparently hard problems.

10.7 Grover’s algorithm

Let’s now consider a search problem, the unstructured database search, with N 1 items. The unstructured database implies no special constrains or symmetries on the database. The task is to locate one particular term amid an unstructured set, with≫N items. Metaphorically, we say the task is to ﬁnd a needle in a haystack, or to catch a particular ﬁsh in the sea.

104 10.7.1 Overview of the problem

Let’s assume that there is a black-box for computing an unknown function fω deﬁned as

n fω x 0, 1 fω x 0, 1 . (10.7.1)

Question: How can we∶ ﬁnd ∀ a∈ unique{ } marked↦ ( term) ∈ {x }ω 0, 1 n such that

1, if x ω;= ∈ { } (10.7.2a) fω x 0, if x ω. (10.7.2b) = ( ) = y Solution 1 With classical deterministic algorithm,≠ we need N 2n steps, in the order of N, to find ω, namely checking the entire database item by item. = y N n−1 Solution 2 With classical random algorithm, we need 2 2 steps, in the order of N, to find ω. = y Solution 3 With Grover’s algorithm which is a quantum algorithm, we only need the order of N steps to find ω, which is a big speed up if considering a huge database, like √N 1010 items. ( ) 10.7.2 Grover’s algorithm= n n n Let’s denote that the size of the database Z2 as N #Z2 2 . In Quantum Mechanics, it is possible to encode the entire database into one state, denoted as state s given by = = 1 s x S ⟩ N x∈{0,1}n S ⟩ ∶= √1 Q 1S ⟩ ω x N N x∈{0,1} x≠ω = √ S ⟩ + √ Q S ⟩ 1 N 1 ω ω⊥ , (10.7.3) N ¾ N − = √ S ⟩ + S ⟩ (10.7.4) where 1 ω⊥ x . (10.7.5) N 1 x≠ω

Rewrite the state s into specialS type⟩ ∶= of√ two-dimensionalQ S ⟩ Hilbert space 2 spanned by ⊥ ⊥ − ω , ω , namely, 2 Span ω , ω , where S ⟩ H S ⟩ S ⟩ H = ωS ω⟩ ⊥S ⟩0, ω⊥ ω⊥ 1. (10.7.6)

As we can show in Figure 10.1, where⟨ S ⟩ = ⟨ S ⟩ = 1 θ arcsin . (10.7.7) N ∶= √ Therefore, we can also rewrite state s in the form of

s S sin⟩ θ ω cos θ ω⊥ . (10.7.8)

S ⟩ = S ⟩ + S ⟩ 105 ω

S ⟩ s

S ⟩ θ ω⊥

Figure 10.1: State s as a state vector in the two-dimensionalS ⟩ Hilbert space 2 Span ω , ω⊥ . S ⟩ H = S ⟩ S ⟩ Question: How to get ω from s ? The quantum search problem becomes the problem of how to rotate s to ω in the two-dimensional Hilbert space 2 Span ω , ω⊥ . The hintS ⟩ is thatS ⟩ a geometric rotation can be decomposed as a product of two reﬂection operations.S ⟩ S ⟩ H = S ⟩ S ⟩ Now, we will show the way to ﬁnd ω from s in the following steps.

○ 1 The first reflection operator US ω⟩is definedS ⟩ as

U ω I 2 2 ω ω . (10.7.9)

And we can get that ∶= − S ⟩⟨ S

U ω ω I 2 2 ω ω ω ω 2 ω ω , (10.7.10) S ⟩ = ( − S ⟩⟨ S) S ⟩ ⊥ ⊥ U ω ω = S I⟩2 − 2Sω⟩ =ω− S ω⟩ ω⊥ . (10.7.11) S ⟩ = ( − S ⟩⟨ S) S ⟩ And we can show that = S ⟩

2 2 U ω I 2 2 ω ω 2 I 2 4 ω ω 2 ω ω = ( − S ⟩⟨ S) I 2. (10.7.12) = − S ⟩⟨ S + ( S ⟩⟨ S)

Thus, with U ω acting on the state= s , we can get

′ s US ω⟩ s ⊥ U ω sin θ ω cos θ ω S ⟩ ∶= S ⟩ sin θ ω cos θ ω⊥ , (10.7.13) = S ⟩ + S ⟩

which can be shown in Figure= 10.2,− and theS ⟩ + geometricS ⟩ implication of operator U ω is reﬂecting the state s around the ω⊥ axis.

○ 2 The second reﬂectionS ⟩ operator U s Sis deﬁned⟩ as

U s 2 s s I 2. (10.7.14)

∶= S ⟩⟨ S −

106 ω

S ⟩ s

S ⟩ θ ω⊥ θ S ⟩ − s′

Figure 10.2: The first reflection operator U ω in the geometricalS ⟩ picture. The state s is reflected around the axis ω⊥ . S ⟩ S ⟩ As we can see that

U s s 2 s s I 2 s 2 s s S ⟩ = ( S ⟩⟨ S − )S ⟩ s , (10.7.15) = S ⟩ − S ⟩ ⊥ ⊥ U s s = S 2⟩ s s I 2 s s⊥ , (10.7.16) S ⟩ = ( S ⟩⟨ S − )S ⟩ where = − S ⟩ s s⊥ 0 and s⊥ s⊥ 1. (10.7.17)

With the second reflection operator⟨ S ⟩ = acting on⟨ theS first⟩ = reflected state, we can get ′ ′ s1 U s s U sU ω s S ⟩ ∶= S ⟩ ⊥ 2 s s I 2 sin θ ω cos θ ω = S ⟩ 2 sin θ s s ω 2 cos θ s s ω⊥ sin θ ω cos θ ω⊥ = ( S ⟩⟨ S − ) − S ⟩ + S ⟩ 2 sin2 θ s 2 cos2 θ s sin θ ω cos θ ω⊥ = − S ⟩⟨ S ⟩ + S ⟩⟨ S ⟩ + S ⟩ − S ⟩ 2 cos 2θ s sin θ ω cos θ ω⊥ = − S ⟩ + S ⟩ + S ⟩ − S ⟩ 2 cos 2θ 1 sin θ ω 2 cos 2θ 1 cos θ ω⊥ = S ⟩ + S ⟩ − S ⟩ cos 2θ 2 cos2 θ sin θ ω cos 2θ 2 sin2 θ cos θ ω⊥ = ( + ) S ⟩ + ( − ) S ⟩ cos 2θ sin θ cos θ sin 2θ ω cos 2θ cos θ sin 2θ sin θ ω⊥ = ( + ) S ⟩ + ( − ) S ⟩ sin 3θ ω cos 3θ ω⊥ . (10.7.18) = ( + )S ⟩ + ( − )S ⟩ And this can= be shownS in⟩ + FigureS 10.3,⟩ as a kind of geometrical interpretation. The second reflection operator is defined as reflecting the state around s axis.

○ 3 Grover’s rotation is deﬁned as S ⟩

Rgrov U s U ω. (10.7.19)

Therefore, we can attain ∶= ○ ′ s1 Rgrov s . (10.7.20)

S ⟩ = S ⟩ 107 ′ ω s1

S ⟩ S ⟩ 3θ s S ⟩ θ ω⊥ θ S ⟩ − s′

′ ′ Figure 10.3: The second reﬂection from the state s toS the⟩ s1 with the second reﬂection operator U s. S ⟩ S ⟩

As we can see that the Rgrov rotates the original state s by the angle 2θ. We can deﬁne further that n S ⟩ Rgrov Rgrov Rgrov Rgrov . (10.7.21) n = ○ ○ ⋯ ○ Therefore, we can infer that ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

n ⊥ Rgrov s sin 2n 1 θ ω cos 2n 1 θ ω . (10.7.22)

This is the so-called GroverS ⟩ = iteration.( + ) S ⟩ + ( + ) S ⟩ As we shall see that, in the Grover problem, the angle θ is very small, since 1 1 sin θ , with n 1. (10.7.23) N 2n−1 = √ = ≫ Suppose that with T -iteration we can get the marked ω , therefore π π 1 2T 1 θ T S ⟩ , 2 4θ 2 1 ⎫ ⎧ 1 ( +sin)θ= ⎪ ⎪ = − ⎪ ⎪ θ , N ⎬ ⇒ ⎨ N ⎪ ⎪ =√ ⎪ ⎪ ≃√ namely ⎭⎪ ⎩⎪ Nπ T 1 O N −1~2 N. (10.7.24) √4 √ = + ( ) ∝ 10.7.3 Example: N = 4 2 fω x 0, 1 fω x 0, 1 , (10.7.25) more explicitly, it can be rewritten∶ ∀ ∈ as{ } ↦ ( ) ∈ { }

0, 0 fω 0, 0 , 0, 1 f 0, 1 , ⎧ ( ) ↦ ω( ) ⎪ (10.7.26) ⎪ 1, 0 fω 1, 0 , ⎪ ( ) ↦ ( ) ⎨ 1, 1 fω 1, 1 . ⎪ ( ) ↦ ( ) ⎪ ⎪ ( ) ↦ ( ) ⎩ 108 2 Suppose that 10 is the marked item we are looking for in the database Z2, i.e.,

S ⟩ fω 1, 0 1, fω 0, 0 f 0, 1 f 1, 1 0, (10.7.27) and the initial state encoded( ) the= database( can) = be( set) as= ( ) =

s sin θ ω cos θ ω⊥ , (10.7.28) where S ⟩ =π S ⟩ + S ⟩ θ , and ω 10 . (10.7.29) 6 With one Grover’s rotation Rgrov=, we can getS the⟩ = initialS ⟩ state rotated counterclockwise π through the angle 2θ 3 , which will lead to the angle between second reﬂected state and the state ω⊥ as π = 2θ θ 3θ , (10.7.30) S ⟩ 2 which implies that the marked state 10 +is found= = through one Rgrov operation. Check the result in algebraic approach S ⟩ 1 3 s 10 10⊥ ; 2 √2 ′ sS ⟩ = USω s⟩ + S ⟩ 1 3 S ⟩ = S10⟩ 10⊥ ; 2 √2 ′ s1 = U− sUS ω⟩s+ S ⟩ π π π π sin 2 10 cos 2 10⊥ ; S ⟩ = 6S ⟩ 6 6 6 = 10 × + S ⟩ + × + S ⟩ (10.7.31) = S ⟩ This can be shown in the Figure 10.4.

′ ω s1

S ⟩ S ⟩

π s 3

π S ⟩ 6 π ⊥ 6 ω

− S ⟩ s′

Figure 10.4: One Grover’s RotationS ⟩ with N 4

Note: The other diagrammatical approach to quantum search algorithm= is shown below.

1 s 2 00 01 10 11 (10.7.32) S ⟩ = (S ⟩ + S ⟩ + S ⟩ + S ⟩) 1 - 2 S ⟩ 00 01 10 11 r r r r = 109 ′ 1 s U ω s 2 00 01 10 11 (10.7.33) S ⟩ = S ⟩ = (S ⟩ +10S ⟩ − S ⟩ + S ⟩) 1 - 2 S ⟩ 00 01 11 r r r r ′ = s1 U sU ω s 10 (10.7.34) S ⟩ = S ⟩ = S ⟩ S - ⟩ 00 01 10 11 r r r r The “needle” is represented= for the state, and the direction of the needle is represented for the relative sign of the state.

10.7.4 Quantum circuit model of Grover’s algorithm 1○ Initial state is prepared in the way as

s H ⊗n 0 ⊗n , (10.7.35)

in quantum circuit model, which canS ⟩ = be shownS ⟩ as

0 H

s S0⟩ H (10.7.36) S ⟩ = S0⟩ H . ⋮ ⋮ 2○ As we know that S ⟩ U ω I 2 2 ω ω , (10.7.37) we can deﬁne the unitary gate U as fω = − S ⟩⟨ S

U fω x y x y fω x , (10.7.38)

with x being a n-qubit state,∶ andS ⟩Sy ⟩being↦ S a⟩S single⊕ ( qubit)⟩ state, i.e.,

S ⟩ S ⟩ n qubits x x (10.7.39) U fω single qubit y y fω x . S ⟩ S ⟩ Therefore, we can get the phase kick-backS ⟩ S ⊕ ( )⟩

0 1 fω(x) 0 1 U f x 1 x . (10.7.40) ω 2 2 S ⟩ − S ⟩ S ⟩ − S ⟩ Lemma 10.7.4.1. S ⟩ √ = (− ) S ⟩ √ 0 1 0 1 U f x U ω I 2 x , (10.7.41) ω 2 2 S ⟩ − S ⟩ S ⟩ − S ⟩ S ⟩ √ = ( ⊗ )S ⟩ √ with the function fω deﬁned as 1, if x ω, (10.7.42a) fω x 0, if x ω. (10.7.42b) = ( ) = ≠

110 This lemma can be represented in the form of Quantum Circuit diagram

0 1

S0⟩2 U ω s H ⊗n U (10.7.43) S ⟩ fω S ⟩

0⋮n ⋮ ⋮ ⋮ S ⟩−S ⟩ S ⟩−S ⟩ 0√S ⟩1 0√ 1 . 2 2

As we shall see that with ω as an unknown state, U fω is actually an oracle.

Proof. S ⟩

• In the case x ω .

S ⟩ = S ⟩ 0 1 f(ω) 0 1 U f ω 1 ω ω 2 2 S ⟩ − S ⟩ S ⟩ − S ⟩ S ⟩ ⊗ √ = (− ) 0S ⟩ ⊗1 √ ω 2 S ⟩ − S ⟩ = −S ⟩ ⊗ √0 1 U ω ω ; (10.7.44) 2 S ⟩ − S ⟩ = S ⟩ ⊗ √ • In the case x ω .

S ⟩ ≠ S ⟩ 0 1 f(x≠ω) 0 1 U f x ω 1 x ω ω 2 2 S ⟩ − S ⟩ S ⟩ − S ⟩ S ≠ ⟩ ⊗ √ = (− ) 0S ≠ 1⟩ ⊗ √ x ω 2 S ⟩ − S ⟩ = S ≠ ⟩ ⊗ √0 1 U ω x ω . (10.7.45) 2 S ⟩ − S ⟩ = S ≠ ⟩ ⊗ √

Note: How to perform Uω has to use of the black box Uf , which suggests that performing Uω takes the main part of the total performing time. 3○ The second reﬂection operator

U s 2 s s I 2, (10.7.46)

is at hand, because its realization does= notS ⟩⟨ involveS − the black-box and only includes elementary quantum gates. And we can derive that

⊗n ⊗n ⊗n ⊗n U s H 2 0 0 I 2 H H ⊗nX ⊗n 2 1 ⊗n 1 I ⊗n X ⊗nH ⊗n = S ⟩ ⟨ S − 2 H ⊗nX ⊗n I ⊗(n−1) H θ(n) I ⊗(n−1) H X ⊗nH ⊗n = H X IS 2⟩ ⟨ S −H θ I 2 H X H , (10.7.47) = − ⊗ ⊗ 111 where θ(n) is the n-qubit Toﬀoli gate, which can be written as

(n) ⊗(n−1) ⊗(n−1) ⊗(n−1) θ I 2 1 1 I 2 1 1 XX. (10.7.48)

Therefore, we can= verify the− S ⟩ E.Q.(10.7.47)⟨ S ⊗ in+ theS ⟩ following⟨ S manner,⊗

⊗(n−1) (n) ⊗(n−1) I 2 H θ I 2 H

⊗(n−1)⊗ ⊗(n−1) ⊗ ⊗(n−1) ⊗(n−1) ⊗(n−1) I 2 H I 2 1 1 I 2 1 1 X I 2 H

= ⊗(n−1)⊗ ⊗ ( n−1) − S ⟩ ⊗⟨(n−S1)⊗ + S ⟩ ⟨ S ⊗ ⊗ I 2 1 1 I 2 1 1 Z ⊗n ⊗(n−1) = I2 1 − S ⟩ 1 ⟨ S2⊗1 +1 S ⟩ ⟨ S ⊗ I ⊗n 2 1 ⊗n 1 . (10.7.49) = 2 + S ⟩ ⟨ S ⊗ − S ⟩⟨ S

Now, we= can make− S use⟩ of⟨ theS above components to construct the quantum circuit model for Grover’s algorithm.

Step 1 The quantum circuit should be

______L 0 1 ______⋯ S0⟩2 H ⊗n ˜ ˜ ˜ (10.7.50) S ⟩ Rgrov Rgrov ⋯ Rgrov

0⋮n ⋮ ⋮ ⋮ ⋮ S ⟩−S ⟩ S ⟩−S ⟩ 0√S ⟩1 ⋯ 0√ 1 . 2 2 ˜ ⋯ Note that Rgrov U s I 2 U fω.

′ Step 2 Suppose that the∶= result( ⊗ we) get in Step 1 is state ω , then we can use the oracle

U fω to check if it is the state ω , i.e., S ⟩ S ⟩ ω′ (10.7.51) U fω S0⟩S−S1⟩ S0⟩−S1⟩ √ √ . 2 2 With the measurement, we can do the check. If we ﬁnd out the result is wrong, then we should start from Step 1 again. If it is the rightful state, namely, the measurement results are 1, then the work is done.

112 Chapter 11

Quantum Circuit Complexity

11.1 Circuit complexity

11.1.1 Definitions Def 11.1.1 (Classical circuit model). A classical circuit model is finite sequence of elementary gates applied to a specified string of input bits. Def 11.1.2 (Circuit Complexity). Circuit complexity is a quantity which quantifies how many resources are exploited in computation, such as the number of elementary gates. Thm 11.1.1.1 (Characterization of Circuit Complexity). Circuit complexity is charac- terized by the size, depth, and width of the smallest circuit to compute a given problem. Def 11.1.3 (Size, Depth, Width). The Size, depth, and width of the circuit can be defined as below: 1. Size: the number of elementary gates used to solve a problem; 2. Depth: the number of time steps to solve a problem; 3. Width: the maximum number of gates that act in any one time step. Example: Construction of the GHZ state is shown in Figure 11.1. There we can get the

x1 H x S 2⟩ ● ● ● x GHZ S 3⟩

Sxn⟩ S ⟩ ⋮ ⋱ ⋮ FigureS ⟩ 11.1: Construction of the GHZ state complexity of this circuit: 1○ Size: there are one Hamard gate H, and n 1 CNOT gates;

○ 2 Depth: there are totally n time steps,which( − is a) polynomial of n and that is a good thing; 3○ Width: since there is only one gate is exploited for every time step, then the depth should be 1.

113 11.1.2 Complexity class We can classify the circuit into different categories according to the size the circuit: size poly n excellent, size Exp n the worst case. ∝ ( ) Def 11.1.4 (Decision problem) . A decision problem is a “yes” or “no” problem. ∝ ( ) Def 11.1.5 (P-problem). P-problem decision problems which can be solved by polynomial size circuit family . Def 11.1.6 (NP-problem). = NP-problem decision problems which can be verified by polynomial size circuit family . Note: verifying a problem is very different from solving a problem. For example, it would = be very difficult to work out a password, but with a password in hand, it would be very easy to verify if it works. Question: What’s the relation between N-problem and NP-problem? It is widely believed, but not solved, that these two categories of problems is not equivalent to each other. But, how can we prove or falsify that kind of consensus? Thm 11.1.2.1 (Cock’s theorem(1971)). Every problem in NP is polynomially reducible to NP-Completeness (NPC) problem. Some relations between the different types of problems expressed in diagram formalism shown in Figure 11.2

Decision problems Decision problems

NPI NPC NP P CO-NP NPC NP P

(a) (b)

Figure 11.2: Decision problem, NP, NPC, NPI and P

To be clariﬁed in the future (2015).

11.1.3 Quantum complexity There are two kinds of classical circuit, namely the deterministic circuits and the probabilistic circuit with probabilistic gates: deterministic circuits, Classical Circuit probabilistic circuits with probabilistic gates, AND with probability ε, Probabilistic gates OR with probability 1 ε, e.g. simulation Monto Carlo Annealing. Classical circuit model is described by Bounded- error Probabilistic Polynomial size (BPP) problem. Quantum circuit,− on the other hand, is described by Bounded-error Quantum Probabilistic Polynomial size (BQP) problem.

114 11.1.4 Accuracy For the exact Quantum Computer, we need inﬁnite number of quantum gates. But, that is impossible to realize in practice. For practical consideration, we would turn to Finite Quantum Approximately Universal Gate set, with which we can get the result in arbitrary accuracy. So the precision is an important topics.

115 Chapter 12

Quantum Simulation

116 Part III

Density Matrix and Quantum Entanglement

117 Chapter 13

Quantum Mechanics (II): Density Matrix

References: [Preskill] Chapter 2: Foundations I: states and ensembles;

● [Nielsen & Chuang] Chapter 2: Introduction to quantum mechanics. ● 13.1 Density matrix as state of quantum open system

Density matrix (operator) describes the state of a quantum open system. It can be introduced in both physical and mathematical approaches. Usually, the quantum computer is an open subsystem, and with the environment by which suggests lots of noises, and they form a closed system. In the mathematical sense, we can view the projective measurement theory in terms of the density operator,

ψ P n ψ ψ P n ak ak ψ ak

⟨ S S ⟩ = Q ⟨akS ψ S ψ⟩⟨P n Sak⟩ ak = trQ ⟨ψ Sψ⟩⟨P nS S ⟩

tr ρPn , (13.1.1) = (S ⟩⟨ S ) with the density operator deﬁned as = ( ) ρ ψ ψ , (13.1.2) and all ak aj A aj aj aj , aj H∶= S ⟩⟨beingS normalized. Therefore, S ⟩ ∈ S ⟩ T S ⟩ = S ⟩A S ⟩ ∈ ψA ψ an ψ P n ψ ⟨ ⟩ = ⟨n S S ⟩ = Q antr⟨ ρPS n S ⟩ n = Qtr ρA (. ) (13.1.3)

We can claim that the probability to obtain= ( the) outcome an when measuring A is actually

Prob an tr ρP n . (13.1.4)

( )118= ( ) In the physical sense, for the quantum closed system with the initial state ψ ,

ψ cn n , S ⟩ n S ⟩ = Q S ⟩ where n is an orthonormal basis for the Hilbert space H . The probability to get the post-measurement states n is S ⟩ n ψ 2 S ⟩ Prob n . (13.1.5) ψ ψ 2 S⟨ S ⟩S (S ⟩) = Question: What is the post-measurement stateS⟨ ofS an⟩S open system?

ρ pn n n , with pn 1 and pn 0. (13.1.6) n n The ρ is the so-called density∶= Q matrix,S ⟩⟨ S which willQ be deﬁned= later.≥ And the density operator ρ is equivalent to state ensemble n , pn , in the Quantum Mechanical sense.

Remark. In the description of theS ⟩ density matrix, the ambiguity of the global phase factor with non-physical meaning is removed,

ψ eiα ψ , (13.1.7)

iα −iα ρ ψ ψ ,S ρ⟩α≅ e S ψ⟩ ψ e ρ. (13.1.8)

13.1.1 State ensemble= formalismS ⟩⟨ S of= densityS ⟩⟨ S matrix=

Def 13.1.1. The state ensemble is a set of state ψi each attached with a probability pi, namely S ⟩ n ψi , pi i 1, . . ., n , with pi 0, 1 and pi 1. (13.1.9) i=1

It means that the physical S ⟩ systemT = has the probability∈ [ of p]i to beQ prepared= in the state ψi .

For any observable A, the expectation value of A is S ⟩ n A pi ψi A ψi i=1 ⟨ ⟩ = dimQ H⟨ n S S ⟩ pi ψi k k A ψi k=1 i=1 = dimQH Qn ⟨ S ⟩⟨ S S ⟩ k A ψi pi ψi k k=1 i=1 = dimQH Q ⟨ S n S ⟩ ⟨ S ⟩ k A ψi pi ψi k k=1 i=1 = dimQH ⟨ S Q S ⟩ ⟨ S S ⟩ k Aρ k k=1 = trQAρ⟨ ,S S ⟩ n with ρ i=1 pi ψi ψi , = ( ) n A p ψ A ψ tr Aρ , (13.1.10) = ∑ S ⟩⟨ S i i i i=1 where k k 1, . . ., dim H ⟨ is⟩ = anQ orthonormal⟨ S S ⟩ basis= ( for) the Hilbert space H .

S ⟩ T = 119 Def 13.1.2. The density matrix for the system described by the state ensemble, as shown in E.Q. (13.1.9), is deﬁned as n ρ pi ψi ψi , (13.1.11) i=1 • if n 1, this represents a pure state,∶= Q andS the⟩⟨ systemS is completely speciﬁed;

• if n = 2, this represents a mixed state, and the system is partially speciﬁed.

Thm 13.1.1.1.≥ A suﬃcient and necessary condition to tell whether the density operator ρ represents a pure state or a mixed state is

pure state ρ2 ρ, trρ2 1; (13.1.12) mixed state ρ2 ρ, trρ2 1. ⇐⇒ = = Proof. As we shall show later that the density⇐⇒ operators≠ are< non-negative and self-adjoint with unit trace. It is always possible to use a set of orthonormal eigenvectors of the density operator ρ as the basis to expand the whole Hilbert space H . In that way, the density matrix ρ is diagonalized, with its eigenvalues λi lying along the diagonal position of the matrix: dim H ρ λi λi λi . (13.1.13) i=1 There we can also get ρ: = Q S ⟩⟨ S

dim H 2 ρ λiλj λi λi λj λj i,j=1 = dimQH S ⟩⟨ S ⟩⟨ S λiλj λi λj δij i,j=1 = dimQH S ⟩⟨ S 2 λi λi λi . (13.1.14) i=1

= Q S ⟩⟨ S 2 dim H dim H 2 2 trρ λi λi 1. (13.1.15) i=1 i=1 Therefore, the suﬃcient and necessary= Q condition≤ Q that ρ is= equal to ρ2 should be

2 λi λi , with i 1, . . ., dim H . (13.1.16)

With ρ being self-adjoint and non-negative= with= unit trace, namely

dim H λi 1, with λi R and λi 0, i 1, . . ., dim H , (13.1.17) i=1 it then can beQ inferred= that E.Q. (13.1.16)∈ is equivalent≥ = to that there should be only one λj being 1 while all other λi with i j are vanishing. And it also says that the same thing as E.Q. (13.1.12), since with only one eigenvalue for the density operator, it is suﬃcient and necessary to get the pure state.≠

Thm 13.1.1.2. The density operator ρ should be subject to the following time-evolution equation ∂ ih ρ t H, ρ t . (13.1.18) ∂t ̵ ( ) = ( ) 120 Notice: We can see that E.Q. (13.1.18) is very similar to the Heisenberg equation, but it’s not a Heisenberg equation at all. This can be shown in the following way.

Proof. To get the time evolution of the physical system, we may start from the state ensemble as given in E.Q. (13.1.9). Every state vector ψi within the state ensemble should follow the Shr¨odingerequation S ⟩ ∂ ih ψ t H ψ t , ∂t i i or we can get the state ensemble at̵ timeS (t )⟩ = S ( ⟩

ψi t , pi i 1, . . ., n , (13.1.19) with S ( )⟩ T = ψi t U t ψi , (13.1.20) where U t is deﬁned in E.Q. (3.1.17).S ( )⟩ Therefore,∶= ( )S the⟩ density operator at time t should be ( ) n ρ t pi ψi t ψi t . (13.1.21) i=1 Now, we can derive the time-evolution( ) = Q equationS ( )⟩ for ⟨ ρ(: )S

∂ n ∂ ∂ ih ρ t pi ih ψi t ψi t ψi t ih ψi t ∂t i=1 ∂t ∂t ̵ ( ) = Qn ̵ S ( )⟩ ⟨ ( )S + S ( )⟩ ̵ ⟨ ( )S pi H ψi t ψi t ψi t ψi t H i=1 = Qρ, H S ( )⟩ ⟨ ( )S + S ( )⟩ ⟨ ( )S (− ) which is equivalent to E.Q.= (13.1.18).[ ]

13.1.2 Operator formalism of density matrix Now, we come to discuss the properties of the density matrix.

Def 13.1.3. The density matrix ρ is a non-negative self-adjoint operator with unit trace, i.e., ρ† ρ self-adjoint operator, ρ 0 non-negative, (13.1.22) ⎧ = ⎪ trρ 1 unit trace. ⎨ ≥ Notice: ρ being non-negative⎪ means that all its eigenvalues are non-negative, which is ⎩⎪ = also equivalent to that given any state its expectation value should be non-negative.

Thm 13.1.2.1. The operator formalism of the density matrix ρ is equivalent to its state ensemble description.

Proof. To derive the three properties of the density operator ρ, we may start from the deﬁnition (13.1.11), for it not only represents the mixed state case, but also the pure state case.

121 • Self-adjoint n † † ρ pi ψi ψi (13.1.23) i=1 n = Q ∗ S ⟩⟨ S pi ψi ψi (13.1.24) i=1 = Qn S ⟩⟨ S pi ψi ψi (13.1.25) i=1 = ρQ, S ⟩⟨ S (13.1.26) since p for i 1, . . ., n. i R = • Non-negative∈ = n φ ρ φ pi φ ψi ψi φ i=1 n ⟨ S S ⟩ = Q ⟨ S ⟩⟨2 S ⟩ pi ψi φ , (13.1.27) i=1

where φ is an arbitrary vector in Hilbert= Q spaceS⟨ SH⟩S. Given pi 0 for all i 1, . . ., n, it’s obvious that φ ρ φ is non-negative, namely ρ is non-negative. S ⟩ ≥ = • Unit trace ⟨ S S ⟩ n trρ tr pi ψi ψi (13.1.28) i=1 = n Q S ⟩⟨ S pitr ψi ψi (13.1.29) i=1 = Qn (S ⟩⟨ S) pi (13.1.30) i=1 = 1Q, (13.1.31) because p with i 1, . . ., n would add up to 1 and all ψ are assumed to be nor- i = i malized. = S ⟩ 13.1.3 Reduced density matrix (State for subsystem) Let’s assume that the state of the composite physical system C consisted of subsystem A and subsystem B, is in state ψ C. Then, we can get the density matrix for the composite physical system C, S ⟩ ρAB ψ AB ψ . (13.1.32) That represents a pure state, as we consider the composite physical system as a whole. = S ⟩ ⟨ S But, now we examine only the subsystem A (Alice) of the composite system. We can get the observable M A for the subsystem of Alice, expressed in Hilbert space HAB of the composite system C as M A IB. Therefore, the expectation value of the observable M is then ⊗ A M A AB ψ M A IB ψ AB

⟨ ⟩ = tr ⟨MS A I⊗B ρABS ⟩ tr M ρ , (13.1.33) = A A⊗A

= 122( ) where ρA trBρAB. (13.1.34) And we have used the facts that = tr AB trA trB AB , (13.1.35) tr M A AB trA M AtrB AB , O = ( O ) which can be proved simply by expressing all the operators in the form of kets and bras. ( O ) = ( O ) The reduced density matrix ρA can describe the subsystem A, given the density operator of the compound physical system is ρAB, in the following sense. If A is an operator corresponding to an arbitrary observable deﬁned in the subsystem A, namely A B HA , then we can express it in the total Hilbert space HAB HA HB asO O ∈ ( ) I . (13.1.36) A B ∶= ⊗ Therefore, the expectation value of the observable should be O ⊗ A tr ρ A AB A OAB tr tr ρ ⟨O ⟩ = A OB A AB tr tr ρ = A A(OB AB ) tr ρ . (13.1.37) = AOA A On the other hand, it really makes a density operator because = O • self-adjoint: † † ρA trBρAB tr ρ† = ( B AB ) trBρAB = ρA; (13.1.38) = • unit trace: =

trAρA trA trBρAB

trABρAB = ( ) 1; (13.1.39) = • nonnegative: =

A ψ ρA ψ A trA ρA ψ A ψ trAB ρAB ψ ψ I B ⟨ S S ⟩ = ( S ⟩ ⟨ AS) namely = [ (S ⟩ ⟨ S ⊗ )] A ψ ρA ψ A 0, ψ A HA (13.1.40) since the probability for the compound system ρ found in state ψ ψ I should ⟨ S S ⟩ ≥ ∀ABS ⟩ ∈ A B be nonnegative. S ⟩ ⟨ S ⊗ E.g. Reduced density matrix ρA of the general bipartite system

ψ AB aiµ i A µ B, (13.1.41) i µ S ⟩ = Q Q S ⟩ ⊗ S ⟩ ∗ ρA trBρAB µ ρAB µ B aiµaiµ i A j , (13.1.42) µ µ i j where the partial trace= ≡ Q⟨ S S ⟩ = Q Q Q S ⟩ ⟨ S trB µ B ν ν µ B δµν. (13.1.43)

(S ⟩ ⟨ S) = ⟨ S ⟩ = 123 13.2 Mixed state formalism of a qubit

Thm 13.2.0.1. Density matrix in the two-dimension Hilbert space H2, can be represented as 1 ρ p I p σ , (13.2.1) 2 2 3 with p R satisfying p 1. And( we⃗) ∶= give the+ ⃗ vector⋅⃗ p the name of the “polarization vector”. ⃗ ∈ Y⃗Y ≤ ⃗ Proof. Firstly, we would prove that any density matrix in the two-dimensional Hilbert space H2 can be expressed in the form as E.Q. (13.2.1). Secondly, we would show that with the constraint p 1, E.Q. (13.2.1) is a density matrix.

a) Any density matrixY⃗Y ≤ in the 2-dimensional Hilbert space H2 can be represented in the form of E.Q. (13.2.1). The most general 2-dimensional matrix would have the form

a b ρ , (13.2.2) c d = which has 2 4 8 degrees of freedom. But to make a density matrix for a physical system, the matrix ρ must satisfy three constraints. × = (i) Self-adjoint: A density matrix must be self-adjoint, i.e., ρ† ρ, which means a∗ a = ∗ ∗ ∗ a c a b ⎧d d ∗ ∗ ⎪ = , (13.2.3) b d c d ⎪b∗ c ⎪ = = ⇐⇒ ⎨c∗ b ⎪ = which actually makes four constraints for ρ, that⎪ would reduce the degrees of ⎩⎪ = freedom from eight to four. (ii) Unit trace: The trace of a density matrix must be equal to one, i.e.,

trρ 1 a d 1, (13.2.4)

which makes a ﬁfth constraint= to ρ.⇐⇒ The degrees+ = of freedom has come down to three now. (iii) Positive: The density matrix must have all its eigenvalues to be non-negative. This would appear as inequalities, namely “holonomic constraints” in the language of Classical Mechanics, which would never reduce a freedom but could absolutely set some kind of range for the parameters.

On the other hand, we know that, the density matrices for the Hilbert space H2 can actually be viewed as a vector space B H2 . The three Pauli matrices plus the identity matrix can be the vector basis for B H2 , because they are linearly independent to each other. Therefore, it would( be) clear that any density matrix can be expressed in the form of E.Q. (13.2.1). ( )

124 b) Now, we set to prove that iﬀ p 1, E.Q.(13.2.1) would be a density matrix. From E.Q.(13.2.1) we get Y⃗Y ≤ 1 1 p3 p1 ip2 ρ with p1, p2, p3 R. (13.2.5) 2 p1 ip2 1 p3 + − = ∈ This deﬁnition implies apparently+ that− ρ is self-adjoint. And we can immediately get the trace and the determinant of ρ: 1 trρ 1 p 1 p 1, (13.2.6) 2 3 3 1 = ( + ) + ( − ) = 1 detρ 1 p 1 p p ip p ip 1 p 2 . (13.2.7) 4 3 3 1 2 1 2 4 Let’s denote= the two( + eigenvalues)( − ) − of(ρ with− λ)(1 and+ λ2), then= ( − Y⃗Y )

trρ λ1 λ2 1, 1 2 ⎧ detρ λ1λ2 1 p , ⎪ = + = 4 ⎨ ⎪ ρ from which we can infer⎩⎪ that ρ being= nonnegative= ( is− equivalentY⃗Y ) to λ 0 1 p 1. (13.2.8) λ2 0 ≥ ¡ ⇐⇒ Y⃗Y ≤ Therefore, ρ deﬁned in E.Q. (13.2.1)≥ is a density matrix iff p 1.

c) We can now claim that, any density operator for two-levelY⃗ systemY ≤ attached with Hilbert space H2, can be written as E.Q. (13.2.1).

13.2.1 Why polarization vector? We ﬁrstly calculate the expectation value σ n , given the physical system is in the state described by the density operator ρ deﬁned in E.Q. (13.2.1), ⟨⃗⋅⃗⟩ σ n tr σ n ρ 1 ⃗ tr ⃗ σ n 1 p σ ⟨ ⋅⃗⟩ = (2 ⋅⃗) 1 = tr(σ⃗⋅n⃗)( tr+ ⃗⋅σ⃗)n p σ . (13.2.9) 2 Amid, we can get = (⃗⋅⃗) + (⃗⋅⃗)(⃗⋅⃗)

tr σ n tr n1σ1 n2σ2 n3σ3

n1trσ1 n2trσ2 n3trσ3 (⃗⋅⃗) = ( + + ) 0, (13.2.10) = + + where we have used the fact that =

trσi 0, with i 1, 2, 3. (13.2.11)

= =

125 And, we can also get

3 tr σ n p σ tr nipjσiσj = ⎛i,j 1 ⎞ (⃗⋅⃗)(⃗⋅⃗) = 3 Q ⎝ nipjtr σiσj ⎠ i,j=1 = Q3 ( ) 2 nipjδij i,j=1 = 2nQp. (13.2.12)

In the above derivation we have used the relation= ⃗⋅⃗

tr σiσj 2δij, (13.2.13) which can be derived in the following manner( ) =

2 trσi trI 2 2 3 3 tr σiσj 2δij, i, j 1, 2, 3. i j tr σiσj tr i k=1 εijkσk i k=1 εijktrσk 0 = = ¡ ⇒ ( ) = = Now,≠ by∶ substituting( ) = ( E.Q.∑ (13.2.10)) = and∑ E.Q. (13.2.12)= into E.Q. (13.2.9), we can conclude that σ n n p, which is equivalent to ⟨⃗⋅⃗⟩ = ⃗⋅⃗ Thm 13.2.1.1. tr σ n ρ p n p, (13.2.14) 3 with n, p R and n 1. (⃗⋅⃗) (⃗) = ⃗⋅⃗ With⃗ this⃗ ∈ theorem,Y⃗ weY = can understand why we call p the “polarization vector”:

(i) σ n tr σ n ρ p 0, for p n; ⃗

(ii) ⟨σ⃗⋅n⃗⟩ = tr(σ⃗⋅n⃗)ρ(p⃗) = 1, for p⃗ n⃗;

(iii) ⟨σ⃗⋅n⃗⟩ = tr(σ⃗⋅n⃗)ρ(p⃗) = 1, for⃗p= ⃗ n;

3 (iv) ⟨σ⃗⋅n⃗⟩ = tr(σ⃗⋅n⃗)ρ(p⃗) = −0, for p ⃗ =0,−⃗n R .

We can⟨⃗⋅ see⃗⟩ = that(p⃗⋅⃗)0 implies(⃗) = no polarization⃗ = ∀⃗ ∈ orientation, which shows “no information”.

⃗ = 13.2.2 Pure state and mixed state in two-dimensional Hilbert space H2

Thm 13.2.2.1. If the state in the two-dimensional Hilber space H2, which means that the state has the density matrix deﬁned in E.Q. (13.2.1). Then the state of the physical system is a pure state, if p is on the Bloch sphere, p 1, a mixed state, if p is inside the Bloch sphere, p 1. ⃗ Y⃗Y = ⃗ Y⃗Y <

126 We can show that 1 1 ρ2 p 1 2p σ p σ 2 1 2p σ p2 , (13.2.15) 4 4 because (⃗) = + ⃗⋅⃗ + (⃗⋅⃗) = + ⃗⋅⃗ + ⃗ 3 3 2 2 2 p σ pi σi pipj σi,σσj i=1 i,j=1 i>j (⃗⋅⃗) = Q + Q 3 2 pi 0 i=1 = Qp 2. + (13.2.16)

Therefore, = Y⃗Y ρ2 p ρ p , if p is on the Bloch sphere, p 1; (13.2.17) ρ2 p ρ p , if p is inside the Bloch sphere, p 1. (⃗) = (⃗) ⃗ Y⃗Y = Thus ρ2 p ρ(⃗p) or≠ p (⃗)1 is suﬃcient⃗ to claim that ρ p is aY⃗ mixedY < state. On the other hand, we can get some examples of case p 1: (⃗) ≠ (⃗) Y⃗Y < (⃗) 1 0 p e , ρ e 0 0 Y⃗Y = ; • z z 0 0 z z

⃗ = ⃗ (⃗ ) = 1 1= S ⟩⟨ S = S↑ ⟩⟨↑ S p e , ρ e 1 1 0 1 0 1 ; • x x 2 1 1 2 x x

⃗ = ⃗ (⃗ ) = 1 i = S ⟩ + S ⟩ ⟨ S + ⟨ S = S+⟩⟨+S = S↑ ⟩⟨↑ S p e , ρ e 1 1 0 i 1 0 i 1 . • y y 2 i 1 2 y y − In general,⃗ = ⃗ if p(⃗ )1,= we can express = pS ⟩as+ pS ⟩ sin⟨θScos− ⟨ϕ,Ssin =θS↑sin⟩⟨ϕ,↑ cosS θ ,

1 1 cos θ sin θ cos ϕ i sin θ sin ϕ Yρ⃗Yp= ⃗ ⃗ = ( ) 2 sin θ cos ϕ i sin θ sin ϕ 1 cos θ + − (⃗) = − 1 1 cos θ +sin θe iϕ − 2 sin θeiϕ 1 cos θ + = 2 θ θ θ −iϕ cos 2 −sin 2 cos 2 e sin θ cos θ eiϕ sin2 θ ⎛ 2 2 2 ⎞ = θ ⎝ cos 2 θ θ −iϕ ⎠ θ iϕ cos 2 sin 2 e sin 2 e = ψ+ θ, ϕ ψ+ θ, ϕ

p⃗ p⃗ . (13.2.18) = S ( )⟩ ⟨ ( )S Therefore, p 1 is= a sufficientS↑ ⟩⟨↑ S condition for that ρ p is a pure state. Given that any state for a two-level physical system, such as spin-1 2 systems, which can be describedY⃗Y by= ρ p as defined in E.Q. (13.2.1), we( can⃗) conclude that now: ~ • p 1 is a sufficient(⃗) and necessary condition for that ρ p is a pure state; • Yp⃗Y = 1 is a sufficient and necessary condition for that ρ(⃗p) is a mixed state. Y⃗Y < (⃗) 127 Pure state vs. Mixed state in 2-dimensional Hilbert space

pure state mixed state

ψ √1 0 1 2 x

S ⟩ = σx S x⟩ + S ⟩x = S↑ ⟩ 1 ρ 2 0 0 1 1 ρ ψ S↑ψ⟩ = S↑x ⟩ x = S ⟩⟨ S + S ⟩⟨ S ρ 1 ρ 2 0 =0 S ⟩⟨0 S1= S↑1⟩⟨↑0 S 1 1

= S ⟩⟨MeasurementS + S ⟩⟨ S + S M⟩⟨x S + Sx⟩⟨ Sx ,NN x x x M N M 1 N 1 M x 1, N x 0= S↑ ⟩⟨↑ S M=xS↓ ⟩⟨2↓, SN x 2 ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = ⟨ ⟩ = 13.3 Convexity of density matrix

Thm 13.3.0.2 (Convexity of density matrix). The density matrices form a convex set, and the pure states are the external points of this convex set.

This is transparent for the two-dimensional Hilbert space H2. As we can show that by using the concept of the Bloch ball, in H2, the most general density matrix can be expressed as 1 ρ p I p σ , with p 1. (13.3.1) 2 2 • If p 1, then ρ p represents(⃗) ∶= a+ pure⃗⋅⃗ state. p Y⃗Y1≤ also means that the state can be represented as one point on the Bloch sphere. So, the pure states is the external pointsY⃗Y = of the Bloch(⃗) ball, which is a convext set.Y⃗Y =

• If p 1, then ρ p represents a mixed state. p 1 also means that the state can be represented as one point inside the Bloch ball. Y⃗Y < (⃗) Y⃗Y < • For any vector p with the constraint p 1, we can express it in the form of

⃗ p n2Y⃗Yλ≤ n1 n2 , (13.3.2)

3 with 0 λ 1 and n1 n2 ⃗ =1,⃗n1+, n2(⃗ R− .⃗ Therefore,) we can get the density matrix ρ p expressed as ≤ ≤ Y⃗ Y = Y⃗ Y = ⃗ ⃗ ∈ 1 (⃗) ρ p I n λ n n σ 2 2 2 1 2 1 (⃗) = λ I+ ⃗n+σ (⃗ −1 ⃗ λ)⋅⃗I n σ 2 2 1 2 2 λ 1 λ = I n+ ⃗σ ⋅⃗ + −I )( n +σ⃗ ,⋅⃗ 2 2 1 2 2 2 − i.e., = + ⃗ ⋅⃗ + + ⃗ ⋅⃗ ρ p λρρ n1 1 λ ρ n2 , (13.3.3) with (⃗) = (⃗ ) 1+ ( − ) (⃗ ) ρ n1 2 I 2 n1 σ , 1 ⎧ ρ n2 I 2 n2 σ . ⎪ (⃗ ) = 2 ( + ⃗ ⋅⃗) ⎨ ⎪ ⎩⎪ (⃗ ) = ( + ⃗ ⋅⃗) 128 Therefore, we can also get

ρ p λ, ψ+ n1 ; 1 λ , ψ+ n2 . (13.3.4)

Remark: The same mixed( state⃗) ⇐⇒ density S matrix(⃗ )⟩ may( have− ) diﬀerentS (⃗ )⟩ interpretations of state ensembles. We can make an easy example to demonstrate this, for instance 1 ρ I. (13.3.5) 2 The density matrix deﬁned in this form can∶= be interpreted as state ensemble consisted of an arbitrary orthonormal basis vector set for the Hilbert space H , with each state vector attached with the probability 1 dim H .

Question: Why can the same mixed state density~ matrices have diﬀerent interpretations of state ensembles? Information encoded in the density matrix of the mixed state (which may denote the subsystem of the entangled pure state), can only be partially speciﬁed. And the entanglement information (as partial information of the composite system) can not be reveled in the subsystem individually, which leads to the ambiguity of the interpretation of state ensembles. As we could see, the state is more entangling, the information is more hidden.

13.4 Two-qubit system and its subsystem

Two-qubit system and its subsystem

entire system subsystem

two-qubit H2 H2 one-qubit H2 density matrix reduced density matrix ∈ ⊗ ∈ pure state (maybe) mixed state (exactly known) (partially known or completely unkonwn)

13.4.1 Example: EPR pair (Bell state) We consider a state of the composite physical system, which is made up of two subsystems A (Alice) and B (Bob), 1 1 ψ AB 00 11 0 A 0 B 1 A 1 B . (13.4.1) 2 AB 2 And the correspondingS ⟩ ∶= √ density S ⟩ + S matrix⟩ should= √ S be⟩ ⊗ S ⟩ + S ⟩ ⊗ S ⟩ 1 ρ ψ ψ 00 00 00 11 11 00 11 11 . (13.4.2) AB AB 2 AB AB AB AB ∶= S ⟩ ⟨ S2= S ⟩ ⟨ S + S ⟩ ⟨ S + S ⟩ ⟨ S + S ⟩ ⟨ S We can check that ρAB ρAB, 2 ρ=AB ψ AB ψ ψ AB ψ ψ AB ψ ρAB, (13.4.3)

= S ⟩ ⟨ S ⟩ 129⟨ S = S ⟩ ⟨ S = amid we have use the fact that 1 ψ ψ 0 0 1 1 0 0 1 1 AB AB 2 A B A A B A B A B 1 ⟨ S ⟩ = ⟨0S ⊗0 ⟨ S +0 0⟨ S ⊗ 0⟨ S1 S ⟩ 0⊗1S ⟩ + S ⟩ ⊗ S ⟩ 2 A A B B A A B B

= ⟨A S1 ⟩0 A ⟨B S1 ⟩0 B+ ⟨A S1 ⟩1 A ⟨B S1 ⟩1 B 1 +1 0⟨ S 0⟩ 1 ⟨ S ⟩ + ⟨ S ⟩ ⟨ S ⟩ 2 1. = + + + Therefore, the entire physical= system is in pure state. Now, we consider the reduced density matrix for the subsystem A (Alice), which is a one-qubit system, ρA trBρAB. (13.4.4)

From above deﬁnition, we can get ∶= 1 ρA B i ρAB i B i=0 1 = Q 0⟨ S 0 S ⟩1 1 2 A A 1 = I S. ⟩ ⟨ S + S ⟩ ⟨ S (13.4.5) 2 2 For an arbitrary orientation, n = sin θ cos ϕ, sin θ sin ϕ, cos θ ,

n⃗σ=A( trA n σA ρA ) 1 ⟨⃗⋅⃗ ⟩ = tr(⃗n⋅⃗σ ) 2 A A = 0. (⃗⋅⃗ ) (13.4.6) Therefore, in this case the polarization vector= for the subsystem A should be a null vector 3 in R , and we can extract “no information” from this subsystem. Remark: Generally, the composite system with pure state, is exactly known. While, for the the subsystem, which is usually in mixed state, is often partially known or nothing known.

1 Def 13.4.1 (Maximally entanglement). If a bipartite ρAB is a pure state with ρA 2I A, then ρAB is called maximally entangled. = 13.4.2 Maximally entangled two-qubit pure states Thm 13.4.2.1. All Bell states are maximally entangled.

Proof. This is transparent from E.Q. (4.3.11), 1 ψ i, j 0i 1 j 1¯i AB 2 AB AB S ( )⟩ = √ S ⟩ + (− ) S ⟩

130 that

ρA trB ψ i, j AB ψ i, j B i ψ i, j ψ i, j i B B ¯i ψ i, j ψ i, j ¯i B = S ( )⟩ AB⟨ ( )S AB 1 1 0 0 1 1 = 2 ⟨ S A ( )⟩2 ⟨A ( )S ⟩ + ⟨ S ( )⟩ ⟨ ( )S ⟩ 1 = IS ⟩, ⟨ S + S ⟩ ⟨ S (13.4.7) 2 A and =

ρB trA ψ i, j AB ψ i, j B 0 ψ i, j ψ i, j 0 B B 1 ψ i, j ψ i, j 1 B = S ( )⟩ AB⟨ ( )S AB 1 1 i i ¯i ¯i = 2 ⟨ SA ( )⟩2 ⟨A ( )S ⟩ + ⟨ S ( )⟩ ⟨ ( )S ⟩ 1 = IS ⟩. ⟨ S + S⟨ S⟩ ⟨ S (13.4.8) 2 B Therefore, we can= conclude now that every Bell state ψ i, j is maximally entangled.

Remark: S ( )⟩ i j (1) Local unitary transformations, e.g. I 2 X Z , preserve the entangling property; 1 (2) With ρA ρB 2I 2, we cannot acquire⊗ any information from local measurement on any one of the two subsystems. = = (3) For product pure state, there is no entanglement. For example,

φ AB n A m B ,

has the corresponding reduced densityS ⟩ = matricesS⃗⟩ ⊗ S ⃗ ⟩

ρA n A n , ρB m B m , = S⃗⟩ ⟨⃗S with n m 1. There is no information= S ⃗ can⟩ ⟨ ⃗ beS hidden in the subsystem in this case. We may say that the more entangled a state is, the more hidden of the quantum informationS⃗S = S ⃗ is.S =

13.4.3 Monogamy of maximal entanglement Alice and Bob share a maximal entangled state (e.g. the Bell state φ+ ). And here comes a third party, Eve. Eve wants to destroy the state shared by Alice and Bob, i.e., destroy φ+ . S ⟩ + S ⟩ φ AB 0 E S separable⟩ ⊗S ⟩ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ ψ 00 e00 01 e01 ↓ ABE AB E AB E entangledS ⟩ = S ⟩ S ⟩ + S ⟩ S ⟩ ´¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¶ 10 AB e10 E 11 AB e11 E. (13.4.9)

But, Alice and Bob can avoid it by Bell+ measurementsS ⟩ S ⟩ +XS A⟩ XS B and⟩ ZA ZB and inform- ing each other their measurements results. ⊗ ⊗ 131 • After measuring ZA ZB, they get the result of 1, then

⊗ ψ ABE

entangledS ⟩ ´¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¶ ′ ψ 00 e00 11 e11 (13.4.10) ↓ ABE AB E AB E Tentangledf = S ⟩ S ⟩ + S ⟩ S ⟩ ´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¶ • After measurement of XA XB, the result is 1, then the ﬁnal state has to be

ψ′ ABE⊗ Tentangledf ´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¶ 1 ψ′′ 00 11 e . (13.4.11) ↓ ABE 2 AB AB E Tseparablef = √ (S ⟩ + S ⟩ ) ⊗ S ⟩ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ + in which the state of Eve e E has been decoupled with φ AB.

S ⟩ S ⟩

132 Chapter 14

Schmidt Decomposition, Puriﬁcation and GHJW Theorem

References:

[Preskill] Chapter 2: Foundations I: states and ensembles;

● [Nielsen & Chuang] Chapter 2: Introduction to quantum mechanics.

● 14.1 Introduction

Let’s consider a bipartite composite system with two subsystems A (Alice) and B (Bob), interacting with each other. The state of the composite system can be described by the pure state ψ AB with the corresponding density matrix ρAB ψ AB ψ . And the density matrix of the subsystem A should be ρA trBρAB trA ψ AB ψ . Before further talking aboutS ⟩ the measurement and quantum operation on the= subsystemS ⟩ ⟨ S of the composite system, in this chapter, we would introduce= some relations= betweenS ⟩ ⟨ S subsystem and composite system. And Figure 14.1 illustrates the plan for this chapter.

Schmidt decomposition

Quantum entanglement Puriﬁcation

GHJW theorem

Figure 14.1: Plan for chapter 14.

14.2 Schmidt decomposition and quantum entanglement

14.2.1 Schmidt decomposition

Thm 14.2.1.1 (Schmidt decomposition). In a bipartite system HA HB, a bipartite pure state ψ AB can be expressed in the form of ⊗ ψ p i i′ , with p 0, p 1, (14.2.1) S ⟩ AB i A B i i i i √ S ⟩ = Q S ⟩ ⊗ T f ≥ Q = 133 ′ where i A and i B are orthonormal basis for HA and HB respectively. This is the so-called Schmidt decomposition. {S ⟩ } {S ⟩ } Example: Bell state (EPR state) 1 ψ AB 00 11 2 AB S ⟩ = √1 1 S ⟩ + S ⟩ ii AB 2 i=0 = √1 Q1 S ⟩ i A i B , (14.2.2) 2 i=0 = √ Q S ⟩ ⊗ S ⟩ with p p √1 . We can also get 0 1 2 √ √ 1 = = ρA I A, 2 (14.2.3) 1 ⎧ ρB I B. ⎪ = 2 ⎨ ′ ⎪ = Remark: For ψ AB i pi i A i B⎩, we may have dim HA dim HB. But ρA and ρB would have the same√ diagonalized formalism, i.e., S ⟩ = ∑ S ⟩ ⊗ S ⟩ ≠ ρ p i i , ρ p i′ i′ . (14.2.4) A i A B i B i i = Q S ⟩ ⟨ S = Q T f a T Def 14.2.1. Schmidt coeﬃcient and Schmidt number:

1. The non-vanishing pi in E.Q. (14.2.1) are called Schmidt coeﬃcients. 2. The number of Schmidt coeﬃcients is called Schmidt number.

14.2.2 Quantum entanglement Def 14.2.2. Separable and Entangled bipartite pure states:

• if the Schmidt number is 1, ψ AB is called separable, which means that the state vector of the composite physical system can be expressed as S ⟩ ψ AB φ A χ B , (14.2.5)

which is the so-called product state;S ⟩ = S ⟩ ⊗ S ⟩

• if the Schmidt number is greater than 1, then the state ψ AB is called entangled state; S ⟩ • if the reduced density matrices for the subsystems are all identity matrices, namely

1 ρA I A, 2 (14.2.6) 1 ⎧ ρB I B, ⎪ = 2 ⎨ then ψ is called maximally entangled.⎪ (We have actually mentioned this at the AB ⎩⎪ = end of the last chapter (Ref. page 130).) S ⟩ Note: We would discuss about the entanglement deﬁnition for bipartite mixed state and multipartite mixed state later.

134 14.2.3 Proof for the theorem of the Schmidt decomposition

Proof. The key point to get the Schmidt decomposition of an arbitrary state ψ AB deﬁned as S ⟩ ψ AB biµ i A µ B , (14.2.7) i,µ S ⟩ ∶= Q S ⟩ ⊗ S ⟩ with i A and µ B being orthonormal basis for HA and HB respectively, is to get both the reduced density matrices ρA and ρB diagonalized. And therefore the corresponding density{S ⟩ matrix} of{S the⟩ } composite physical system should be

∗ ρAB ψ AB ψ biµbjν i A j µ B ν . (14.2.8) i,j,µ,ν ∶= S ⟩ ⟨ S = Q S ⟩ a T ⊗ S ⟩ ⟨ S Now, we can go on step by step. Step 1: Compute ρA by the taking the partial trace of ρAB over HB:

ρA trBρAB µ′ ρ µ′ B AB B = µ′ = Q a bT b∗ T if j µ′ µ ν µ′ iµ jν A B i,µ,j,ν,µ′ = Q ∗ S ⟩ a T ⊗ a T f a T f biµbjν i A j B ν µ B i,µ,j,ν = ∗ S ⟩ a T ⟨ S ⟩ Qbiµbjµ i A j . (14.2.9) i,µ,j = Q S ⟩ a T Step 2:

Diagonalize ρA by choosing another orthonormal basis i A , all vectors in which are all eigenvectors of ρA, such that {S ⟩ } ρA pi i A i , with pi 1 and pi 0, (14.2.10) i i = Q S ⟩ ⟨ S Q = ≥ since ρA is the density matrix for the subsystem A, which should be nonnegative and self-adjoint operator with unit trace. And i A and i A are associated with a unitary transformation. {S ⟩ } {S ⟩ } Step 3:

With the basis i A for HA and the basis µ B for HB, we can reformulate the state vector for the composite system, {S ⟩ } {S ⟩ } ψ AB aiµ i A µ B . (14.2.11) i,µ S ⟩ = Q S ⟩ ⊗ S ⟩ Now, we can choose another basis for HB to simplify the expression of ψ AB: ˜i i ψ a i j µ a µ , (14.2.12) B A AB jµA A B iµ B S ⟩ j,µ µ T f ∶= ⟨ S ⟩ = Q ⟨ S ⟩ ⊗ S ⟩ = Q S ⟩ with which we can rewrite the state vector ψ AB again, ψ i ˜i . (14.2.13) AB S A⟩ B i S ⟩ = Q S ⟩ ⊗ T f 135 Correspondingly, the density matrix ρAB should take the form of

ρ i j ˜i ˜j . (14.2.14) AB A B i,j = Q S ⟩ ⟨ S ⊗ T f a T Step 4: We can prove that the newly chosen basis ˜i is orthogonal and we also could B normalize it. To accomplish those, again we can compute ρA through partial trace calculation of ρAB over HB: T f

ρA B µ ρAB µ B µ = Q ⟨ iS j S ⟩ µ ˜i ˜j µ A B B B i,j,µ = Q i S ⟩ j⟨ S ⊗ ˜ja ˜iT f .a T f (14.2.15) A B B i,j = Q S ⟩ ⟨ S ⊗ a T f On the other hand, we know from E.Q. (14.2.10) that

ρA piδij i A j . (14.2.16) i,j = Q S ⟩ ⟨ S By comparing E.Q. (14.2.15) with E.Q. (14.2.16), we can get that

˜j ˜i p δ , (14.2.17) B B i ij

which means that the basis ˜i isa actuallyT f = orthogonal. B Now we can normalize ˜i by deﬁning BT f T f 1 i′ ˜i , (14.2.18) B B pi T f ∶= √ T f where pi 0, which means that we are discussing in the subspaces of HA and HB which ensures pi 0. ′ Now, with> the basis i for HA and i for HB we can rewrite the state vector for the third time> {S ⟩} ψ {S p⟩} i i′ , AB i A B i √ which is exactly the same asS E.Q.⟩ (14.2.1).= Q S ⟩ ⊗ T f

14.3 Example for the Schmidt decomposition

As an example of the Schmidt decomposition, here we present the solution of the exercise 2.2, Chapter 2 of John Preskill’s online lecture notes.

Problem: For the two-qubit state

1 1 3 1 3 1 Φ A B B A B B (14.3.1) 2 2 √2 2 √2 2 S ⟩ = √ S ↑⟩ S ↑⟩ + S ↓⟩ + √ S ↓⟩ S ↑⟩ + S ↓⟩ (a) Compute ρA trB Φ Φ and ρB trA Φ Φ .

= (S ⟩⟨ S) = (S ⟩⟨ S) 136 (b) Find the Schmidt decomposition of Φ .

Notation: S ⟩ (1) 0 , 1 .

(2)2S ↑⟩ 2= matrix:S ⟩ S ↓⟩ = S ⟩ 1 × N i Nij j , (14.3.2) i,j=0 where i is the row index and j is the= columnQ S ⟩ index.⟨ S

(3)4 4 matrix: 1 × M ij Mij,kl kl , (14.3.3) i,j,k,l=0 where i, j are the row indexes and= kQ, l areS the⟩ column⟨ S indexes.

(4) Product basis in Hilbert space H2 H2:

1 0 ⊗ 0 0 0 1 0 0 00 ; 01 ; 10 ; 11 . (14.3.4) ⎛ 0 ⎞ ⎛ 0 ⎞ ⎛ 1 ⎞ ⎛ 0 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ S ⟩ = ⎜ 0 ⎟ S ⟩ = ⎜ 0 ⎟ S ⟩ = ⎜ 0 ⎟ S ⟩ = ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Rewrite the state Ψ AB as

S ⟩1 1 3 1 3 1 Φ AB 0 A 0 B 1 B 1 A 0 B 1 B 2 2 √2 2 √2 2 S ⟩ = √1 S ⟩ S ⟩ + S ⟩ + √ S ⟩ S ⟩ + S ⟩ 00 AB 3 01 AB 3 10 AB 11 AB 8 √ √ = √1 S1 ⟩ + S ⟩ + S ⟩ + S ⟩ aij ij AB. (14.3.5) 8 i,j=0 = √ Q S ⟩ And in the matrix expression, we have

1 1 3 Φ AB . (14.3.6) 8 ⎛ √3 ⎞ ⎜ ⎟ ⎜ 1 ⎟ S ⟩ = √ ⎜ √ ⎟ ⎜ ⎟ Note: Ψ AB is symmetric under the exchange⎝ of system⎠ A and system B, which implies ρA ρB. S ⟩ =

137 ρAB Ψ AB Ψ 1 ∗ = S ⟩ ⟨ijS AB kl aijakl i,j,k,l=0 = Q1 S ⟩ ⟨ S ij AB kl ρ˜ij,kl i,j,k,l=0 = 1 Q S ⟩ ⟨ S 3 3 1 00 AB 00 00 AB 01 00 AB 10 00 AB 11 (14.3.7) 8 √8 √8 8 = S3 ⟩ ⟨ S + 3S ⟩ ⟨ S + 3 S ⟩ ⟨ S + 3S ⟩ ⟨ S 01 AB 00 01 AB 01 01 AB 10 01 AB 11 (14.3.8) √8 8 8 √8 + 3S ⟩ ⟨ S + 3S ⟩ ⟨ S + 3S ⟩ ⟨ S + 3S ⟩ ⟨ S 10 AB 00 10 AB 01 10 AB 10 10 AB 11 (14.3.9) √8 8 8 √8 + 1 S ⟩ ⟨ S + 3S ⟩ ⟨ S + S3 ⟩ ⟨ S + 1S ⟩ ⟨ S 11 AB 00 11 AB 01 11 AB 10 11 AB 11 . (14.3.10) 8 √8 √8 8 + S ⟩ ⟨ S + S ⟩ ⟨ S + S ⟩ ⟨ S + S ⟩ ⟨ S (14.3.11) And in the matrix expression, we have

1 1 3 ρAB 1 3 3 1 8 ⎛ √3 ⎞ ⎜ ⎟ √ √ ⎜ 1 ⎟ = ⎜ √ ⎟ ⎜ ⎟ 1 3 3 1 ⎝ ⎠ 1 3 3 3 3 √ √ . (14.3.12) 8 ⎛ √3 3 3 √3 ⎞ ⎜ ⎟ ⎜ 1 3 3 1 ⎟ = ⎜ √ √ ⎟ ⎜ √ √ ⎟ (a) ⎝ ⎠

ρA trBρAB 1 = B m ρAB m B m=0 = Q1 ⟨ 1 S S ⟩ B m ij ABρ˜ij,kl kl m B m=0 i,j,k,l=0 = Q1 1Q ⟨ S ⟩ ⟨ S ⟩ i A k ρ˜im,km m=0 i,k=0

= Qρ˜00Q,00 S ⟩ρ˜01⟨ ,01S ρ˜00,10 ρ˜01,11 ρ˜10,00 ρ˜11,01 ρ˜10,10 ρ˜11,11 + + = 1 1 3 3 3 + + 8 3 3√ 3 √1 + √ + = 1 √1 √3 √ + 2 + 2 3 1 ⎛ 2 ⎞ = 1 3 11⎝2 X, ⎠ (14.3.13) 2 √4 = + 138 where X is the Pauli matrix, 0 1 X σ . (14.3.14) x 1 0 = = ρB trAρAB 1 = A m ρAB m A m=0 = Q1 ⟨ 1 S S ⟩ A m ij ABρ˜ij,kl kl m A m=0 i,j,k,l=0 = Q1 1Q ⟨ S ⟩ ⟨ S ⟩ j B l ρ˜mj,ml m=0 i,k=0

= Qρ˜00Q,00 S ⟩ρ˜10⟨,10S ρ˜00,01 ρ˜10,11 ρ˜01,00 ρ˜11,10 ρ˜01,01 ρ˜11,11 + + = 1 1 3 3 3 + + 8 3 3√ 3 √1 + √ + = 1 √1 √3 √ + 2 + 2 3 1 ⎛ 2 ⎞ = 1 3 11⎝2 X ⎠ 2 √4 = ρA. + (14.3.15) (b) Eigenstate of Pauli gate Z=. Z 0 0 ,ZZ 1 1 . (14.3.16) Eigenstate of Pauli gate X. S ⟩ = S ⟩ S ⟩ = −S ⟩ X . (14.3.17) √1 0 1 ; 2 S±⟩ = ±S±⟩ (14.3.18) ⎧ 1, 0; ⎪ S±⟩ = (S ⟩ ± S ⟩) ⎪ I 2 . ⎨ Find the eigenstate of reduced⟨+S+⟩ = density⟨−S−⟩ = matrix⟨+Sρ−⟩ =and⟨−S+ρ⟩ =. ⎪ A B ⎩⎪ = S+⟩⟨+S + S−⟩⟨−S 1 3 ρA,B λ± , λ± , (14.3.19) 2 √4 ρA,B S±λ⟩ += A,BS±⟩ λ−= A,B± . (14.3.20) Therefore = S+⟩ ⟨+S + S−⟩ ⟨−S Ψ AB A Ψ AB A Ψ AB

A ϕ˜1 B A ϕ˜2 B, (14.3.21) S ⟩ = S+⟩ ⟨+S ⟩ + S−⟩ ⟨−S ⟩ where = S+⟩ S ⟩ + S−⟩ S ⟩ 6 2 ϕ˜1 B A Ψ AB B λ+ B, √ 4 √ + » (14.3.22) S ⟩ = ⟨+S ⟩ = 6 2S+⟩ = S+⟩ ϕ˜2 B A Ψ AB B λ− B. √ 4 √ − » Thus, the SchmidtS decomposition⟩ = ⟨−S ⟩ of= state− Ψ ABS−shows⟩ = − as S−⟩

Ψ AB λ+ A SB ⟩ λ− A B. (14.3.23) » » S ⟩ = S+⟩ S+⟩ − S−⟩ S−⟩ 139 14.4 The puriﬁcation theorem and GHJW theorem

14.4.1 Puriﬁcation

Thm 14.4.1.1 (Puriﬁcation). For any given mixed state ρA for system A, there is a bipartite pure state Φ AB such that

S ⟩ ρA trB Φ AB Φ . (14.4.1)

Then Φ AB is called the puriﬁcation of= ρA. S ⟩ ⟨ S For example, the density matrix S ⟩ 1 ρ I , A 2 A as we have shown in E.Q. (13.4.5), has the= puriﬁcation 1 ψ 00 11 . AB 2 AB AB Proof. The key point is the SchmidtS ⟩ = decomposition.√ S ⟩ + S Suppose⟩ that

ρA pi ϕi A ϕi , with pi 1, and pi 0, (14.4.2) i i where ϕi is an orthonormal∶= Q S basis⟩ ⟨ ofS HA. Q = > Now, we introduce another system B and construct the following state for the composite physical{S system⟩} HA HB, Φ p ϕ α , with p 1, (14.4.3) ⊗ AB i i A i B i i i √ S ⟩ = S ⟩ ⊗ S ⟩ = amid, αi B is the orhonormalQ basis of HB. And we can showQ that Φ AB is a pure state, which is self-convincing, and it is also normalized since {S ⟩ } S ⟩ AB Φ Φ AB pjpi A ϕj B αj ϕi A αi B i,j √ ⟨ S ⟩ = ⟨ S ⊗ ⟨ S S ⟩ ⊗ S ⟩ Q pipjA ϕj ϕi A B αj αi B i,j √ = Q pipjδij⟨δij S ⟩ ⟨ S ⟩ i,j √ = Q pi i = 1Q. We can also verify that Φ defined in E.Q. (14.4.3) is truly purification of the state ρ AB = A as defined in E.Q. (14.4.2). As we can see that S ⟩ trBρAB B αi ρAB αi B i = Q ⟨ S S ⟩ pjpkB αi ϕj A αj B A ϕk B αk αi B i,j,k √ = Q ⟨ S S ⟩ ⊗ S ⟩ ⟨ S ⊗ ⟨ S S ⟩ pjpk ϕj A ϕk B αi αj B αk αi B i,j,k √ = Q S ⟩ ⟨ S ⊗ ⟨ S ⟩ ⟨ S ⟩ pjpk ϕj A ϕk δijδki i,j,k √ = S ⟩ ⟨ S Qpi ϕi A ϕi i = ρQA, S ⟩ ⟨ S

= 140 i.e., ρA trBρAB. (14.4.4) ρ Therefore, Φ AB is a puriﬁcation of ρA. = There are someS ⟩ things that we should remark:

• ρA can be prepared as ensembles of pure states in many different ways. This can be inferred directly from the theorem of purification, since based on any state ensemble interpretation of the ρA, we can always construct the corresponding purification, which can be different from each other.

• All these puriﬁcation are experimentally indistinguishable, if one only observes the system A. Because for any measurement M A on the subsystem A, we have

M A trA M AρA .

⟨ ⟩ = • Diﬀerent puriﬁcations are associated with a local unitary transformation U B on the subsystem B, which we will prove in the next subsection and introduce the GHJW theorem.

14.4.2 The GHJW theorem

Thm 14.4.2.1 (GHJW). Consider many ensembles that realising ρA. There is a Hilbert space HB and a bipartite pure state Φ AB, by which any one of these ensembles can be realized measuring a suitable observable of B. S ⟩ Proof. Firstly, we would show that two arbitrary purification of ρA, e.g. Φ 1 and Φ 2, are connected with a local unitary transformation on the subsystem B. Let’s denote that S ⟩ S ⟩ Φ1 i pi ϕi αi , A A B (14.4.5) ⎧ Φ2 i √qi ψi βi , ⎪ S ⟩A ∶= ∑ S ⟩A ⊗ S ⟩B ⎨ √ ⎪ ρ where pi, ϕi A and ⎩⎪qi,S ψi⟩ A ∶=are∑ two stateS ⟩ ensemble⊗ S ⟩ interpretations of ρA and αi B and βi B are two orthonormal basis for HB. Assume{( thatS ρA⟩ has)} the following{( S ⟩ eigenvectors)} and the associated eigenvalues {S ⟩ } {S ⟩ } ρA k A λk k A , with λk 1, and λk 0. (14.4.6) k S ⟩ = S ⟩ Q = ≥ Since Φ1 AB and Φ2 AB are two different purification of ρA, it’s apparent that

S ⟩ S ⟩ trA Φ1 AB trA Φ2 AB ρA. (14.4.7)

With E.Q. (14.4.7) in mind, we canS ⟩ see= that theS ⟩ respective= Schmidt decompositions for Φ1 AB and Φ2 AB are

′ S ⟩ S ⟩ Φ1 k λk k k1 , AB B (14.4.8) ′ ⎧ Φ2 k √λk k k . ⎪ S ⟩AB = ∑ S ⟩ ⊗ S 2⟩B ⎨ √ ⎪ Therefore, Φ1 AB and Φ2 AB⎩⎪ Sare⟩ connected= ∑ withS the⟩ ⊗ localS ⟩ unitary translation

S ⟩ S ⟩ Φ1 AB I A U B Φ2 AB , (14.4.9)

S ⟩ = ( 141⊗ )S ⟩ with U k′ k′ . (14.4.10) B 1 B 2 k Substitute E.Q. (14.4.5) into E.Q. (14.4.9),= Q T andf wea canT get that

Φ1 AB qi ψi A U B βi B , i √ S ⟩ = Q S ⟩ ⊗ S ⟩ namely

Φ1 AB qi ψi A γi B , (14.4.11) i √ where S ⟩ = Q S ⟩ ⊗ S ⟩

γi B U B βi B . (14.4.12)

Surely, γi B makes an orthonormalS basis⟩ ∶= for SHB⟩ , too. Secondly, let’s consider two observables B and B deﬁned in the subsystem B. And we {S ⟩ } consider that case the two vector basis αi B and γi B respectively are eigenvectors of B and B, namely M N {S ⟩ } {S ⟩ } M N B αi B Mi αi B , with i, j 1, . . ., dim HB. B γj B Nj γj B , M S ⟩ = S ⟩ = Therefore, we can concludeN S ⟩ that:= S ⟩

• After the measurement of B, there would be a probability of pi for the subsystem B in the state αi B, namely the subsystem A in the state ϕi A. Hence, measuring B in the subsystem B wouldM prepare the subsystem A in the mixed state ρA with S ⟩ S ⟩ the state ensemble pi, ϕi A . M • After the measurement{( ofS ⟩ B)}, there would be a probability of qj for the subsystem B in the state γj B, namely the subsystem A in the state of ψi A. Thus, the measurement of B in theN subsystem B would prepare the subsystem A in the S ⟩ S ⟩ mixed state ρA with the state ensemble pi, ψi A . N Therefore, we get the GHJW theorem. {( S ⟩ )}

14.5 Information is physics

Ambiguities in the concept of density matrix

(a) A density matrix has many diﬀerent interpretations of state ensemble.

(b) A density matrix can have many diﬀerent interpretations of puriﬁcation.

Here we discuss an example of bipartite system which is made up of two subsystems Alice and Bob. Alice and Bob share a pure state of the form 1 ψ 00 11 . (14.5.1) AB 2 AB AB S ⟩ = √ S ⟩ + S ⟩

142 We can easily get the reduced density matrices for Alice and Bob respectively,

ρA trBρAB 1 = B i ρAB i B i=0 1 = Q1 ⟨ S S ⟩ ′ ′ B i j A j B A j B j i B 2 i,j,j′=0 = 1 Q1 ⟨ S (S ⟩ ⊗ S ⟩ )( a T ⊗ a T)S ⟩ j i j j′ j′ i A B B A B B 2 i,j,j′=0 = 1 (S ⟩ ⊗ ⟨ S ⟩ )( a T ⊗ a T f ) 1 Q ′ j A j δijδj′i 2 i,j,j′=0 = 1 1Q S ⟩ a T i A i , 2 i=0 i.e., = Q S ⟩ ⟨ S 1 ρ I . (14.5.2) A 2 2 Reasoning in the same manner, we can get= 1 ρ I . (14.5.3) B 2 2 Now, let’s consider the following processes:= • The ﬁrst process: – Bob performs a measurement along the z-direction on his system but doesn’t tell Alice via classical communication (e.g. phone call). Therefore, no information 1 transfer between Alice and Bob. In this case, we can get ρA 2I 2, which is a mixed state. – Bob performs a measurement along the z-direction on his system= and phone

calls Alice about his measurement result (e.g. 0 B). There is information transferred between Alice and Bob. And, in this case ρA 0 A 0 , which is a pure state. S ⟩ = S ⟩ ⟨ S From this process, we can conclude that information (via classical communication) changes the physics of the system of Alice.

• The second process: – Bob measures his qubit along the z-axis, and phone calls Alice but only tells her that he has measured his qubit along the z-direction. Therefore, Alice’s system is prepared in state ensemble 1 1 , 0 , , 1 . (14.5.4) 2 2 – Bob measures his qubit along the Sx⟩-axis, andS ⟩ phone calls Alice but only tells her that he has measured his qubit along the x-direction. Therefore, Alice’s system is prepared in state ensemble 1 1 , , , . (14.5.5) 2 2 S+⟩ S−⟩ 143 The two state ensembles described by (14.5.4) and (14.5.5) respectively, are not distinguished by any conceivable measurement on the subsystem A only. The reason is that both these two state ensemble are corresponding to the same density operator 1 ρA 2I 2. =

144 Chapter 15

Mixed State Entanglement and Multi-partite Entanglement

entanglement is a fundamentally new resource in the world that goes essentially beyond classical resources; iron to the classical world’s bronze age. ⋯ —Nielsen & Chuang

A major task of quantum computation and quantum information is to be exploit this new resource (quantum entanglement) to do information processing tasks impossibility or much more diﬃcult with classical information. —Nielsen & Chuang

A lot of mileage in quantum computation, and especially quantum information, has come from asking the simple question: “what would some entanglement buy me in this problem?”. —Nielsen & Chuang

References:

[Preskill] Chapter 4: Quantum entanglement;

● [Nielsen & Chuang] Chapter 12: Quantum information theory.

● 15.1 Bipartite mixed state entanglement

15.1.1 Separability 15.1.1.1 Bipartite pure states A bipartite pure state is separable if it is a tensor product of two pure states, i.e.,

ψ AB α A β B , (15.1.1) which is equivalent to say that theS bipartite⟩ = S ⟩ pure⊗ S state⟩ has the Schmidt number as 1 is separable.

145 15.1.1.2 Bipartite mixed state

A bipartite mixed state ρAB is separable if it admits the ensemble description given by

ρAB pijρA,i ρB,j, (15.1.2) i,j = Q ⊗ where pij 1, pij 0 i,j Q = ≥ and ρA,i i are density operators for subsystem A, while ρB,j j for subsystem B. Example: This is a separable mixed state { S ∀ } { S ∀ } ρAB pi αi αi qj βj βj (15.1.3) i j = Q S ⟩⟨ S⊗Q S ⟩⟨ S where pij piqj pi 1, with pi 0, i, = i and Q = ≥ ∀ qj 1, with qj 0, j. i Q = ≥ ∀ 15.1.2 Quantum bipartite entanglement A bipartite state is entangled if it is not a separable state. Here is one separable state 1 ρ I I . (15.1.4) 1 4 2 2 And another separable state shows as = ⊗

1 1 1 ρ2 i A i j B j 4 i=0 j=0 = 1Q1S ⟩ ⟨ S⊗QS ⟩ ⟨ S ij AB ij 4 i,j=0 = 1 Q S ⟩ ⟨ S 00 00 01 01 10 10 11 11 . (15.1.5) 4 AB AB AB AB

And for the state ρ=3, isS expressed⟩ ⟨ asS + S ⟩ ⟨ S + S ⟩ ⟨ S + S ⟩ ⟨ S 1 ρ φ+ φ+ φ− φ− ψ+ ψ+ ψ− ψ− , (15.1.6) 3 4 which is hard to determine= S whether⟩⟨ S + itS is⟩⟨ separableS + S or⟩⟨ entangled.S + S ⟩⟨ S However, as we shall see that ρ1, ρ2 and ρ3 are actually same, since both four product states ij i, j 0, 1 and four Bell states ψ ij i, j 0, 1 can span the Hilbert space H2 H2, namely {S ⟩S = } ρ1 {Sρ2 ( ρ3)⟩. S = } (15.1.7) ⊗ Remarks: Quantum entanglement is a= very= important but diﬃcult issue in quantum physics. How to quantify multi-particle entanglement is an open problem up to now.

146 15.1.3 Positive-partial transpose (PPT) criterion for quantum bipartite separability

Thm 15.1.3.1. If ρAB is separable, namely be of the form (15.1.2), then the partial transpose of ρAB, deﬁned as

PT ρAB I2 T ρAB T = p⊗ijρA,i ρB,j , (15.1.8) i,j = Q ⊗ is still non-negative, where T denotes the transpose of matrix, namely

T i j j i . (15.1.9)

Proof. As we shall know that ρB,j j S ⟩⟨areS density= S ⟩⟨ S operators of the subsystem B. There- fore, for an arbitrary ρB,j, the eigenvectors can span the local Hilbert space HB. If we can diagonalize ρB,j with the{ basisS ∀ being} the complete set of independent orthonormal eigenvectors of ρB,j. After ρB,j being diagonalized, we can get that its transpose is diagonalized, too. It would obvious that, that the transpose would preserve the eigenvalues of T ρB,j. In another worlds, ρB,j should also be nonnegative, namely

T ( ) ρB,j 0, j. (15.1.10)

On the other hand, the fact that ρA(,i ) i ≥being∀ the density operators of the subsystem PT A, ensures that they are all nonnegative. Therefore, ρ is nonnegative. { S ∀ } AB PT Corollary 15.1.3.1. If ρAB is not positive, then ρAB is an entangled state.

This is a direct corollary of the theorem 15.1.3.1.

15.1.4 Example: the Werner state and the PPT criterion Werner State, firstly presented in 1989 by Reinhard F. Werner, is defined as 1 ρ λ 1 λ I λ φ+ φ+ (15.1.11) 4 4 with λ being a real number. We( can) ∶= show − that if+ρ λS with⟩⟨ definitionS (15.1.11) is a density matrix, iff ( ) • Unit trace. 1 tr ρ λ tr 1 λ I λ φ+ φ+ 4 4 1 [ ( )] = 1 λ−tr I + Sλtr⟩⟨φ+S φ+ 4 4 = 1( λ− ) λ ( ) + (S ⟩⟨ S) 1. = − + =

147 • Self-adjoint.

1 † ρ λ † 1 λ I λ φ+ φ+ 4 4 1 † [ ( )] = (1 − λ) I + S λ⟩⟨φ+ Sφ+ † 4 4 1 ∗ = 1( −λ )I† λ+∗( φS + ⟩⟨φ+ S)† 4 4 1 = 1 − λI +λ φ+(S φ+⟩⟨ S) 4 4 = ρ λ −, + S ⟩⟨ S

since λ R. = ( )

• Nonnegative.∈

ρ λ 0 the eigenvalues of ρ λ 0.

And it would be easy( to) ≥ verify⇐⇒ that the Bell’s states ψ( i,) j≥ i, j 1, 2 are four eigenstates of ρ λ . {S ( )⟩ S = } ( ρ) λ ψ 0, 0 ρ λ φ+ 1 1 λ I λ ψ− ψ− φ+ ( )S ( )⟩ = (4 )S ⟩ 4 1 = 1 −3λ φ++ . S ⟩⟨ S S ⟩ (15.1.12) 4 and with i 0 and j 0, we can= get + S ⟩

≠ρ λ ψ i,≠ j ρ λ ψ i, j 1 1 λ I λ ψ 0, 0 ψ 0, 0 ψ i, j ( )S ( )⟩ = (4 )S ( 4)⟩ 1 = 1 −λ ψ i,+ j S ,( )⟩ ⟨ ( )S S ( )⟩ (15.1.13) 4 as we know that ψ i, j ,= with i,− j S0,(1 ,)⟩ are mutually independent. Therefore, we’ve gotten the complete set of mutually independent eigenvectors of ρ λ . S ( )⟩ ( = ) ( ) Therefore, for the constrain of nonnegative:

1 1 3λ 0 4 1 ρ λ 0 λ 1. (15.1.14) ⎪⎧ 1 ( + ) ≥ 3 ⎪ 4 1 λ 0 ( ) ≥ ⇐⇒ ⎨ ⇐⇒ − ≤ ≤ ⎪ 1 ⎪ ( − ) ≥ When λ 3 ⎩ 1 1 1 = −ρ I φ+ φ+ φ− φ− ψ+ ψ+ ψ− ψ− , 3 3 4 3 therefore − = ( − S ⟩⟨ S) = (S ⟩⟨ S + S ⟩⟨ S + S ⟩⟨ S) 1 1 ρ φ+ I φ+ φ+ φ+ 0 φ+ , (15.1.15) 3 3 4 − S ⟩ = − S ⟩⟨ SS ⟩ = S ⟩

148 and 1 1 1 ρ ψ i, j I ψ 0, 0 ψ 0, 0 ψ i, j ψ i, j (15.1.16) 3 3 4 3 with i −0 and S j( 0.)⟩ = ( − S ( )⟩ ⟨ ( )S) S ( )⟩ = S ( )⟩ When λ 1 ≠ ≠ ρ 1 φ+ φ+ , = which is a pure state with the relations ( ) = S ⟩⟨ S ρ 1 φ+ φ+ φ+ φ+ φ+ . (15.1.17)

and ( )S ⟩ = S ⟩⟨ SS ⟩ = S ⟩ ρ 1 ψ i, j φ+ φ+ ψ i, j 0 ψ i, j (15.1.18) with i 0 and j 0. ( )S ( )⟩ = S ⟩⟨ SS ( )⟩ = S ( )⟩ Calculate the partial transpose of Werner state ρ λ . ≠ ≠ PT ρ λ I2 T ρ λ ( ) 1 + + I2 T 1 λ I4 λ φ φ ( ) = ⊗ (4 ) 1 = 1⊗ λ I −T I +λ SI ⟩⟨T φS+ φ+ 4 2 4 2 1 1 = (1 −λ)I ⊗ λI +T 00⊗ 00S ⟩⟨00 S 11 11 00 11 11 4 4 2 2 1 1 = 1 − λI + λ 00⊗ 00 (S 01⟩⟨ 10S + S 10⟩⟨ 01S + S 11⟩⟨ 11S + S , ⟩⟨ (15.1.19)S) 4 4 2 from which we can= see that− + (S ⟩⟨ S + S ⟩⟨ S + S ⟩⟨ S + S ⟩⟨ S) PT PT 00 11 ρ λ φ+ ρ λ 2 1 S ⟩1+ S ⟩ ( ) S ⟩ = ( 1) λ √λ φ+ 4 2 1 = 1( −λ )φ++ , S ⟩ (15.1.20) 4 PT PT 00 11 ρ λ φ− = ρ(λ+ )S ⟩ 2 1 S ⟩1− S ⟩ ( ) S ⟩ = ( 1) λ √λ φ− 4 2 1 = 1( −λ )φ+− , S ⟩ (15.1.21) 4 PT PT 01 10 ρ λ ψ+ = ρ(λ+ )S ⟩ 2 1 S ⟩1+ S ⟩ ( ) S ⟩ = ( 1) λ √λ ψ+ 4 2 1 = 1( −λ )ψ++ , S ⟩ (15.1.22) 4 PT PT 01 10 ρ λ ψ− = ρ(λ+ )S ⟩ 2 1 S ⟩1− S ⟩ ( ) S ⟩ = ( 1) λ √λ ψ− 4 2 1 = 1( −3λ) −ψ− . S ⟩ (15.1.23) 4 = ( − )S ⟩ 149 In another viewpoint, we can see 1 + + 1 I 2 T φ φ I 2 T ii jj 2 i,j=0 ( ⊗ )S ⟩⟨ S = (1 1⊗ ) Q S ⟩⟨ S I 2 T i j i j 2 i,j=0 = 1 Q1 ( ⊗ )(S ⟩⟨ S ⊗ S ⟩⟨ S) i j T i j 2 i,j=0 = 1 Q1 S ⟩⟨ S ⊗ (S ⟩⟨ S) i j j i 2 i,j=0 = 1 Q1 S ⟩⟨ S ⊗ S ⟩⟨ S ij ji 2 i,j=0 = 1 Q S ⟩⟨ S SWAP (15.1.24) 2 where the Swap gate is deﬁned in (7.2.4)= and has the matrix expression 1 0 0 0 1 0 0 1 0 SWAP ij ji . (15.1.25) i,j=0 ⎛ 0 1 0 0 ⎞ ⎜ ⎟ ≡ S ⟩⟨ S = ⎜ 0 0 0 1 ⎟ Q ⎜ ⎟ Note that the SWAP gate is identical with its inverse,⎝ namely⎠

SWAP SWAP I 4, (15.1.26) which implies the characteristic equation ○ = 2 SWAP I 4 0. (15.1.27) Thus, we can see the eigenvalue of the SWAP gate is 1. In two electrons system, i.e., two 1 − = spin- 2 system, triplet states are eigenstate of SWAP gate with eigenvalue 1, expressed as ± φ+ φ+ SWAP φ− φ− . (15.1.28) S +⟩ ⎫ S +⟩ ⎫ ψ ⎪ ψ ⎪ S ⟩ ⎬ = S ⟩ ⎬ For singlet state, it is the eigenstate of SWAP⎪ gate with⎪ eigenvalue 1, S ⟩ ⎭⎪ S ⟩ ⎭⎪ SWAP ψ− ψ− . (15.1.29) − Therefore, rewrite the SWAP gate as S ⟩ = −S ⟩ + + − − + + − − − − SWAP φ φ φ φ ψ ψ ψ ψ I 4 2 ψ ψ . (15.1.30) And the partial transpose of Werner state ρ λ can be calculated as = S ⟩⟨ S + S ⟩⟨ S + S ⟩⟨ S − S ⟩⟨ S = − S ⟩⟨ S PT ρ λ I2 T ρ(λ) 1 + + I2 T 1 λ I4 λ φ φ ( ) = ⊗ (4 ) 1 λ = 1⊗ λ I −SWAP + S ⟩⟨ S 4 4 2 1 1 = (1 − λ)I + λII λ ψ− ψ− 4 4 2 4 1 = (1 − λ)I + λ ψ− −ψ−S . ⟩⟨ S (15.1.31) 4 4 = ( + ) − S ⟩⟨ S 150 And the eigenvalues can be obtained by the way

PT 1 1 3λ ρ λ ψ− 1 λ λ ψ− ψ− , (15.1.32) 4 4 − ( ) S ⟩ = (φ+( + ) − )S ⟩ =φ+ S ⟩ PT 1 ρ λ φ− 1 λ φ− . (15.1.33) S +⟩ ⎫ 4 S +⟩ ⎫ ψ ⎪ ψ ⎪ ( ) S ⟩ ⎬ = ( + ) S ⟩ ⎬ Therefore, ρ λ has the eigenvalues 1 1 ⎪λ , 1 1 3λ , and⎪ the eigenstates S 4 ⟩ ⎭⎪ 4 S ⟩ ⎭⎪ ( ) φ+ , φ− , ψ(+ +, ) ( −ψ−) . (15.1.34)

1 (1+λ), triplet state 1 (1−3λ), singlet state 4S ⟩ S ⟩ S ⟩ 4 S ⟩ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ ± For the partial transpose criterion, we have the constrain for the parameter λ

1 1 λ 0 4 1 ρ λ PT 0 1 λ . (15.1.35) ⎪⎧ 1 ( + ) ≥ 3 ⎪ 4 1 3λ 0 [ ( )] ≥ ⇐⇒ ⎨ ⇐⇒ − ≤ ≤ ⎪ From inequality (15.1.14) and inequality⎩⎪ ( (15.1.35),− ) ≥ we can infer that 1 ρ λ 0 λ 1, 3 1 ρ λ (PT) ≥ 0 ⇐⇒ −1 ≤λ ≤ . 3 That means [ ( )] ≥ ⇐⇒ − ≤ ≤ 1 1 ρ λ is separable λ , 3 3 1 ρ (λ ) is entangled ⇐⇒ − ≤ λ ≤1. 3 ( ) ⇐⇒ < ≤ 15.1.5 Example: the Werner state and the CHSH inequality Now let’s consider the CHSH inequality for the bipartite Werner states. We may set the four observables to be A σA a1 and B σA a2 in subsystem Alice, and C σB b1 and D σB b2 in subsystem Bob. Firstly, we can calculate = ⋅⃗ = ⋅⃗ = ⋅⃗ = ⋅⃗ σA a σB b trAB σA a σB b ρ λ 1 + + a ⃗f trAB σA a σB ⃗b 1 λ I 4 λ φ φ ( ⋅⃗) ⋅ = ( ⋅⃗) ⋅ (4 ) 1 λ ⃗ + + = (tr ⋅σ⃗) a tr⋅ σ (b − λ) φ+ σS a⟩⟨ σSb φ 4 A A B B A B − + + = λ φ σA( a ⋅⃗σ)B b φ ⋅⃗ + ⟨ S( ⋅⃗) ⋅⃗ S ⟩ λa b. (15.1.36) = ⟨ S( ⋅⃗) ⋅⃗ S ⟩ Therefore, we can get = ⃗⋅⃗

AC BC AD BD λ a1 b1 a2 b1 a1 b2 a2 b2 λ a a b a a b . (15.1.37) S⟨ ⟩ + ⟨ ⟩ + ⟨ ⟩ − ⟨ ⟩S = S S T⃗ ⋅1⃗ +2⃗ ⋅⃗1 + ⃗ 1⋅⃗ −2⃗ ⋅⃗2 T ⃗ ⃗ 151= S S T(⃗ + ⃗ )⋅ + (⃗ − ⃗ )⋅ T ′ ′ ′ ′ Since a1 a2 and a1 a2 are mutually orthogonal,therefore

(⃗ + ⃗ ) (⃗ − 0⃗ ) a1 a2 b1 a1 a2 b2 2 2. (15.1.38) √ To ensure the CHSH inequality≤ T(⃗ + ⃗ )⋅⃗ + (⃗ − ⃗ )⋅⃗ T ≤

AC BC AD BD 2 (15.1.39) being correct, we only needS⟨ ⟩ + ⟨ ⟩ + ⟨ ⟩ − ⟨ ⟩S ≤

AC BC AD BD max 2, (15.1.40) i.e., S⟨ ⟩ + ⟨ ⟩ + ⟨ ⟩ − ⟨ ⟩S ≤ 1 1 2 2 λ 2, λ . (15.1.41) 2 2 √ Combine above constrain of parameterS S ≤ λ⇐⇒with the− √ inequality≤ ≤ √ (15.1.14) which ensures ρ λ nonnegative namely a density operator, and we can get ( ) 1 1 λ . (15.1.42) 3 2 As we shall see that with λ 1 3, 1 ,− the≤ operator≤ √ ρ λ is ensured to be a possible density operator, and ∈ [− ~ ] ( ) • if λ 1 3, 1 , the Werner state ρ λ is entangled;

• if λ ∈ ( ~1 3,]1 2 ,the Werner state( )ρ λ obeys the CHSH inequality; √ • if λ ∈ [1− 3~, 1 ~ 2 ,] the Werner state ρ λ( )is entangled but compatible with the CHSH inequality. √ ∈ [ ~ ~ ] ( ) Since the CHSH inequality is actually a kind of Bell’s inequality, which is the criterion of local hidden variable theorem, we can conclude that entanglement is not always contra- dicted with the local hidden variable theorem, namely the entanglement can exist in the local hidden variable theory in some case.

15.2 Multi-partite entanglement

15.2.1 Definition Def 15.2.1. We define the multi-partite separability and entanglement in the following. 1○ The n-partite state is fully separable iff

• Pure state: ψ ψ ψ ψ ; (15.2.1) A1A2⋯An A1 A2 An • Mixed state: S ⟩ = S ⟩ ⊗ S ⟩ ⊗⋯⊗ S ⟩ ρ ρi ρi ρi ρA1A2⋯An piρA1 ρA2 ρAn . (15.2.2) i=1 2○ The n-partite state is fully entangled= iﬀQ it is not⊗ fully⊗⋯⊗ seperable.

3○ The n-partite state is genuinely entangled iﬀ all bipartite partitions are entangled. Example:

152 A1

A2 ○ A3

Figure 15.1: Diagrammatical representation○ ○ of fully separable tripartite state

A1 A1 A1

A2 ○ A3 A2 ○ A3 A2 ○ A3

○ (a) ○ ○ (b) ○ ○ (c) ○ Figure 15.2: Diagrammatical representation of fullly entangled tripartite state in three cases

1○ A fully separable tripartite state is shown in Figure 15.1, where the dashed lines represent separable states and Ai denotes the i-th particle, with i 1, 2, 3.

○ 2 As shown in Figure 15.2, there are three cases of the fully entangled= tripartite state, where the solid lines represent the entangled states.

3○ See Figure 15.3 for diagrammatical representation of genuinely entangled tripartite state.

A2 ○ A3

Figure 15.3: Diagrammatical representation○ ○ of genuinely entangled tripartite state

15.2.2 The GHZ state Def 15.2.2 (n-qubit GHZ states). The state deﬁned as 1 GHZ x x¯ (15.2.3) n 2 S ±⟩ = √ S ⟩ ± S ⟩ is called n-qubit GHZ state, where x is an n-bit string of 0 and 1 with x x¯ 1 1 . binary n + = ⋯ From this deﬁnition we can see the examples as ±

1○ n 1: 1 0 1 , = 2 are eigenstates of X matrix withS±⟩ eigenvalues= √ (S ⟩ ± S 1,⟩) and form an orthogonal basis of Hilbert space H2. ±

153 2○ n 2: 1 ψ 0, 0 00 11 = 2 ⎫ S ( )⟩=√1 (S ⟩ + S ⟩) ⎪ ψ 0, 1 00 11 ⎪ 2 ⎪ ± ± ⎪ Bell states φ , ψ , S ( )⟩=√1 (S ⟩ − S ⟩) ⎪ ψ 1, 0 01 10 ⎪ 2 ⎬ ⇐⇒ {S ⟩ S ⟩} ⎪ S ( )⟩=√1 (S ⟩ + S ⟩) ⎪ ψ 1, 1 01 10 ⎪ 2 ⎪ ⎪ are eigenvectorsS ( of)⟩= the√ parity-bit(S ⟩ − S operator⟩) ⎪ Z Z and the phase-bit operator X X, ⎭⎪ and form an orthogonal basis of Hilbert space H2 H2. ⊗ ⊗ 3○ n 3: 1 1⊗ Φ1 000 111 , Φ8 000 111 , = 2 2 ⎧ ⎪ S ⟩=√1 S ⟩ + S ⟩ S ⟩=√1 S ⟩ − S ⟩ ⎪ Φ2 001 110 , Φ7 001 110 , ⎪ 2 2 ⎪ ⎪ S ⟩=√1 S ⟩ + S ⟩ S ⟩=√1 S ⟩ − S ⟩ ⎪ Φ3 010 101 , Φ6 010 101 , ⎨ 2 2 ⎪ ⎪ S ⟩=√1 S ⟩ + S ⟩ S ⟩=√1 S ⟩ − S ⟩ ⎪ Φ4 011 100 , Φ5 011 100 , ⎪ 2 2 ⎪ as the eigenstates⎪ S of⟩= the√ phase-bit S ⟩ + S operator⟩ SX ⟩X=√X Sand⟩ − S ⟩ ⎩⎪ • 1st parity-bit operator: Z Z I 2, ⊗ ⊗ • 2nd parity-bit operator: I 2 Z Z. ⊗ ⊗ 4○ n-qubit: GHZ √1 00 0 11 1 , with n observables n 2 ⊗ ⊗ n n S ⟩ = S ⋯ ⟩ + S ⋯ ⟩ ²phase-bit² operator X X X, ⎧ n ⎪ ∶ Z ⊗Z ⊗⋯⊗I I ⎪ 1st parity-bit operator Z´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Z I 2 I 2, ⎪ ⎪ n−2 ⎪ ∶ I ⊗ Z⊗ Z ⊗⋯⊗I I ⎪ 2nd parity-bit operator I 2 Z ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶Z I 2 I 2, ⎪ ⎪ n−3 ⎨ ∶ ⊗ ⊗ ⊗ ⊗⋯⊗ ⎪ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ ⎪ ⎪ n 1 -th parity-bit operator I 2 I 2 I 2 Z ZZ. ⎪ ⋮ ⎪ n−2 ⎪ ( − ) ∶ ⊗ ⊗⋯⊗ ⊗ ⊗ ⎪ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ ⎩⎪

X 1 X 2 X n GHZ n

1 x1 (X 1⊗X 2⊗⋯⊗X n)S 0⟩x2x3 xn 1 1¯x2x¯3 x¯n 2 1 = ( ⊗ ⊗⋯⊗ ) √ [Sx1 ⋯ ⟩ + (− ) S ⋯ ⟩] 1¯x2x¯3 x¯n 1 0x2x3 xn 2 1 = √ xS 1 ⋯ ⟩ + (− ) S x1 ⋯ ⟩ 1 0x2x3 xn 1 1¯x2x¯3 x¯n 2 x1 = (−1) √GHZ S n . ⋯ ⟩ + (− ) S ⋯ ⟩

= (− ) S ⟩ 154 Z 1 Z 2 I 2 I 2 GHZ n n−2 ⊗ ⊗ ⊗⋯⊗ S ⟩ 1 x1 Z 1 Z 2 I´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶2 I 2 0x2x3 xn 1 1¯x2x¯3 x¯n 2 n−2 = ⊗ ⊗ ⊗⋯⊗ √ S ⋯ ⟩ + (− ) S ⋯ ⟩ 1 x2 x1 x2+1 1 ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶0x2x3 xn 1 1 1 1¯x2x¯3 x¯n 2 1 = √ (x2− ) S ⋯ ⟩ + (− )x1 (− )(− ) S ⋯ ⟩ 1 0x2x3 xn 1 1¯x2x¯3 x¯n 2 x2 = (−1) √GHZ S n . ⋯ ⟩ + (− ) S ⋯ ⟩

= (− ) S ⟩

I 2 Z 2 Z 3 I 2 I 2 GHZ n n−3 ⊗ ⊗ ⊗ ⊗ S ⟩ 1 x1 I 2 Z 2 Z 3 I´¹¹¹¹¸¹¹¹¹¶2 I 2 0x2x3 xn 1 1¯x2x¯3 x¯n 2 n−3 = ⊗ ⊗ ⊗ ⊗ √ S ⋯ ⟩ + (− ) S ⋯ ⟩ 1 x2+x3 x1 x2+1 x3+1 1 ´¹¹¹¹¸¹¹¹¹¶0x2x3 xn 1 1 1 1¯x2x¯3 x¯n 2 1 = √ (x2−+x)3 S ⋯ ⟩ + (− )x1 (− ) (− ) S ⋯ ⟩ 1 0x2x3 xn 1 1¯x2x¯3 x¯n 2 x2+x3 = (−1) √GHZ S n . ⋯ ⟩ + (− ) S ⋯ ⟩

= (− ) S ⟩

I 2 I 2 Z 3 Z 4 I 2 I 2 GHZ n n−4 ⊗ ⊗ ⊗ ⊗ ⊗⋯⊗ S ⟩ 1 x1 I 2 I 2 Z 3 Z 4 I´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶2 I 2 0x2x3 xn 1 1¯x2x¯3 x¯n 2 n−4 = ⊗ ⊗ ⊗ ⊗ ⊗⋯⊗ √ S ⋯ ⟩ + (− ) S ⋯ ⟩ 1 x3+x4 x1 x3+1 x4+1 1 0x2´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶x3 xn 1 1 1 1¯x2x¯3 x¯n 2 1 = √ (x3−+x)4 S ⋯ ⟩ + (− )x1 (− ) (− ) S ⋯ ⟩ 1 0x2x3 xn 1 1¯x2x¯3 x¯n 2 x3+x4 = (−1) √GHZ S n . ⋯ ⟩ + (− ) S ⋯ ⟩

= (− ) S ⟩

I 2 I 2 Z n−1 Z n GHZ n n−2 ⊗⋯⊗ ⊗ ⊗ S ⟩ 1 x1 I´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶2 I 2 Z n−1 Z n 0x2x3 xn 1 1¯x2x¯3 x¯n 2 n−2 = ⊗⋯⊗ ⊗ ⊗ √ S ⋯ ⟩ + (− ) S ⋯ ⟩ 1 xn−1+xn x1 xn−1+1 xn+1 ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶1 0x2x3 xn 1 1 1 1¯x2x¯3 x¯n 2 1 = √ (xn−−1)+xn S ⋯ ⟩ + (− )x1 (− ) (− ) S ⋯ ⟩ 1 0x2x3 xn 1 1¯x2x¯3 x¯n 2 xn−1+xn = (−1) √GHZ S n . ⋯ ⟩ + (− ) S ⋯ ⟩

= (− ) S ⟩ 155 We can then conclude that the n-qubit GHZ state, i.e., state GHZ n, is the common eigenvector of the phase-bit operator, the 1st parity-bit operator, the 2nd parity-bit operator, and the n 1 -th parity bit operator, with the correspondingS ⟩ eigenvalues x1, x2, x2 x3, , xn−1 xn . ⋯ ( − ) ○ 5 (As n large+ number,⋯ the+ n)-qubit GHZ state deﬁned as 1 → GHZ live dead n 2 is also called the cat state, whichS ⟩ is= the√ macroscopically(S ⟩ + S ⟩) distinguished states used in Schr¨odinger’scat experiment.

15.2.3 Properties of GHZ states 1○ For the GHZ states GHZ , each qubit is maximally entangled with the other n 1 qubits, e.g. S ±⟩ − ρA1 trA2A3⋯An GHZ n GHZ 1 tr ⋯ 0y 1¯y 0y 1¯y , ∶= 2 A2A3 ASn ±n⟩ ⟨ n ±Sn n with y as a n 1 -bit= string of 0 and 1,S thus⟩ ± S ⟩ ⟨ S ± ⟨ S 1 ρ y′ 0y 1¯y 0y 1¯y y′ A1 ( − ) A2A3⋯An n n n n A2A3⋯An 2 y′ = 1 Q ⟨ S S ⟩ ± S ⟩ ⟨ S ± ⟨ S S ⟩ 0 δy′y 1 δy′y¯ 0 δyy′ 1 δyy¯ ′ 2 y′ = 1 Q S ⟩ ± S ⟩ ⟨ S ± ⟨ S 0 0 1 1 , 2 i.e., = S ⟩⟨ S + S ⟩⟨ S 1 ρ I . (15.2.4) A1 2 2 2○ Cutting one qubit gives rise to a separable= n 1 -qubit mixed state with rank 2. ρ tr GHZ GHZ A2⋯An A1 n ( − ) 1 1 = S i ±⟩0y⟨ 1¯±yS 0y 1 y i A1 n n n n A1 2 i=0 1 = Qy ⟨ S yS ⟩y¯ + S y¯⟩ , ⟨ S + ⟨ ⟨ SS S ⟩ 2 n−1 n−1 = S ⟩ ⟨ S + S ⟩ ⟨ S) from which we can ﬁnd that ρA2⋯An can be rewritten in the form as illustrated in

E.Q. (15.2.2), namely ρA2⋯An is separable. Note that we can apply the PPT criterion namely ρPT ρ ρ ρA2⋯An Id T ρA2⋯An ρA2⋯An 0, (15.2.5) which is the necessary condition for quantum separability. = ( ⊗ ) = ≥ 3○ n-qubit GHZ states GHZ √1 x x¯ form an orthonormal basis of n-qubit n 2 n Hilber space, see Table 15.1. S ±⟩ = S ⟩ ± S ⟩ ○ n 4 Given an n-qubit GHZ n state, other 2 1 GHZ n states can be generated via LOCC, e.g. in the case of n 2, the other Bell states can be obtained from φ+ state via LOCC, asS shown⟩ in Figure 15.4.( − )S ±⟩ = S ⟩

156 Table 15.1: GHZ n states form an orthonormal basis of n-qubit Hilbert space.

S ±⟩ n GHZ states (orthnormal basis) n-qubit Hilbert space

n 1 H2 n 2 ψ i, j , i, j 0, 1 = S±⟩ H2 H2 n 3 Φ , i 1, 2, , 8 = S (i )⟩ = H2 H⊗2 H2 = S ⟩ = ⋯ ⊗ ⊗ n GHZ n ⋮ H2 H2 H2 n S ±⟩ ⊗ ⊗⋯⊗ ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

+ φ I I2 I 4 I XZ Z 2 S ⟩ X I 2 ⊗ ⊗ ⊗ φ+ φ− ψ+ ψ−

S ⟩ Figure 15.4:S Bell⟩ states generatedS ⟩ from φ+ S ⟩

S ⟩

157 Chapter 16

Entanglement Measures and Entropy: Bi-Partite System

The main problem is that we do not understand fully what entanglement is. —Horodecki

Then the usefulness of entanglement emerges because it allows us to overcome a particular constraint that will be the LOCC constraint. —Plenio and Virmani

It is the constraint to LOCC operations that entanglement to the status of a resource. —Plenio and Virmani

We will simply accept the non-uniqueness of entanglement measures as an expression of the fact that they correspond to different operational tasks under which different forms of entanglement may have different degree of usefulness. —Plenio and Virmani

Reference:

[Preskill] New Chapter 4: Quantum entanglement;

● 16.1 Pure states and mixed states

1○ A quantum system is described by the density matrix which is

1 linear & non-negative operators, i.e., ρ 0; 2 Hermitian: ρ† ρ; ( ) ≥ 3 Trace 1: tr ρ 1. ( ) = ○ 2 2 2 ρ(is) called pure( state) = iﬀ ρ ρ or tr ρ 1. Note: ρ is a pure state ρ ψ ψ . = ( ) = ⇐⇒ = S ⟩⟨ S

158 3○ ρ is called mixed state iﬀ ρ2 ρ or tr ρ2 1. Note: The mixed state has the formalism ≠ ( ) ≤

ρ pi ψi ψi (16.1.1) i = Q S ⟩⟨ S with i 2, pi 0 and i pi 1. It stands for a state ensemble pi, ψi . Every mixed state density matrix has inﬁnity number of state ensembles, which is called the ambiguity≥ of≥ the density∑ matrix.= { S ⟩}

16.2 Bipartite entangled pure states

Def 16.2.1. The bipartite pure state χ AB is called separable or factorizable if it can be written as a product of two states, namely S ⟩ χ AB ϕ A ψ B. (16.2.1)

Def 16.2.2. The bipartite pure stateS ⟩ χ AB= S is⟩ entangled⊗ S ⟩ if it is not separable.

Thm 16.2.0.1. With the Schmidt decompositionS ⟩ of χ AB, the Schmidt number is 1, then χ AB is separable, and the Schmidt number is greater than 1, then χ AB is entangled. S ⟩ SThm⟩ 16.2.0.2. The density matrix of bipartite system has the formalismS ⟩ expressed as ρAB χ AB χ . ○ 1 ∶=WhenS ⟩ ρ⟨A S trBρAB or ρB trAρAB is pure state, then χ AB is separable; ○ 2 When ρA =or ρB is mixed state,= then χ AB is entangled;S ⟩ ○ 1 3 When ρA ρB dId, then χ AB is maximalS ⟩ entangled. Proof. = = S ⟩ 1○ When χ AB ϕ A ψ B, (16.2.2)

then S ⟩ ∶= S ⟩ ⊗ S ⟩ ρAB ϕ A ϕ ψ B ψ . (16.2.3)

Therefore = S ⟩ ⟨ S ⊗ S ⟩ ⟨ S ρA ϕ A ϕ , ρB ψ B ψ , (16.2.4)

which are pure states. = S ⟩ ⟨ S = S ⟩ ⟨ S ○ 2 With the Schmidt decomposition of state χ AB, density matrix ρA or ρB is mixed state which is directly associated with the Schmidt decomposition number of state χ AB greater than 1. S ⟩

○ 3 MaximallyS ⟩ entangled state means our maximal ignorance on subsystem, i.e., ρ Id.

e.g.1. Bell state: ∝ 1 Ψ AB 00 11 , (16.2.5) 2 S ⟩ ∶= √ S 1⟩ + S ⟩ ρ ρ I . (16.2.6) A B 2 2 = = 159 e.g.2. Higher dimensional Bell state:

1 d−1 Ψ AB ii , (16.2.7) d i=0 S ⟩ ∶= √ 1Q S ⟩ ρ ρ IdId. (16.2.8) A B d = =

Def 16.2.3 (Von-Neumann entropy). The von-Neumann entropy is a distinguished measure of bipartite entangled pure state ρAB deﬁned as

S ρA tr ρA logρA , (16.2.9)

S ρB tr ρB logρB . (16.2.10) ( ) ∶= − ( ) Properties: ( ) ∶= − ( ) ○ 1 S ρA 0 implies ρA is a pure state. ○ 2 S(ρA) =is in the domain 0 S ρA log d where d min dim HA, dim HB . ○ 3 In( the) diagonalized form of≤ density( ) ≤ matrix ρ, = ( )

ρ λi ψi ψi , (16.2.11) i von-Neumann entropy has the formulation= Q S ⟩⟨ S

S ρ tr ρ logρ λi log λi. (16.2.12) i ( ) = − ( ) = − Q Note 1: The von-Neumann entropy is deﬁned only for pure states not for mixed states.

Note 2: The von-Neumann entropy is a measure of our ignorance of the quantum state, which is associated with the Shannon entropy in information theory or thermodynamics.

Examples: e.g.1. For the separable pure state

1 0 ρA ϕ A ϕ , (16.2.13) ⎛ ⎞ ⎜ ⎟ ∶= S ⟩ ⟨ S = ⎜ 0 ⎟ ⎜ ⋱ ⎟ the von-Neumann entropy is ⎝ ⎠

1 1 0 0 S ρA tr log 0, (16.2.14) ⎛⎛ ⎞ ⎛ ⎞⎞ ⎜⎜ ⎟ ⎜ ⎟⎟ ( ) = − ⎜⎜ 0 ⎟ ⎜ 0 ⎟⎟ = ⎜⎜ ⋱ ⎟ ⎜ ⋱ ⎟⎟ where the relations ⎝⎝ ⎠ ⎝ ⎠⎠ 1 log 1 0, 0 log 0 0 (16.2.15)

have been applied. ⋅ = ⋅ =

160 e.g.2. For the higher dimensional Bell state:

1 d−1 Ψ AB ii , (16.2.16) d i=0

S ⟩ ∶= 1√ Q S ⟩ d 1 1 d ρA Id , (16.2.17) d ⎛ ⎞ ⎜ 1 ⎟ = = ⎜ d ⎟ ⎜ ⋱ ⎟ ⎝ ⎠ 1 1 d d 1 1 d d S ρA tr log ⎛⎛ ⎞ ⎛ ⎞⎞ ⎜⎜ 1 ⎟ ⎜ 1 ⎟⎟ ( ) = − ⎜⎜ d ⎟ ⎜ d ⎟⎟ ⎜⎜1 1 ⋱ ⎟ ⎜ ⋱ ⎟⎟ ⎝⎝d log d ⎠ ⎝ ⎠⎠ 1 log 1 tr d d ⎛ ⎞ ⎜ 1 1 ⎟ = − ⎜ d log d ⎟ ⎜1 1 ⋱ ⎟ d ⎝ log log d. ⎠ (16.2.18) d d So the higher dimensional= − Bell⋅ state is= the maximal entangled states.

Def 16.2.4. Arbitrary two-qubit pure state has the formalism

ψ α1 00 α2 01 α3 10 α4 11 (16.2.19) with the matrix formulationS ⟩ expressed∶= S ⟩ as+ S ⟩ + S ⟩ + S ⟩

α1 α ψ 2 . (16.2.20) ⎛ α3 ⎞ ⎜ ⎟ S ⟩ ∶= ⎜ α4 ⎟ ⎜ ⎟ The Pauli σy is deﬁned as ⎝ ⎠ 0 i σ . (16.2.21) y i 0 − And we have = ∶=

∗ ψ˜ σy σy ψ ∗ 1 α1 S ⟩ ∶= ( ⊗ )S ⟩ ∗ 1 α2 − ∗ ⎛ 1 ⎞ ⎛ α3 ⎞ ⎜ ⎟ ⎜ ∗ ⎟ = ⎜ 1 ⎟ ⎜ α4 ⎟ ⎜ ∗ ⎟ ⎜ ⎟ ⎝ α4 ⎠ ⎝ ⎠ − ∗ α3 − ∗ , (16.2.22) ⎛ α2 ⎞ ⎜ ∗ ⎟ = ⎜ α1 ⎟ ⎜ ⎟ where state ψ∗ is denoted as the complex⎝ − conjugate⎠ of the state ψ .

S ⟩ 161 S ⟩ The concurrence of two-qubit pure state is deﬁned as

C ψ ψ ψ˜ ∗ α4 (S ⟩) ∶= S⟨ S ⟩S ∗ ∗ ∗ ∗ ∗ α3 α1 α2 α3 α4 − ∗ ⎛ α2 ⎞ ⎜ ∗ ⎟ = S ⎜ α1 ⎟ S ⎜ ⎟ 2 α2α3 α1α4 . (16.2.23) ⎝ − ⎠ = S − S 16.3 Example for two-qubit pure entangled state

For arbitrary two-qubit pure state

ψ AB α1 00 α2 01 α3 10 α4 11 , (16.3.1) the concurrence is S ⟩ ∶= S ⟩ + S ⟩ + S ⟩ + S ⟩ C ψ AB 2 α1α4 α2α3 . (16.3.2) The density matrix of subsystem A is (S ⟩ ) = S − S

ρA trB ψ AB ψ

B 0 ρAB 0 B B 1 ρAB 1 B = S∗ ⟩ ⟨ ∗S ∗ ∗ α1α1 α2α2 α3α1 α4α2 = ⟨ S∗ S ⟩ ∗+ ⟨ S ∗ S ⟩ ∗ (16.3.3) α1α3 α2α4 α3α3 α4α4 + + = with the eigenvalue + +

∗ ∗ ∗ ∗ 1 1 4 α1α4 α2α3 α1α4 α2α3 λ± ρ . (16.3.4) A » 2 ± − ( − )( − ) According to the formula( ) (16.2.12),= the von-Neumann entropy of subsystem A is

S ρA λ+ ln λ+ λ− ln λ−. (16.3.5)

1○ Bell states. ( ) = − − 1 1 φ± 00 11 , ψ± 01 10 . (16.3.6) 2 2 Von-Neumann entropy:S ⟩ ∶= √ S ⟩ ± S ⟩ S ⟩ ∶= √ S ⟩ ± S ⟩

S φ± ln 2,S ψ± ln 2. (16.3.7)

Concurrence: (S ⟩) = (S ⟩) = C φ± 1,C ψ± 1. (16.3.8) Bell states are maximal entangled states. (S ⟩) = (S ⟩) = 2○ Separable states. ψ AB x1 0 x2 1 x3 0 x4 1 . (16.3.9) Von-Neumann entropy: S ⟩ ∶= ( S ⟩ + S ⟩) ⊗ ( S ⟩ ⊗ S ⟩) S ψ AB 0. (16.3.10) Concurrence: (S ⟩ ) = C ψ AB 0. (16.3.11)

(S ⟩ ) = 162 3○ Evolution of Bell states. π π ψ cos θ 00 eiϕ sin θ 11 , (16.3.12) AB 4 AB 4 AB π S ⟩3π ∶= ( − )S ⟩ − ( − )S ⟩ where 4 θ 4 and 0 ϕ 2π.

Von-Neumann− ≤ ≤ entropy: ≤ ≤

The density matrix of subsystem A denoted as ρA has the eigenvalues 1 sin 2θ λ± . (16.3.13) 2 ± Thus = 1 sin 2θ 1 sin 2θ 1 sin 2θ 1 sin 2θ S ρ ln ln . (16.3.14) A 2 2 2 2 + + − − When θ 0,( state) = −ψ AB evolves into Bell− states with the von-Neumann entropy 1 1 1 1 = S ⟩ S ρ ln ln ln 2. (16.3.15) A θ=0 2 2 2 2

π ( )S = − − = When θ 4 , state ψ AB evolves into separable states with the von-Neumann entropy

= S ⟩ S ρA = π 0. (16.3.16) θ 4 ( )S = Concurrence:

π π C ψ 2 cos θ sin θ cos 2θ . (16.3.17) AB 4 4 Remarks: S ψ and C (Sψ ⟩ are) = diﬀerentV ( measures.− ) ( − )V = S S

○ 1 Degree of(S entanglement⟩) (S ⟩) measure of separable state is zero. 2○ Bell states are maximally entanglements.

3○ The conﬁgurations of C ψ and S ψ look the same in shape.

(S ⟩) (S ⟩) 16.4 Axioms of entanglement measure

Def 16.4.1. An entanglement measure or degree of entanglement is a real valued function E ρ to quantify the amount of entanglement in a given quantum state ρ.

Necessary( ) requirements on E ρ . ○ 1 E ρ is a mapping from a bipartite( ) state ρ to a positive valued real number, namely + ( ) ρ E ρ R . (16.4.1)

2○ Normalization. → ( ) ∈ E ρ E Ψ AB Ψ ln d, (16.4.2)

where Ψ is Bell state. ( ) ≤ (S ⟩ ⟨ S) =

S ⟩ 163 3○ E ρ 0 for separable states.

○ 4 E( ψ) =AB ψ S trB ψ AB ψ for bipartite pure states. ○ 5 E(Sρ ⟩is not⟨ S) increased= ( S by⟩ LOCC.⟨ S)

( ) LOCC ′ † † ρ ρ AiρρAAi , AiAi IdId. (16.4.3) i i Ð→ = = QE ρ E ρ′Q. (16.4.4)

6○ Convexity. ( ) ≥ ( ) E λρρ 1 λ σ λE ρ 1 λ E σ . (16.4.5)

7○ Additivity. ( + ( − ) ) ≤ ( ) + ( − ) ( ) E ρ⊗n nE ρ . (16.4.6)

8○ Subadditivity. ( ) = ( ) E ρ σ E ρ E σ . (16.4.7)

9○ Continuity. ( ⊗ ) ≤ ( ) + ( ) E ρ E σ 0, if ρ σ 0. (16.4.8)

( ) − ( ) → Y − Y → 16.5 Example: entanglement of formation

Def 16.5.1. The entanglement of formation of the mixed state ρ i pi ψi ψi is deﬁned as = ∑ S ⟩⟨ S EF ρ Inf piE ψi ψi , (16.5.1) i where Inf stands for Inﬁmum, namely( ) ∶= greatQ lowest(S bound.⟩⟨ S)¡ E.g. For a bipartite mixed state ρ,

1 1 C2 ρ E ρ S , (16.5.2) F » 2 ⎛ + − ( )⎞ ( ) = where C ρ is the concurrence and the function⎝ S is deﬁned⎠ as

( ) S x x log2 x 1 x log2 1 x . (16.5.3)

With the new deﬁned density( ) matrix= − denoted− ( as− ρ˜) ( − )

ρ˜ σy σy ρ σy σy , (16.5.4) the non-Hermitian matrix ρρ˜ has the∶= ( diagonized⊗ ) ( formalism⊗ ) 4 2 ρρ˜ λi ψi ψi , (16.5.5) i=1 = Q S ⟩⟨ S where λ1 λ2 λ3 λ4 0.

> > > ≥ C ρ max 0, λ1 λ2 λ3 λ4 . (16.5.6)

Note: Entanglement of formation( ) = is the{ entanglement− − − measure} for ρ neither the concurrence nor the von-Neumann entropy.

164 16.6 PPT criterion for quantum separability

PPT is the abbreviation of positive partial transpose criterion or Peres-Horodecki criterion for quantum separability.

Thm 16.6.0.3. For a quantum state ρAB, when ρAB is separable, then T ρAB Id T ρAB 0, (16.6.1) which is only a necessary condition for∶= separability.( ⊗ ) ≥ Thm 16.6.0.4. When dim HA 2 and dim HB 2 or 3, then T ρAB= is separable = ρAB 0. (16.6.2) T Corollary 16.6.0.1. When ρAB is not positive,⇐⇒ then ρA≥is entangled. E.g. Werner state in dimension 2. 1 ρ λ 1 λ I λ φ+ φ+ (16.6.3) 4 4 with ( ) ∶= ( − ) + S ⟩⟨ S 1 φ+ 00 11 . (16.6.4) 2 Note: S ⟩ = √ S ⟩ + S ⟩ AACC ρ (16.6.5) C † D where A, B, C and D are 2 2 matrix.= AACC† ×ρTA T I ρ ; (16.6.6) 2 CCBB

∶= ( ⊗ ) = AT C T ρTB I T ρ . (16.6.7) 2 C ∗ BT ∶= ( ⊗ ) = The Werner state has the matrix formalism 1 λ 4 1 λ 0 0 2 0 0 0 0 ρ ρ λ ⎛ ( − ) 1 ⎞ , (16.6.8) 0 0 4 1 λ 0 ⎜ λ 1 ⎟ ⎜ 0 0 1 λ ⎟ ( ) = ⎜ 2 4 ⎟ ⎜ ( − ) ⎟ TA 1 1 and the its partial transpose⎝ matrix has the eigenvalues(ρ − λ) ⎠ 4 1 λ , 4 1 3λ . 1 Therefore, when λ 3 , according to the PPT criterion, Werner state is entangled. Detailed calculations refer to the section 15.1.4. ( ) = ( + ) ( − ) > 16.7 Example: Werner state in d-dimension

The Werner state in dimension d is deﬁned as 1 ρ λ 1 λ I 2 λ Φ Φ (16.7.1) d2 d with ( ) ∶= ( − ) + S ⟩⟨ S 1 d−1 Φ ii , (16.7.2) d i=0 which is a maximally entangled state,S ⟩ namely∶= √ SQ ΦS ⟩ ln d.

165 (S ⟩) = 16.7.1 ρ(λ) is a density matrix 1○ Trace 1. 1 trρ λ λ 1 λ trI 2 λ 1 λ 1. (16.7.3) d2 d 2○ Hermitian. ( ) = + ( − ) = + − = † 1 ρ λ 1 λ I 2 λ Φ Φ ρ λ . (16.7.4) d2 d 3○ Positivity. ( ) = ( − ) + S ⟩⟨ S = ( ) 1 ρ λ Φ λ 1 λ Φ . (16.7.5) d2 1 ρ λ has the eigenstate Φ with( )S the⟩ eigenvalue= ( + ( denoted− ))S as⟩ τΦ, where τΦ λ d2 1 λ . 1 ( ) S ⟩ ρ λ Φ 1 λ Φ , = + (16.7.6)( − ) d2 where state Φ is orthogonal( to)S the⟩ = state( Φ− . )Sρ λ⟩ has the eigenstate Φ with 1 the eigenvalue denoted as τΦ , where τΦ d2 1 λ . Positivity requires τΦ, τΦ 0, therefore, whenS ⟩ S ⟩ ( ) S ⟩ 1 = ( − ) ≥ λ 1, (16.7.7) 1 d2 ρ λ is a density matrix. ≤ ≤ − 16.7.2( ) Partial transpose of ρ(λ)

1 d−1 Φ Φ TB Id T ii jj d i,j=0 (S ⟩⟨ S) = (1 d⊗ ) Q S ⟩⟨ S ij ji d i,j=1 = 1 Q S ⟩⟨ S SWAP. (16.7.8) d Construct the symmetric eigenstate of= the SWAP gate denoted as S ij

S ij ij ji , S ( )⟩ (16.7.9) from which we have S ( )⟩ ∶= S ⟩ + S ⟩ SWAP S ij S ij . (16.7.10) Construct the anti-symmetric eigenstate of the SWAP gate denoted as A ij S ( )⟩ = S ( )⟩ A ij ij ji , S ( )⟩ (16.7.11) from which we have S ( )⟩ ∶= S ⟩ − S ⟩ SWAP A ij A ij . (16.7.12) With the new-deﬁned operators denoted as E and E expressed as S ( )⟩sym= −S ( )⟩anti 1 d E sym S ij S ij , 4 i,j=1 = 1 Qd S ( )⟩⟨ ( )S E anti A ij A ij , (16.7.13) 4 i,j=1 = Q S ( )⟩⟨ ( )S 166 the SWAP gate has the formalism

SWAP E sym E anti I d2 2E anti, (16.7.14) where the relation E sym E anti =I d2 has− been applied.= − 1 1 +trE = d2 d d2 d d d 1 , (16.7.15) sym 4 2 1 1 trE = (d2 + d + d2 + d) = d(d + 1), (16.7.16) anti 4 2 1 1 trE =trE( − +d d− 1) = d( d− 1) . (16.7.17) sym anti 2 2 + = ( + ) + ( − ) 16.7.3 Positivity of partial transpose ρ(λ) The partial transpose of the Werner state is denoted as 1 1 TB ρ λ 1 λ I 2 λ I 2 2E d2 d d d anti 1 1 λ 2λ ( ( )) = ( − ) λ I+ 2 ( E− , ) (16.7.18) d d d d anti − which has the eigenvalues denoted= as τ(sym and+ τ)anti − 1 1 λ τ λ , (16.7.19) sym d d 1 1 − λ 2λ τ = ( + λ) . (16.7.20) anti d d d − The positivity requires the parameter=λ in( the region+ ) − 1 1 λ , (16.7.21) 1 d 1 d therefore, with the constrain in E.Q. (16.7.7),≤ ≤ we can infer that ρ λ is entangled when − + 1 λ 1. ( ) (16.7.22) 1 d Note: We do not know whether ρ λ is separable≤ ≤ or not when 1 λ 1 , since PPT + 1−d2 1+d criterion is only a necessary condition. ( ) ≤ ≤ 16.8 Example: the other form of the Werner state 1 1 ρ λ 1 λ I 2 λ E . (16.8.1) anti d2 d 1 anti 2 d d 1 ( ) ∶= ( − ) + ( − ) 16.8.1 ρanti is a density matrix 1○ Trace 1. 1 1 trρ λ 1 λ tr I 2 λ tr E anti d2 d 1 anti 2 d d 1 ( ) = 1 (λ − λ) ( ) + ( ) ( − ) 1. (16.8.2) = − + = 167 2○ Hermitian.

† 1 1 † ρ λ 1 λ I 2 λ E anti d2 d 1 anti 2 d d 1 ( ) = 1 ( − ) + 1 1 λ I 2 λ E d2 d 1 ( − ) anti 2 d d 1 = ρanti( −λ .) + (16.8.3) ( − ) 3○ Positivity. = ( ) 1 τ 1 λ , (16.8.4) sym d2 1 1 τ = λ ( − ) 1 λ . (16.8.5) anti 1 d2 2 d d 1 = + ( − ) ( −1 ) d ρ 0 λ 1. (16.8.6) anti 1 d − ≥ ⇐⇒ ≤ ≤ 16.8.2 Partial transpose of ρanti(λ) +

PT 1 Φ Φ I 2 2E , d d anti 1 PT (S ⟩⟨ ΦS) Φ = (I 2 − 2E ), (16.8.7) d d anti therefore S ⟩⟨ S = ( − ) PT 1 E I 2 d Φ Φ . (16.8.8) anti 2 d = ( − S ⟩⟨ S) PT 1 λ PT ρ 1 λ I 2 E anti d2 d 1 anti 2 d d 1 = ( 1− ) + λ λd d ( 1− 1) λ I d2 Φ Φ , (16.8.9) d2 d 1 d 1 = ( + ( − )( − )) − S ⟩⟨ S which has the eigenvalus ( − ) − 1 λ τ λd d 1 1 λ , Φ d2 d 1 d 1 = 1 ( + ( − )( − )) − τ λd d 1 1 λ . (16.8.10) Φ d2(d − 1) − = ( + ( − )( − 1)) ρPT( −0 ) 1 d λ . (16.8.11) anti 1 d 1 When 1+d λ 1, ρanti is entangled.≥ However,⇐⇒ − we≤ do≤ not know ρanti is separable or not, 1−d 1 + when + λ + . 1 d ≤ ≤1 d ≤ ≤

168 Part IV

Quantum Open System and Quantum Error Correction Codes

169 Chapter 17

Quantum Mechanics (III): Quantum Open System

Real systems suﬀer from unwanted interactions with the outside world. These unwanted interactions show up as noise in quantum information processing systems. We need to understand and control such noise processes in order to build useful quantum information processing systems. —Nielsen & Chuang No quantum systems are ever perfectly closed, and especially not quantum computers, which must be delicately programmed by an external system to perform some desired set of operations. —Nielsen & Chuang An open system is nothing more than one which has interactions with some other environment system, whose dynamics we wish to neglect or average over. —Nielsen & Chuang The mathematical formalism of quantum operations is the key tool for our description of the dynamics of open quantum systems. —Nielsen & Chuang Another advantage of quantum operations in applications to quantum computation and quantum information is that they are especially well adapted to describe discrete state changes, that is, transformations between an initial state ρ and ρ′, without explicit reference to the passage of time. —Nielsen & Chuang We will construct an artiﬁcial example of a system whose evolution is not described by a quantum operation, . It is an interesting problem for further study to study quantum information processing beyond the quantum operation formalism. ⋯ —Nielsen & Chuang References: [Preskill] Chapter 2: Foundations II: measurement and evolution;

● [Nielsen & Chuang] Chapter 8: Quantum noise and quantum operation. ● 170 17.1 Introduction

17.1.1 Why we talk about quantum open system Reason 1: Quantum Computer is an open system. Quantum computer will interact with the environment (noise). The observers (control) will also interact with the Quantum Computer.

Reason 2: Quantum Computer is a many-qubit system, and it consists of a set of subsystems which interact with each other.

17.1.2 Closed system and open system

Table 17.1: Comparison between closed system and open system

closed system open system state pure state/Ray density matrix observable self-adjoint self-adjoint measurement projective measurement general measurement evolution Shcr¨odingerEquation Quantum operation/Superoperator

Table 17.2: State and quantum gate in closed and open system

state quantum Gate cos θ closed system pure state: ψ 2 SU 2 θ iϕ ⎛sin 2 e ⎞ S ⟩ = ⎜ ⎟ ( ) mixed state: ρ 1 1 p σ open system ⎝2 ⎠ superoperator: ρ ρ′ (density matrix)= ( + ⃗⋅⃗) →

17.2 Projective measurement

The following three points are equivalent:

1○ An arbitrary observable can be expressed as

O anΠn, (17.2.1) n O = where Q Πn n n (17.2.2)

with an being the eigenvalue of the observable∶= S ⟩⟨ S associated with the eigenvector n . O S ⟩

171 The mean value of the observable is

ψ ψ

ψ anΠn ψ ⟨O⟩ = ⟨ SOn S ⟩ = ⟨ SaQn ψ n S n⟩ ψ n = Q anP⟨ n,S ⟩⟨ S ⟩ (17.2.3) n with = Q 2 Pn ψ n . (17.2.4)

2○ In the orthonormal basis n , observables∶= S⟨ S can⟩S be determined by the basis. The probability, that the post-measurement state is n , should be {S ⟩} 2 Prob n n ψ S ⟩ ψ n n ψ ( ) = S⟨ S ⟩S ψ Πn ψ = ⟨ S ⟩⟨ S ⟩ tr ρΠn . (17.2.5) = ⟨ S S ⟩ The post-measurement state is = ( ) n ψ ψ n , (17.2.6) Prob n ⟨ S ⟩ or S ⟩ → S ⟩ » ( ) Π ψ ψ Π† Π ρΠ ρ ψ ψ n n n n . (17.2.7) Prob n Prob n S ⟩⟨ S The expression is for the∶= pureS ⟩⟨ state.S → = ( ) ( ) ○ 3 The complete set of orthogonal projectors Πn are deﬁned as

ΠnΠm δmn, Orthonormal;{ } ⎧ Π† Π , Hermition; ⎪ n = n ⎪ (17.2.8) ⎪ 2 ⎪ Πn = Πn, Projection; ⎪ ⎨ ⎪ Πn = Id, Completeness. ⎪ n ⎪ ⎪ = Now, if we consider a mixed⎪ Q state described by the density matrix, then after the measurement, ⎩ ′ ΠnρΠn ρ ρ Prob n ΠnρΠn (17.2.9) n Prob n n which gives rise to the ensemble→ = Q ( ) = Q ( ) Π ρΠ ρ′ Prob n , n n . Prob n = { ( ) } Example: A single-qubit density matrix is defined( ) as 1 ρ p 1 p σ , with p 3, and 0 p 1. (17.2.10) 2 R (⃗) ∶= ( + ⃗⋅⃗) ⃗ ∈ ≤ Y⃗Y ≤ 172 We can verify the projectors defined as 1 Π 1 n σ n n , (17.2.11a) 1 2 ⎧ 1 ⎪ ∶= ( + ⃗⋅⃗) = S⃗⟩⟨⃗S ⎪ Π2 1 n σ n n , (17.2.11b) ⎨ 2 ⎪ ⎪ ∶= ( − ⃗⋅⃗) = S−⃗⟩⟨−⃗S satisfying the definition ⎩⎪ Π1 Π2 Id, Π Π δ Id, ⎧ i j ij (17.2.12) ⎪ + Π2 = Π , ⎪ i i ⎪ † = ⎨ Πi Πi. ⎪ = ⎪ The probabilities that the post-measurement⎪ = state is n and n can be calculated respectively ⎩ S⃗⟩ S−⃗⟩ 1 P tr ρΠ 1 p n , (17.2.13a) 1 1 2 ⎧ 1 ⎪ = ( ) = ( + ⃗⋅⃗) ⎪ P2 tr ρΠ2 1 p n . (17.2.13b) ⎨ 2 ⎪ ⎪ = ( ) = ( − ⃗⋅⃗) 17.3 General measurement⎩ theory

1○ Measurement operators

• measurement operators M m ; • the measurements can be represented by measurement operators; { } • measurement operator M m is labeled by m; † • completeness relation m M mM m Id.

The projector is deﬁned as ∑ = Πn n n . (17.3.1)

The measurement operator is deﬁned as∶= S ⟩⟨ S

M n Πn. (17.3.2)

Therefore, ∶= † M nM n Πn, (17.3.3)

namely = † M nM n Πn n n Q = Q n n n = QId.S ⟩⟨ S (17.3.4)

2○ State =

• pure state ψ ; • mixed state ρ. S ⟩ 3○ Measurement statistics

173 † • pure state: Prob m ψ M mM m ψ ; mixed state: Prob m tr M † M ρ . • ( ) = ⟨ S m mS ⟩ 4○ Total probability: ( ) =

• pure state:

† Prob m ψ M mM m ψ m m † Q ( ) = Qψ ⟨ SM mM m Sψ⟩ m = ⟨ψSψQ S ⟩ 1; (17.3.5) = ⟨ S ⟩ • mixed state: =

† Prob m tr M mM mρ m m

Q ( ) = Q † tr M mM mρ m = trρQ 1. (17.3.6) = 5○ Post-measurement state: =

• pure state: M ψ ψ m ; Prob m S ⟩ S ⟩ → » • mixed state: ( ) M ρρMM † ρ m m . Prob m → 17.4 Deﬁnition of POVM ( )

POVM is the abbreviation of “Positive Operator-Valued Measurement”. Usually, we do not care for the post measurement state, because the post measurement states may be destroyed in experiments immediately. Therefore, only the measurement probability matters.

Def 17.4.1. General Measurement with negligible post-measurement state,

† F a M aM a, (17.4.1) with ∶=

† † † † F a M a M a M †M = a a F a, (17.4.2) =

=F a 0, (17.4.3)

≥ 174 and F a Id. (17.4.4) a

F a a is the set of positive operator-valuedQ = measurements, and M a a is the set of the general measurements. { S ∀ } { S ∀ } The probability is determined by F a, and M a is connected with the post measurement state, which we have no interest.

Examples: (1) Projective measurement is the special case of POVM.

F a 0, F a Πa (17.4.5) a F a Id. ≥ = Ô⇒ (2) Construct the non-orthogonal projective measurement∑ = denoted as

Πa φa φa (17.4.6)

with ∶= S ⟩⟨ S φa φa 1, φa φb δab. (17.4.7)

Construct the POVM denoted⟨ asS ⟩ = ⟨ S ⟩ ≠

F a λa φa φa φ˜a φ˜a , (17.4.8)

where λa is a real number with∶= 0 λSa ⟩⟨1 andS ∶= S ⟩⟨ S ˜ <φa ≤ λa φa . (17.4.9) » Note that the measurement operatorS ⟩F∶=a satisﬁesS ⟩ the relations F 2 F , a a (17.4.10) F a F b 0 with a b. ≠ The probability of measurement result⋅ ≠ associated with≠ the POVM F a is

Pa tr F aρ λa φa ρ φa φ˜a ρ φ˜a 0. (17.4.11)

The post-measurement state= ( is ) = ⟨ S S ⟩ = ⟨ S S ⟩ ≥

′ ρ ρ F aρ F a Pa φa φa , (17.4.12) a » » a where → = Q = Q S ⟩⟨ S F a λa φa φa . (17.4.13) » » (3) Construct the POVM denoted as F 1∶=,FF 2,FFS3 ,⟩⟨ expressedS as 2 1 F { n n } 1 n σ , (17.4.14a) 1 3 1 1 3 1 ⎧ 2 1 ⎪ ⃗ ⎪ F 2 ∶= Sn⃗2⟩⟨n⃗2S ∶= (1 + n⃗2⋅σ) , (17.4.14b) ⎪ 3 3 ⎪ 2 1 ⎨ F 3 ∶= Sn⃗3⟩⟨n⃗3S ∶= (1 + n⃗3⋅σ⃗) , (17.4.14c) ⎪ 3 3 ⎪ ⎪ ∶= S⃗ ⟩⟨⃗ S ∶= ( + ⃗ ⋅⃗) ⎩⎪ 175 n1

⃗

2π 2π 3 3 2π 3

n2 n3

Figure 17.1: Diagrammatical⃗ representation of unit⃗ vectors n1, n2 and n3.

⃗ ⃗ ⃗ where n1, n2 and n3 are three coplanar unit vectors and the angle between any pair 2π 3 of them is 3 in R , as shown in Figure 17.1. And we⃗ can⃗ see F⃗i with i 1, 2, 3 satisfy the relations

3 = F i Id; (17.4.15a) i=1 ⎧ 2 ⎪ QF a = F a, with a 1, 2, 3; (17.4.15b) ⎪ ⎪ 4 ⎨ F aF b ≠ na nb na nb 0, with a,= b 1, 2, 3. (17.4.15c) ⎪ 9 ⎪ We can verify these⎪ relations= ⟨ one⃗ S by⃗ ⟩S one.⃗ ⟩⟨⃗ S ≠ = ⎩⎪ a)

3 1 1 1 F i 1 n1 σ 1 n2 σ 1 n3 σ i=1 3 3 3 1 ⃗ ⃗ ⃗ Q = (3 + ⃗n⋅ )n+ (n + ⃗σ ⋅ ) + ( + ⃗ ⋅ ) 3 1 2 3 1 = [3 + (0⃗ + ⃗ + ⃗ ) ⋅⃗] 3 = Id(. + ) (17.4.16) b) =

2 2 F F n n n n a a 3 a a a a 2 2 = Sn⃗ ⟩⟨n⃗ S ⃗ ⟩⟨⃗ S 3 a a 2 = F .S⃗ ⟩⟨⃗ S (17.4.17) 3 a

c) = 1 F F 1 n σ 1 n σ a b 9 a b 1 = (1 + ⃗n⋅⃗)(n +σ⃗ ⋅⃗)n σ n σ 9 a b a b 1 = [1 + (n⃗ + n⃗ ) ⋅σ⃗ + n(⃗ n⋅⃗)(, ⃗ ⋅⃗)] (17.4.18) 9 a b a b = [ + (⃗ + ⃗ ) ⋅⃗ + ⃗ ⋅⃗ ] 176 therefore F aF b 0, (17.4.19) since F F vanishes iﬀ a b ≠ nb na. (17.4.20)

⃗ = −⃗ 17.5 More on POVM

17.5.1 Dimensional analysis For a Hilbert space H , the number of projective measurements should be

# Πn dim H , (17.5.1) while the number of POVM should satisfy{ } =

# F n dim H . (17.5.2)

For example, in the case of qubit, i.e.,{ #H} 2≥ 2, we have

# Π1,Π= 2 2, # F 1,FF 2,FF 3 3. (17.5.3) { } = 17.5.2 POVM on subsystem can{ be viewed} = as projective measurements on the entire system A POVM can be viewed as projective measurement on a larger Hilbert space. With E a A being the projective measurement on the original Hilbert space HA and F a being the A POVM on the HA. There is a larger Hilbert space HA′ with HA HA′ , such that F a A′ on Hilbert space HA can be viewed as projective measurements E a on HA′ . ⊆ A A A′ # E a # F a # E a . (17.5.4)

17.5.3 Tensor product realization ≤ of POVM≤ A Thm 17.5.3.1. Given one dimension non-negative POVM F a on HA, there exists HB such that AB A Prob a trAB E a ρA ρB trA F a ρA , (17.5.5) AB where E a are the projective( ) = measurements ( ⊗ on H)A= HB. A AB We can construct F a from E a , namely ⊗ A ρ A A Pa (17.5.6) E a ρA F A a ρB = = ⊗ B , where the circles mean traces.

Let’s assume that i A j A is an orthonormal basis of the Hilbert space HA, and µ B ν B is an orthonormal basis of the Hilbert space HB. Then iµ AB jν AB is an orthonormal basis{S ⟩ } of= the{S ⟩ Hilbert} space HA HB. {S ⟩ } = {S ⟩ } {S ⟩ } = {S ⟩ } 177 ⊗ 17.5.3.1 Operator formula of F a From A AB trA F a ρA trAB E a ρA ρB tr tr E AB I ρ ρ , (17.5.7) = A B (a ⊗ A )B A we can see A = AB ( ⊗ ) F a trB E a I A ρB . (17.5.8)

= ( ⊗ ) 17.5.3.2 Matrix formalism of Fa

trA F aρA j F aρA j j

( ) = Q ⟨j SF a i S i⟩ρA j , (17.5.9) i,j = Q ⟨ S S ⟩⟨ S S ⟩ AB AB trAB E a ρA ρB jν E a iµ iµ ρA ρB jν j,ν i,µ ( ⊗ ) = Q a T T f ⟨ S ⊗ S ⟩ AB jν E a iµ i ρA j µ ρB ν . (17.5.10) j,ν i,µ = Q a T T f ⟨ S S ⟩⟨ S S ⟩ Therefor, we can get A AB j F a i jν E a iµ µ ρB ν . (17.5.11) µ,ν a T T f = Q a T T f ⟨ S S ⟩ A 17.5.3.3 The properties of F a 1 Hermitian. With E.Q.(17.5.8) and E.Q.(17.5.11), from

AB † AB † ) E a E a and ρB ρB, we can see = = † F a F a. (17.5.12) A 2 Positivity, i.e., F a 0. From the diagonal= form of ρB, i.e., ρ ) ρB P≥µ µ B µ , with Pµ 1, and 0 Pµ 1, (17.5.13) µ µ combine it with= Q E.Q.(17.5.8),S ⟩ ⟨ S and we canQ get= ≤ ≤ ψ F A ψ ψ tr E AB I ρ ψ A a A A B a A B A ψ ν E AB µ ψ µ ρ ν A B a B A B B B a T T f = µ,νa T ( ⊗ ) T f = Q ⟨ψS aν TE AB T µf Sψ⟩ P⟨δ S S ⟩ A B a B A µ µν µ,ν = Q ⟨ψS aµ TE AB T µf Sψ⟩ P . (17.5.14) A B a B A µ µ Thus = Q ⟨ S a T T f S ⟩ ψ F A ψ 0, (17.5.15) A a A since ABa T T f ≥ E a 0, and Pµ 0.

≥ ≥ 178 3 Completeness.

A AB ) F a trB E a I A ρB a a

Q = Q AB( ⊗ ) trB E a I A ρB a = trB IQA ρB ( ⊗ ) I A trBρB = ( ⊗ ) I A. (17.5.16) = ⊗ 17.5.3.4 Example = We are going to talk about an example that has been discussed at the beginning of A section 17.4. The POVM denoted as F a a 1, 2, 3 , have the explicit expression 2 F A n n S, =with a 1, 2, 3. (17.5.17) a 3 a a = S⃗ ⟩⟨⃗ S = And the vectors na a 1, 2, 3. are shown in Figure 17.1, from which we can see

⃗ S = n1 n2 n3 0. (17.5.18)

Now, we introduce an auxiliary Hilbert⃗ + space⃗ + ⃗H=B and the density matrix ρB on HB is deﬁned as

ρB 0 B 0 . (17.5.19)

Therefore the density matrix of the compound∶= S ⟩ Hilbert⟨ S space HA HB is

ρAB ρA ρB ⊗

ρA 0 0 . (17.5.20) ∶= ⊗ B We can construct the two-qubit state as= ⊗ S ⟩ ⟨ S

2 1 Φ n 0 0 1 , with a 1, 2, 3. (17.5.21) a AB ¾3 a A B ¾3 A B S ⟩ ∶= S⃗ ⟩ S ⟩ + S ⟩ S ⟩ = It can be easily veriﬁed that Φa AB a 1, 2, 3. makes an orthonormal basis for the compound Hilbert space HA HB. S ⟩ S = AB Φa′ Φa AB⊗ 2 1 2 1 ⟨ S n ′⟩ 0 0 1 n 0 0 1 ¾3 A a B ¾3 A B ¾3 a A B ¾3 A B ⎛ ⎞ ⎛ ⎞ = 2 ⟨⃗ S ⟨ 1S + ⟨ S ⟨ S S⃗ ⟩ S ⟩ + S ⟩ S ⟩ ⎝ A na′ na , ⎠ ⎝ ⎠ 3 A 3 where = ⟨⃗ S ⃗ ⟩ + 1 , if a a′; 2 A na′ na A (17.5.22) ⎪⎧ ′ ⎪ − 1, if a ≠ a . ⟨⃗ S ⃗ ⟩ = ⎨ ⎪ With another two-qubit state deﬁned as ⎪ = ⎩⎪ Φ0 AB 1 A 1 B , (17.5.23)

S ⟩ ∶=179S ⟩ ⊗ S ⟩ we can get ′ Φa′ Φa δaa′ , a, a 0, 1, 2, 3 . (17.5.24)

With the orthonormal basis⟨ ΦSa AB⟩ =a 0, 1,∀2, 3. ,∈ we{ can construct} the projectors in the following manner {S ⟩ S = } AB E a Φa AB Φa 2 1 2 1 = S ⟩ n⟨ S 0 0 1 n 0 0 1 ¾3 a A B ¾3 A B ¾3 A a B ¾3 A B ⎛ ⎞ ⎛ ⎞ = 2 S⃗ ⟩ S ⟩ + S ⟩ S1⟩ ⟨⃗ S ⟨ S + ⟨ S ⟨ S ⎝ na 0 A na B 0 0⎠ ⎝1 A 0 B 1 ⎠ 3 A B 3 A B = 2S⃗ ⟩ S ⟩ ⟨⃗ S ⟨ S + 2S ⟩ S ⟩ ⟨ S ⟨ S na 0 A 0 B 1 0 1 A na B 0 , √3 A B √3 A B namely + S⃗ ⟩ S ⟩ ⟨ S ⟨ S + S ⟩ S ⟩ ⟨⃗ S ⟨ S 2 1 E AB n n 0 0 0 0 1 1 a 3 a A a B 3 A B = S⃗2⟩ ⟨⃗ S ⊗ S ⟩ ⟨ S + S ⟩2 ⟨ S ⊗ S ⟩ ⟨ S na 0 0 1 0 na 1 0 , (17.5.25) √3 A B √3 A B AB + S⃗ ⟩ ⟨ S ⊗ S ⟩ ⟨ S + S ⟩ ⟨⃗ S ⊗ S ⟩ ⟨ S for a 1, 2, 3. And E 0 on the other hand, is deﬁned as

AB = E 0 Φ0 AB Φ0 1 1 1 1 . (17.5.26) ∶= S A⟩ ⟨ S B Therefore, = S ⟩ ⟨ S ⊗ S ⟩ ⟨ S

AB E a I A ρB AB E I A 0 0 a ( ⊗ )B 2 2 = na( n⊗aS ⟩ 0⟨ S) 0 0 na 1 0 , (17.5.27) 3 A B √3 A B for a 1, 2, 3, and = S⃗ ⟩ ⟨⃗ S ⊗ S ⟩ ⟨ S + S ⟩ ⟨⃗ S ⊗ S ⟩ ⟨ S AB E a I A ρB 0. (17.5.28) = Thus, ( ⊗ ) = 2 F A tr E AB I ρ n n , (17.5.29) a B a A B 3 a A a for a 1, 2, 3, and = ( ⊗ ) = S⃗ ⟩ ⟨⃗ S A F 0 0. (17.5.30) = Therefore, the relation 17.5.8 has been veriﬁed.=

A ρA A Φ Φ (17.5.31) ρ 2 a AB a A 3 na na 0 B 0 = ⊗ S ⟩ ⟨ S S⃗ ⟩⟨⃗ S B S ⟩ ⟨ S

180 17.5.4 Direct-sum realization of POVM

Thm 17.5.4.1 (Neumark’s theorem). Given one-dimensional non-negative POVM F a . ⊥ It can be realized by extending HA to HA′ , HA HA′ , such that HA′ HA HA , where ⊥ { } HA is orthogonal to HA. ⊆ = ⊕ Proof. See Preskill’s lecture notes, Chapter 3.1.4.

17.5.5 POVM as quantum operation (superoperator) A Thm 17.5.5.1. Given one-dimensional non-negative POVM F a on HA. It can be realized by a unitary transformation on HA HB denoted as U AB followed with a orthogonal B { } measurement Πµ on HB, namely ⊗ A B ′ trA ρAF a trAB ΠµρAB , (17.5.32) where ( ) = ( ) ′ † ρAB U ABρABU AB. (17.5.33) = A ρA A (17.5.34) U AB ρA F A a 0 0 ΠB = ⊗B µ B S ⟩ ⟨ S

Proof. The basis of the Hilbert space HB is denoted as µ B ν B .

B ′ B ′ trB ΠµρAB B µ ΠµρAB µ B {S ⟩ } = {S ⟩ } µ ( ) = ⟨ S B S ⟩ ′ Q B µ Πµ ν B ν ρAB µ B µ,ν ′ = Qδµν⟨ BS ν ρS AB⟩ ⟨µ SB S ⟩ ν ′ = BQµ ρAB⟨ µS B S ⟩ † B µ U AB ρA 0 B 0 U µ B = ⟨ S S ⟩ AB µ U 0 ρ 0 U † µ = B⟨ S AB( B ⊗ AS ⟩B ⟨ S) AB S B⟩ M ρ M † , (17.5.35) = ⟨ µS A Sµ ⟩ ( ) ⟨ S S ⟩ where = M µ B µ U AB 0 B. (17.5.36)

B ′ ∶= ⟨ S S ⟩ † trAB ΠµρAB trA M µρAM µ trA F AρA , (17.5.37) ( ) = ( ) where the POVM is deﬁned as = ( ) A † F a MµMµ. (17.5.38)

∶=

181 † Check: Due to MµMµ 0, F µ 0.

† † B≥ 0 U AB≥µ B µ U AB 0 B B 0 UABUAB 0 B µ

Q ⟨ S S ⟩ ⟨ S S ⟩ = B⟨0SI AB 0 B S ⟩

I A, (17.5.39) = ⟨ S S ⟩ therefore = † F µ Id M µM µ IdId. (17.5.40) µ µ Remark: Quantum operation.Q = ⇐⇒ Q =

S ′ † ρA ρA S ρA M µρAM µ, (17.5.41) µ → = ( ) = where Q † M µM µ IdId. (17.5.42) µ It is called the Kraus representation ofQ quantum operation,= and details see following section 17.6.4.

17.5.6 Realization of a POVM Problem description1:

Consider the POVM deﬁned by the four positive operators 1 1 P ,PP , 1 2 z z 2 2 z z 1 1 P = S ↑ ⟩⟨↑ S,PP = S ↓ ⟩⟨↓ S . (17.5.43) 3 2 x x 4 2 x x Show how this POVM can be realized= S as↑ ⟩⟨ an↑ orthogonalS = measurementS ↓ ⟩⟨↓ S in a two-qubit Hilbert space, if one ancilla spin is introduced. For convenience, we apply the notations

0 z , 1 z , x , x , (17.5.44) in the following discussions.S ⟩ = S ↑ ⟩ S ⟩ = S ↓ ⟩ S+⟩ = S ↑ ⟩ S−⟩ = S ↓ ⟩

○ 1 Check P i i 1, 2, 3, 4 indeed as a POVM. 1 { S = } P 2 P (17.5.45) i 4 i

implies that P i is not a projector. = With the new notation 1 P k k , (17.5.46) k 2 where k 0, 1, , . For ψ , we have ∶= S ⟩⟨ S 1 = + − ∀S ⟩ ψ P ψ ψ k 2 0. (17.5.47) k 2 1Originated from the exercise 3.1, Chapter⟨ S 3 ofS ⟩ John= Preskill’sS⟨ S ⟩S ≥ online lecture notes.

182 P i sum to the identity, namely 1 P P P P 0 0 1 1 1 2 3 4 2 1 1 + + + = I(S ⟩⟨ SI+ S ⟩⟨ S + S+⟩⟨+S + S−⟩⟨−S) 2 2 2 2 = I 2. + (17.5.48)

○ = 2 Construct the projective measurement E i i 1, 2, 3, 4 . 1 1 E 00 00{,ES E= 10 }10 , 1 2 2 2 1 1 E = S 1⟩⟨ S1 ,EE = S ⟩⟨1 S 1 . (17.5.49) 3 2 4 2 = S + ⟩⟨+ S = S − ⟩⟨− S Check E i i 1, 2, 3, 4 as projective measurement.

2 { S = } E i E i; (17.5.50)

† E i = E i; (17.5.51)

E iE j =0, i j; (17.5.52) E 1 E 2 E 3 E 4 00 00 10 =10 ≠1 1 1 1 I 4. (17.5.53) ○ 3 Realize POVM+ + P i+on Hilbert= S ⟩⟨ spaceS + S H⟩⟨A byS + S projective+ ⟩⟨+ S + measurementS − ⟩⟨− S = E i on Hilbert space HA HB.

In Hilbert space HA HB, construct the density matrix ρAB as ⊗ 1 ⊗ ρ ρ I . (17.5.54) AB A 2 B ∶= ⊗ Check the relation trAB E iρAB trA P iρA one by one.

trAB( E 1ρAB) = ( AB 00) ρAB 00 AB 1 00 ρ I 00 ( ) = AB⟨ S A S 2⟩ B AB 1 = ⟨ 0 ρS 0⊗ S ⟩ 2 A A A 1 = tr ⟨ S 0 S 0⟩ρ A 2 A = trA(PS1ρ⟩⟨A S; ) (17.5.55) 1 tr E ρ 10 ρ = I (10 ) tr P ρ ; (17.5.56) AB 2 AB AB A 2 B AB A 2 A 1 ( tr )E= ρ⟨ S ⊗ ρS ⟩ = tr (P ρ) . (17.5.57) AB 3,4 AB 2 A A A A 3,4 A ( ) = ⟨ ± S S±⟩ = ( )

183 17.6 Quantum operation (superoperator)

17.6.1 Deﬁnition of the superoperator Question: What’s the time evolution equation of density matrix?

y As we know, for the pure state ψ , ρ ψ ψ , ψ satisﬁes the Shr¨odingerEquation, which implies it will evolve under unitary transformation, i.e., S ⟩ = S ⟩⟨ S S ⟩ U ψ U ψ , (17.6.1)

U † † ρ ψ S ψ⟩ → U Sψ⟩ ψ U UUρρρUU . (17.6.2)

y In the case of mixed state∶= describedS ⟩⟨ S → by ρ,S we⟩⟨ canS assume=

S ρ ρ′ S ρ , (17.6.3)

where S is called “superoperator”. We→ are= now( going) to explore some properties of the superoperator S. As for sure both ρ and ρ′ are density matrices, therefore S is a mapping between density matrices. The supperoperator has the following constrains:

• a linear mapping on density matrices, i.e.,

S λ1ρ1 λ2ρ2 λ1S ρ1 λ2S ρ2 , with λ1, λ2 C; (17.6.4)

• trace-preserving( + mapping) = (TP),( ) namely+ ( ) ∈ tr S ρ tr ρ 1; (17.6.5)

• positive mapping (P), i.e., [ ( )] = ( ) = S ρ 0. (17.6.6)

Therefore, S should be a linear positive trace-preserving( ) ≥ operator.

17.6.2 The wonderful theorem The following statements are equivalent. 1 S ρ is a CPTP mapping (Complete-Positive-trace-preserving).

) ( ) S ρ 0, Positive; (17.6.7) ⎧ S Id ρAB 0, Complete Positive. ⎪ ( ) ≥ ⎨ ⎪ 2 The Kraus representation⎩⎪( theorem.⊗ ) ≥

) Thm 17.6.2.1 (Kraus representation). A CPTP mapping S ρ has the operator sum representation, † ( ) S ρ M µρρMM µ, (17.6.8) µ

with the Kraus operators M µ satisfying( ) = Q

† M µM µ Id. (17.6.9) µ Q = 184 3 The Stinespring representation

) † S ρA trB U AB ρA 0 B 0 U AB (17.6.10)

with the diagram representation( ) = ( ⊗ S ⟩ ⟨ S)

A ρA S ρA U AB B 0 0 (17.6.11) ∶ B ( ) . ∶ S ⟩ ⟨ S

The symbol stands for the recycle bin, which means that the corresponding output qubit is out of our interest. Math tool: Function Analysis.

4 The Choi-Jamiolkowski representation

′ ) ρAB S Id Ψ AB Ψ , (17.6.12)

where Ψ AB is the maximal bipartite∶= ( entangled⊗ )(S ⟩ state⟨ S) deﬁned as

S ⟩ 1 Ψ AB ii AB (17.6.13) N i S ⟩ = √ Q S ⟩ with N as the dimension of the Hilbert space H . The superoperator can be uniquely determined in the way

′ T S ρA NtrB ρAB I A ρA , (17.6.14)

T where ρA is deﬁned as the transposition( ) = of the( density⊗ ) matrix ρA.

17.6.3 The CPTP mapping

S ρA is positive, and S I B ρAB is also positive.

Remarks( ) : ( ⊗ )

• Separable system ρAB ρA ρB. ρA S S ρA 0 (17.6.15) = ⊗ ρB ρB. ( ) ≥ • Entangled state. S S ρA 0 ρAB (17.6.16) ρ . B( ) ≥ The CPTP is required for the entangled system after the superoperator acting on the subsystem to ensure the validity of the density matrix.

• Diﬀerence between CPTP mapping and PTP mapping. For example, transpose is PTP mapping, but not CPTP mapping. We can see that

T ρ 0 ρT 0, (17.6.17)

∶ ≥ ↦ ≥ 185 i.e., transpose is PTP mapping. On the other hand, we can consider a density matrix ρAB, 1 ρAB ii AB jj . (17.6.18) N i,j = Q S ⟩ ⟨ S If we denote the transpose as T , then T I B means the partial transpose acting on subsystem A. ′ ⊗ ρAB T I B ρAB 1 = ( ⊗ T) I B ii AB jj N i,j = 1 Q( ⊗ )S ⟩ ⟨ S ji AB ij N i,j = 1 Q S ⟩ ⟨ S SWAP (17.6.19) N As we know that = SWAP 2 Id SWAP Id SWAP Id 0, (17.6.20)

thus SWAP( has eigenvalues) = ⇒ 1 and( 1. It− suggests)( SWAP+ )is= not positive, namely ′ ρAB T I B ρAB is not positive. Therefore, T I B is not CPTP mapping. − 17.6.4 The= ( ⊗ Kraus) representation ⊗

′ † ρ S ρ M µρρMM µ, (17.6.21) µ with ∶= ( ) = Q † M µM µ Id. (17.6.22) µ Examples: Q = (i) Unitary evolution of closed system, ρ UUρρρUU †, (17.6.23)

with ↦ † † M 1 UU,MM 1 U ; † (17.6.24) ⎧M 2 0,MM 0. ⎪ = 2 = ⎨ It can be easily veriﬁed that ⎪ ⎩⎪ = = † † M 1M 1 M 2M 2 Id. (17.6.25)

(ii) Projective measurement. + =

† Πn n n , ΠnΠn Id. (17.6.26) n

= S {⟩⟨Π }S Q = n ′ † ρ ψ ψ ρ S ρ ΠnρΠn. (17.6.27) n ∶= S ⟩⟨ S Ð→ = ( ) = Q (iii) POVM with one-dimensional operators F a . ′ ρ ρ F aρ F a, with{F a} 0, and F a 1. (17.6.28) a » » a ↦ = Q ≥ Q = 186 17.6.5 The Stinespring representation

A ρA S ρA U AB B 0 0 (17.6.29) ∶ B ( ) . ∶ S ⟩ ⟨ S The Stinespring representation is actually equivalent with Kraus representation.

′ † ρAB ρA 0 B 0 ρAB U ABρABU AB, (17.6.30) therefore ∶= ⊗ S ⟩ ⟨ S ↦ ∶=

′ ′ ρA trBρAB tr U ρ U † = B AB AB AB † = B µ U ABρABUAB µ B µ = Q ⟨ S S ⟩ † B µ U AB ρA 0 B 0 U AB µ B µ = Q ⟨ S ⊗ S ⟩ ⟨ S † S ⟩ B µ U AB 0 B ρA B 0 UAB µ B µ = ⟨ S †S ⟩ ( ) ⟨ S S ⟩ QM µρAM µ, (17.6.31) µ = Q where † † M µ B µ U AB 0 B ,MM µ B 0 U AB µ B . (17.6.32) Then = ⟨ S S ⟩ = ⟨ S S ⟩ † † M µM µ B 0 U AB µ B µ U AB 0 B µ µ Q = Q ⟨ †S S ⟩ ⟨ S S ⟩ B 0 U ABU AB 0 B I A. (17.6.33) = ⟨ S S ⟩ 17.6.6 Remarks on quantum= operation Remarks: 1○ The Kraus representation is not unique and is base-dependent. In the construction of POVM as quantum operation, it has the Kraus representation

A † F a MµMµ (17.6.34)

with ∶= M µ B µ U AB 0 B. (17.6.35)

With the new basis deﬁned as = ⟨ S S ⟩

B ν UνµB µ UνµM µ, (17.6.36) µ µ ⟨ S = Q ⟨ S = Q the superoperator has the Kraus representation

† S ρA N νρAN ν, (17.6.37) ν ( ) = Q 187 where N ν B ν U AB 0 B UνµM µ. (17.6.38) µ

The Kraus representations of= the⟨ superoperatorS S ⟩ = QS, namely M µ and N ν, are equivalent, i.e., M µ N ν . ○ 2 The dimension{ } of≅ { Hilbert} space HA is N, namely dim HA N. Kraus operator as 2 the mapping from Hilbert space HA to HA, there at most N Kraus operators, i.e., 2 # M µ N . = ○ 3 { } = S ρ ρ′ S ρ . (17.6.39) In general, S −1 ρ′ dose not exist! Open system changes usually for decoherence or → = ( ) information loss. ( ) Note: In closed system, S −1 ρ′ exists, since

′ ′ (ρ ) UUρρρUU †, ρ U †ρ UU. (17.6.40)

= = 17.7 Quantum channel

Example for Quantum Operation: ρ S ρ′ S ρ , (17.7.1) where the superoperator S is also called a quantum channel. = ( ) 17.7.1 The bit-ﬂip channel The quantum bit-ﬂip channel:

X X 0 1 , 1 0 , with X σx. (17.7.2)

It stands for the quantumS ⟩ error-model:↦ S ⟩ S ⟩ ↦ S ⟩ =

p 1 p 0

0 S ⟩ , 1 S ⟩ , (17.7.3) 1 p 1 p S ⟩ 0 S ⟩ 1 − − where p is the error-probability, 0 pS ⟩1, namely S ⟩

S 0 ≤0 ≤ 1 p 0 0 p 1 1 ; (17.7.4a) S 1 1 p 0 0 1 p 1 1 . (17.7.4b) (S ⟩⟨ S) = ( − )S ⟩⟨ S + S ⟩⟨ S Therefore (S ⟩⟨ S) = S ⟩⟨ S + ( − )S ⟩⟨ S ρ ρ′ S ρ 1 p ρ pXXXρρρXX (17.7.5) We can then ﬁnd the Kraus operators ↦ ∶= ( ) = ( − ) + 1 0 M 1 pII 1 p ; 0 2 0 1 ⎪⎧ » » (17.7.6) ⎪ = − = −0 1 ⎪M 1 pXX p , ⎨ 1 0 ⎪ √ √ ⎪ = = ⎩⎪ 188 and verify that

† † 2 2 M 0M 0 M 1M 1 1 p I 2 pXX 1 p I 2 pII 2 + = ( − )( ) + I 2. (17.7.7) = ( − ) + 1) Unitary Representation (Stinespring representation).=

A ψ A S ψ A ψ U AB B 0 ∶ S ⟩B (S ⟩ ⟨ S) (17.7.8) , ∶ S ⟩ where

U AB ψ A 0 B M 0 I 2 iMM 1 X ψ A 0 B 1 p ψ 0 i pXX ψ 1 . (17.7.9) (S ⟩ ⊗ S ⟩ ) ∶= ( ⊗ +A ⊗B )S ⟩ ⊗ S ⟩A B » √ Then = − S ⟩ ⊗ S ⟩ + S ⟩ ⊗ S ⟩ ′ ρA S ρA tr U ψ ψ 0 0 U † = (B ) AB A B AB

= trB 1 Sp ⟩ψ A⟨ S ⊗0S B⟩ ⟨i S pXX ψA 1 B » √ = ( 1− pSA ⟩ψ ⊗ SB ⟩0 + i pA Sψ⟩X⊗ SB ⟩1) » √ trB (1 p− ψ A⟨ Sψ⊗ ⟨0S B− 0 pX⟨X Sψ ⊗A ψ⟨ XS) 1 B 1

= i (p 1− )Sp tr⟩B ⟨ S ψ⊗ SA⟩ ψ⟨ XS + 0 BS ⟩1 ⟨XS ψ⊗AS ⟩ψ ⟨ S1 B 0 1»p ψ ψ pXX ψ ψ XX, (17.7.10) + ( − A) − S ⟩ ⟨A S ⊗ S ⟩ ⟨ S + S ⟩ ⟨ S ⊗ S ⟩ ⟨ S and = ( − )S ⟩ ⟨ S + S ⟩ ⟨ S † † U ABU AB M 0 I 2 iMM 1 X M 0 I 2 iMM 1 X M †M I M †M X 2 i M †M X M †M X = ( 0 ⊗0 +2 ⊗1 )1 ( ⊗ + 1 0⊗ ) 0 1 M †M M †M I i M M X M M X = 0 ⊗0 +1 1 ⊗2 − 1 0 ⊗ − 0 1 ⊗ I I i M M X M M X = (2 2 + 1 0 )⊗ − 1 0 ⊗ − ⊗ I 4, (17.7.11) = ⊗ − ⊗ − ⊗ where we have applied= the relations of the Kraus operators expressed as † M 0 M 0, (17.7.12a) † ⎧ M M 1, (17.7.12b) ⎪ 1 = ⎨ as is shown in E.Q.(17.7.6). Thus, we⎪ can also verify ⎩⎪ = † U ABU AB I 4. (17.7.13) Therefore, = U AB M 0 I 2 iMM 1 X (17.7.14) is unitary. ∶= ⊗ + ⊗

189 2) The Bloch sphere representation. 1 ρ Id p Z (17.7.15) 2 3 with = ( + ) p3 1, p p3eˆ3, (17.7.16)

and S S ≤ ⃗ = Z σ3. (17.7.17)

Therefore, = S ρ 1 p ρ pXXXρρρXX 1 Id p′ Z , (17.7.18) ( ) = (2 − ) +3 with = ( + ) ′ p3 1 2p p3, (17.7.19)

since ∶= ( − ) 1 p p 1 p ρ pXXXρρρXX Id p Z X Id p Z X 2 3 2 3 1 − p p ( − ) + = (Id + p Z) + Id( p+ Z ) 2 3 2 3 1 − = Id( 1+ 2p )p+Z (. − ) (17.7.20) 2 3 Qubit-flip error model means contraction= [ + in( the− Bloch) ] ball instead of inflation, i.e., ′ p3 p3. We may conclude here that the inflation is not permitted here in Quantum Mechanics. ≤ 17.7.2 The phase-flip channel Next, we are going to discuss the phase-flip error, which exists in Quantum Mechanics and has no classical analogy. Phase-flip error: 0 0 , 1 1 . (17.7.21)

Therefore, the phase-ﬂip channelS ⟩ should→ S ⟩ haveS the⟩ → form− S ⟩ S ρ 1 p ρ pZZZρρρZZZ. (17.7.22)

And we can deﬁne the bit-phase-ﬂip( ) error= ( − as ) +

S ρ 1 p ρ pYY ρρYYY, with Y σy. (17.7.23)

17.7.3 Depolarizing( channel) = ( − ) + = Depolarization suggests the process

Polarization ρ Non polarization ρ (17.7.24) with the formalism ( ) ↦p − ( ) S ρ 1 p ρ XXρρρXX Y ρρYY ZZρρρZZ , (17.7.25) 3 ( ) = ( − ) + ( + + ) 190 and the Kraus operators deﬁned as

p p p M 1 pII ,MM XX,MM YY,MM ZZ. (17.7.26) 0 2 1 3 2 3 3 3 » ½ ½ ½ We can verify∶= that − ∶= ∶= ∶=

3 † † † † † M iM i M 0M 0 M 1M 1 M 2M 2 M 3M 3 i=0 p p p Q = 1 p I+ I + I I+ 2 3 2 3 2 3 2 I = I(2.− ) + + + The Bloch Sphere representation.= 1 ρ Id p σ , with p 1 and p 3. (17.7.27) 2 R Therefore, = ( + ⃗⋅⃗) Y⃗Y ≤ ⃗ ∈ 1 S ρ S 1 p σ 2 1 ( ) = S( I+ ⃗⋅⃗S) p σ 2 2 1 p = I ( )1+ p(⃗p⋅⃗σ) Xp σσXX Y p σσYY Zp σσZZ 2 2 3 1 p = I + (1− p)⃗p⋅⃗σ + ( p⃗⋅⃗ p + p ⃗⋅⃗X + p⃗⋅⃗ p) p Y p p p Z 2 2 3 1 1 1 2 2 2 3 3 3 1 2p = I + (1 −p)p⃗⋅σ⃗ + p( σ − − ) + (− + − ) + (− − + ) 2 2 3 1 = I + (p′ −σ ,)⃗⋅⃗ − ⃗⋅⃗ (17.7.28) 2 2 with = + ⃗ ⋅⃗ 4p p′ 1 p. (17.7.29) 3 ⃗ ∶= − ⃗ ′ 3 Depolarization is also a contraction process, due to p p . When p 4 , 1 S ρ I , S⃗ S < S⃗S = (17.7.30) 2 2 indicated no-polarization at all. ( ) =

17.7.4 The phase-damping channel 2 S † ρ S ρ M iρρMM i , (17.7.31) i=0 with ↦ ( ) = Q M 0 1 pII 2,MM 1 p 0 0 ,MM 2 p 1 1 . (17.7.32) » √ √ Therefore, with the matrix∶= expression− of∶= the densityS ⟩⟨ S matrix∶=ρ, S ⟩⟨ S ρ ρ ρ 00 01 , (17.7.33) ρ10 ρ11 ∶= 191 we can get ρ 0 S ρ 1 p ρ p 00 0 ρ11 ( ) = ( − ρ) + 1 p ρ 00 01 . (17.7.34) 1 p ρ10 ρ11 ( − ) = As we shall see that from ρ to S ρ , ( − ) • the diagonal terms are unchanged;( ) • the non-diagonal terms decay with the factor 1 p .

Therefore, in the manner shown as ( − ) n S ρ S 0S 0 S 0 ρ n ( ) ∶= ⋯ n ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶ρ00 1 p ρ01 n , (17.7.35) 1 p ρ10 ρ11 ( − ) = with the new parameters deﬁned as ( − ) p Γ∆t, t n∆t, (17.7.36) which have the formalism = = ∆t→0 1 p n 1 Γ∆t t~∆t e−Γt, (17.7.37) we can ﬁnd immediately that( with− ) n= ( −, namely) tÐ→ , −Γt n ρ00→ ∞e t→∞ →ρ∞00 0 S ρ −Γt , (17.7.38) e ρ11 0 ρ11 i.e., the non-diagonal terms( vanish) = or the coherent Ð→ terms vanish. The geometric picture can be described as the Bloch ball contracted into a line on the poles, namely the contraction has the preferred bases 0 and 1 .

Note: In the Kraus operatorsS ⟩ ofS ⟩ phase-damping channel M i i 1, 2, 3 , we can rewrite the operators as { S = } p p M 1 pII ,MM I Z ,MM I Z . (17.7.39) 0 2 1 2 2 2 2 2 » √ √ Then ∶= − ∶= ( + ) ∶= ( − ) 2 † S ρ M iρρMM i i=0 p p ( ) = Q1 p ρ I Z ρ I Z I Z ρ I Z 4 2 2 4 2 2 1 1 = (1 − )p ρ+ ( pZZZρ+ρρZZZ.) ( + ) + ( − ) ( − ) (17.7.40) 2 2 And now we can redeﬁne= ( the− Karus) + formalism of phase-damping channel as

1 1 N 1 pII ,NN pZZZ. (17.7.41) 0 ¾ 2 2 1 ¾2

The Karus operator is not unique,∶= N−0,NN 1 M 0,M∶=M 1,MM 2 .

{ 192} ≅ { } 17.7.5 The amplitude-damping channel

S † † ρ S ρ M 0ρρMM 0 M 1ρρMM 1, (17.7.42) with ↦ ( ) = + 1 0 0 p M 0 ,MM 1 . (17.7.43) 0 1 p 0√ 0 ∶= √ ∶= We can verify that −

† † 1 0 1 0 0 0 0 p M 0M 0 M 1M 0 0 1 p 0 1 p p 0 0√ 0 + = 1√ 0 0 0√ + √ − − 0 1 p 0 p = 1 0 + − 0 1

= I2. (17.7.44)

And the amplitude-damping= channel works as

1 0 ρ ρ 1 0 S ρ 00 01 0 1 p ρ10 ρ11 0 1 p ( ) = 0 √ p ρ ρ 0√ 0 − 00 01 − 0√ 0 ρ10 ρ11 p 0 + ρ00 1 pρ01 √ pρ11 0 1 pρ10 √1 p ρ11 0 0 − = √ρ pρ 1 pρ + 00 − 11 ( − ) 01 , (17.7.45) 1 pρ10 √1 p ρ11 + − = √ i.e., − ( − ) ρ ρ 1 p ρ 1 pρ S ρ 00 11 11 01 . (17.7.46) 1 pρ10 √1 p ρ11 + − ( − ) − Thus ( ) = √ − n ( − )n~2 n ρ00 ρ11 1 p ρ11 1 p ρ01 S ρ n~2 n , (17.7.47) 1 p ρ10 1 p ρ11 + − ( − ) ( − ) n ( ) = with n , 1 p 0, i.e., ( − ) ( − )

→∞ ( − ) → n→∞ ρ ρ 0 S n ρ 00 11 . (17.7.48) 0 0 + ( ) Ð→ Example: If ρ a 2 0 0 b 2 1 1 , with a 2 b 2 1, (17.7.49) we can get ∶= S S S ⟩⟨ S + S 2S S ⟩⟨2 S S S + S S = n→∞ a b 0 1 0 S n ρ 0 0 . (17.7.50) 0 0 0 0 S S + S S Therefore, we can get the( pure) Ð→ state 0 0 from the= mixed = stateS ⟩⟨ ρS a 2 0 0 b 2 1 1 through the amplitude-damping channel. The geometric picture can be described as the Bloch ball contracted toward to the pointS ⟩⟨ S0 0 . = S S S ⟩⟨ S + S S S ⟩⟨ S

S ⟩⟨ S 193 17.7.6 The two-Pauli channel p p M 1 pII ,MM XX,MM ZZ. (17.7.51) 0 2 1 2 2 2 » ½ ½ In the Bloch sphere representation∶= − of the single-qubit∶= density∶= matrix 1 ρ I p σ , (17.7.52) 2 2 after the two-Pauli channel, ∶= ( + ⃗⋅ ⃗) p p S ρ 1 p ρ XXρρρXX ZZρρρZZ 2 2 1 ( ) = ( I− ) 1+ p p X+ 1 2p p Y 1 p p Z . (17.7.53) 2 2 1 2 3 And the parameter vector= p(changes+ ( − in) the way+ ( − ) + ( − ) )

(1) (1) (1) p1 1 ⃗ p p1, p2 1 2p p2, p3 1 p p3, (17.7.54) thus we can see = ( − ) = ( − ) = ( − ) 1 S (n) ρ I (17.7.55) 2 2 as n . The polarized state is transformed( ) → into unpolarized state, which implies that the information is completely lost. → +∞ 17.8 The master equation

The master Equation is also called Lindblud’s Equation. In general, we apply the CPTP mapping to describe the quantum operation. In special cases, we perform the approximation and view the evolution as Markovin process:

tm−1 tm tm+1 , (17.8.1) where state at time ti+1 only depends on state at time ti. Therefore, in Makovin process, ρA t dt depends only on ρA t . From Figure 17.2 we can see there are three time scales,

( + ) ( ) interaction open system A with ∆tA interaction

environment ∆tE observer ∆tO

Figure 17.2: Three time scale i.e., ∆tA, ∆tO and ∆tE. In Markovin process, we assume

∆tA ∆tO ∆tE, (17.8.2) which means that for the observer, the≫ physical≫ system is changing very slowly while the environment is changing very fast. This ensures that ρA t dt depends on ρA t only, so the diﬀerential equation is possible. ( + ) ( ) 194 1○ First-order approximation.

ρ dt ρ 0 O dt ρ 0 dρ 0 . (17.8.3)

2○ Kraus representation. ( ) = ( ) + ( ) = ( ) + ( )

† ρ dt S ρ 0 M µ dt ρ 0 M µ dt , (17.8.4) µ ( ) = ( ( )) = Q ( ) ( ) ( ) with † M µ dt M µ dt Id. µ Q ( ) ( ) = 3○ Assume

† † M 0 Id iHH K dt, with H HH,KK K; (17.8.5a) M L dt, with k 1, 2, 3,.... (17.8.5b) k = k + (− + ) = = √ 4○ = =

† ρ 0 dρ 0 Id iHH K dt ρ 0 Id iHH K dt Lkρ 0 Lkdt, (17.8.6) k ( ) + ( ) = [ + (− + ) ] ( )[ + ( + ) ] + Q ( ) then † dρ iHH K ρdt ρ iHH K dt LkρρLLkdt, (17.8.7) ν=1 i.e., = (− + ) + ( + ) + Q dρ 1 † HH,ρρ KK,ρρ LkρρLLk. (17.8.8) dt i ν With the requirement relation= [ ] + + Q

† M µM µ Id, (17.8.9) µ Q = we have the constrain relation

† iHH K iHH K LνLν 0, (17.8.10) ν (− + ) + ( + ) + = i.e., Q 1 † K LνLν. (17.8.11) 2 ν Hence, we can obtain = − Q

dρ 1 † 1 † HH,ρρ LνρρLLν LνLν,ρρ . (17.8.12) dt i ν 2 = [ ] + Q − Note: For the special case of K 0 and Lk 0, k 1, 2, 3,..., we can get dρ 1 = HH,ρ=ρ , = (17.8.13) dt i which is the time evolution of closed system= [ governed] by the Shcr¨odingerEquation.

195 Chapter 18

Notes on Finite Group Theory

196 Chapter 19

Notes on Stabilizer Formalism of Quantum Error Correction Codes

197 Part V

Fault-tolerant Quantum Computation

198 Chapter 20

Measurement-Based Quantum Computation

199 Chapter 21

Topological Quantum Computation

200 Chapter 22

Integrable Quantum Computation

Interested readers are invited to refer to two documents on the same homepage: the one is the published paper with the title of “Integrable Quantum Computation”; and the other one is the associated .ppt ﬁle of interpreting the published one.

201 Part VI

Research Projects

202 Chapter 23

Research Projects

203