Lecture Notes on Quantum Information and Computation

Transcription

1 Draft for Internal Circulations: v: Fall Semester, 0; v: Fall Semester, 03; v3: Fall Semester, 04; Lecture Notes on Quantum Information and Computation Yong Zhang School of Physics and Technology, Wuhan University (Fall 04) 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. 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 Peng Undergraduate student (Id: ), Participant in the Fall semester, 03. * The third version of these notes are typeset by Kun Zhang 3 Graduate student (Id: ), Participant in the Fall semester, yong zhang@whu.edu.cn pengyi py@whu.edu.cn 3 kun zhang@whu.edu.cn

2 Acknowledgements Draft for Internal Circulations: v3: Fall Semester, 04 I thank all participants in class including advanced undergraduate students, first-year graduate students and French students for their patience and persistence 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 (997-present), preskill/ph9/ preskill/ph9/ lecture * [Nilesen & Chuang] Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information (Cambridge, 000&00) * [KLM] Phillip Kaye, Raymond Laflamme and Michele Mosca, An Introduction to Quantum Computing (Oxford, 007). * [Zhang] Yong-De Zhang (University of Science & Technology of China), Principles of Quantum Information Physics (in Chinese) (Science Press, 009). Main References to Homeworks * [Preskill] John Preskill (Caltech): online lecture notes on QIC (997-present), preskill/ph9/ lecture

3 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

4 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 a quantum computer. 4

5 Contents I Introduction to Quantum Information and Computation 0 Introduction. Reasons to learn Quantum Information and Computation What s Quantum Information and Computation? Research topics Quantum Mechanics (I): Axioms 4. Axioms of quantum mechanics for closed system State Observable Projective measurement Schrödinger equation Composite system Single-Qubit and Two-Qubit Systems 9 3. Overview Pure state formalism of a qubit Single-qubit gate SU() in the spin-/ case Spin-/ operator and Pauli matrices Spinor representation of SU() group Properties Physical realization of qubit Bell states Notation Parity-bit (i) and Phase-bit (j) Orthonormal basis of the two-qubit Hilbert space How to distinguish (create) ψ(i, j) No-Cloning, Dense Coding, Teleportation and Cryptography 5 4. No-cloning theorem Dense coding Quantum teleportation The quantum teleportation using continuous variables Quantum cryptography (information security) Classical cryptography Quantum key distribution (QKD) BB84 quantum key distribution

6 5 Bell Inequalities 4 5. Einstein s quantum mechanics: local hidden variable theory (LHV) What hidden variable (HV) theory? What local theory? The rule to justify the rightful theory Bell s inequality in the local hidden variable theory Bell s inequality in quantum mechanics The CHSH inequality Hints for the violation of the Bell inequality Hardy s theorem The GHZ theorem II Quantum Computing and Quantum Algorithm 57 6 Classical Circuit and Quantum Circuit Classical circuit Universal gate set Elementary logical gates Universal gate set Reversible classical computation Irreversible computation Classical reversible gate Three-bit Toffoli gate Three-bit Fredkin gate The construction of an n-bit Toffoli gate using the 3-bit Toffoli gate Quantum circuit model Definition of quantum circuit One-qubit gates Hadamard gate: H Controlled two-qubit gates and controlled three-qubit gates CNOT gate Quantum Toffoli gate and Fredkin gate Quantum circuit model of Bell states Quantum circuit model of GHZ states Universal quantum computation Quantum universal gate set Universal quantum gate set of two-qubit gates Deutsch s gate is a universal quantum gate Quantum Algorithms 9 7. Classical and quantum algorithm Oracle model Deutsch s algorithm Definitions Classical algorithm Deutsch s problem Phase kick-back Deutsch s algorithm

7 7.4 Deutsch-Jozsa s algorithm Constant and balanced function in n-qubit Constant or balanced function? Notation and lemma Deutsch-Jozsa s algorithm Bernstein-Vazirani s algorithm Simon s algorithm Grover s algorithm Overview of the problem Grover s algorithm Example: N Quantum circuit model of Grover s algorithm III Density Matrix and Quantum Entanglement 8 Quantum Mechanics (II): Density Matrix 8. Density matrix as state of quantum open system State ensemble formalism of density matrix Operator formalism of density matrix Reduced density matrix (State for subsystem) Mixed state formalism of a qubit Why polarization vector? Pure state and mixed state in two-dimensional Hilbert space H Convexity of density matrix Two-qubit system and its subsystem Example: EPR pair (Bell state) Maximally entangled two-qubit pure states Monogamy of maximal entanglement Schmidt Decomposition, Purification and GHJW Theorem 7 9. Introduction Schmidt decomposition and quantum entanglement Schmidt decomposition Quantum entanglement Proof for the theorem of the Schmidt decomposition Example for the Schmidt decomposition The purification theorem and GHJW theorem Purification The GHJW theorem Information is physics Mixed State Entanglement and Multi-partite Entanglement Bipartite mixed state entanglement Separability Bipartite pure states Bipartite mixed state Quantum bipartite entanglement Positive-partial transpose (PPT) criterion for quantum bipartite separability

8 0..4 Example: the Werner state and the PPT criterion Example: the Werner state and the CHSH inequality Multi-partite entanglement Definition The GHZ state Properties of GHZ states IV Quantum Open System and Quantum Error Correction Codes 5 Quantum Mechanics (III): Quantum Open System 53. Introduction Why we talk about quantum open system Closed system and open system Projective measurement General measurement theory Definition of POVM More on POVM Dimensional analysis POVM on subsystem can be viewed as projective measurements on the entire system Tensor product realization of POVM Operator formula of F a Matrix formalism of F a The properties of F A a Example Direct-sum realization of POVM POVM as quantum operation (superoperator) Quantum operation (superoperator) Definition of the superoperator The wonderful theorem The CPTP mapping The Kraus representation The Stinespring representation Remarks on quantum operation Quantum channel The bit-flip channel The phase-flip channel Depolarizing channel The phase-damping channel The amplitude-damping channel The master equation Notes on Finite Group Theory 74 3 Notes on Stabilizer Formalism of Quantum Error Correction Codes 75 8

9 V Selected Topics 76 4 Entanglement Measures and Entropy: Bi-Partite System 77 5 Quantum Circuit Complexity Circuit complexity Definitions Complexity class Quantum complexity Accuracy Integrable Quantum Computation 8 9

10 Part I Introduction to Quantum Information and Computation 0

11 Chapter Introduction References: [Preskill] Chapter : Introduction and overview; [Nielsen & Chuang] Chapter : Introduction and overview.. Reasons to learn Quantum Information and Computation For an advanced undergraduate major in physics, he or she has to learn Quantum Information 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 Quantum 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.. 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)

12 According to Rolf Landauer (96), Information is physical which says that information is something that is encoded in the states of physical systems. According to David Deutsch (985), 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.. Table.: Computers vs. Physical Systems Computer Computation Input Rules Output Physical System Motion Initial State Laws of Motion Final State.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. 03: 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.

13 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

14 Chapter 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 : Foundations I: states and ensembles; [Nielsen & Chuang] Chapter : Introduction to quantum mechanics.. 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 Measurement. 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... State Axiom... A state ψ or a ray {e iα ψ }, from the Hilbert space H, can make a complete description of a physical system (with no hidden variable). 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 + is physically different from 0 + e iα with e iα. Note that the superposition principle is defined only for state vectors, not for state rays. 4

15 Axioms of Quantum Mechanics Space State Closed Systems Eg: the entire universe pure state vector φ H pure state ray {e iα φ } H Open Systems Eg: subsystems of a composite system Hilbert space H density matrix (operator) ρ called mixed state Observable Measurement Dynamics self-adjoint operator A n a np n with {a n} R and {P n} being orthogonal projections projective measurement (orthogonal) A φ A φ unitary evolution Eg: i h d φ(t) H φ(t) Eg: dt general measurement (non-orthogonal) A tr(aρ) non-unitary evolution via superoperator i h d ρ(t) [H, ρ(t)] dt.. Observable An observable in physics is a property of a physical system that can be measured. Axiom... 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. (..) Let a n be one of the eigenvalues of A and a n is the associated eigenvector, A a n a n a n. (..) All the eigenvalues a n of A are real, while the eigenvectors { a n } form a complete orthogonal basis of the Hilbert space H. It would be easy to prove that the eigenvalues of the observable A is real, and the its eigenvectors are orthogonal. Proof. (i) The eigenvalues of the observable A are real. i.e., a n is real. A a n a n a n ( a n A a n ) a n A a n a n A a n } a n a n, (..3) (ii) The eigenvectors of the observable A are orthogonal. In the case of two eigenvectors with different eigenvalues, namely { A a k a k a k A a l a l a l (..4) with k l and a k a n. Therefore, a k A a l a l a k a l a k A a l a k a k a l } (a k a l ) a k a l 0 a k a l 0. (..5) 5

16 In the circumstance of degeneration, namely, there are at least two mutually independent eigenvectors of A, but with the same eigenvalue. The set of all these eigenvectors of A associated with the specific eigenvalue would span a subspace of the Hilbert space H. Thus, we can employ the Schmidt orthogonalization 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 a n of the Hilbert space H, namely as H an {A φ a n φ φ H }, (..6) P n Degeneracy k a (k) n a (k) n, (..7) with { a (k) n k,..., Degeneracy} be an orthonormal basis for the subspace H an. It s easy to see that Hence, we can expand the operator A as which is the spectral theorem...3 Projective measurement P n P n, P n P n, P n P m 0, if n m. (..8) A a n P n, (..9) n Axiom..3. For an observable A with eigenvalues a n and eigenvectors a n, given the system is in the state ψ, the probability of obtaining a n (in the non-degeneration case) as the outcome of the measurement of A is Prob(a n ) a n ψ. (..0) And the expectation value (mean value) of the observable A would be A ψ A ψ ψ ψ. (..) After the measurement, the system is left in the state within the subspace corresponding to the eigenvalue a n (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 a n a n, itself is a self-adjoint operator, which has the mean value, ψ P n ψ ψ ( a n a n ) ψ ψ a n a n ψ a n ψ Prob(a n ), (..) 6

17 where ψ is the normalized state vector of the physical system. P n ψ ψ P n ψ a n a n ψ a n ψ, (..3) ψ P n an an which is the post-measurement state. 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 a k with k n, in the same time. ψ (). Information acquirement: P n ψ. (3). Ture randomness: Projector ψ P n ψ with probability. P n ψ (). Information loss // wavepacket collapse: P m ψ killed (m n). (4). Irreversible, non-unitary process: ψ P n ψ, P n ψ / ψ...4 Schrödinger equation Axiom..4. The time evolution of a closed state is unitary. Equivalently, the time evolution of the state vector ψ(t) is governed by the Schrödinger equation: i h ψ(t) H(t) ψ(t), (..4) t where H(t) is the Hamiltonian operator of the physical system. From the Schrödinger equation we can see that the time evolution operator U (t + t, t) should be U (t + dt, t) ihdt h. (..5) As we shall see that U (t + dt, t)u (t + dt, t) U (t + dt, t)u (t + dt, t) I + O(dt ). 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 dt) U (dt, 0) ψ(0). (..6) In the special case of time-independent Hamiltonian H, we can write down the explicit expression of the time evolution operator U (t, 0), for simplicity which we denote as U (t), U (t) e iht/ h. (..7) 7

18 In the case of time-dependent Hamiltonian H(t), we have to use the Dyson s formula U (t) I + n n ( n! i h) see Google & Wiki for more details. t t t dt n dt n dt H(t n )H(t n ) H(t ), (..8) t n>t n > >t Remark: Why is the Schrödinger equation 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?..5 Composite system For system A, ψ A H A, and system B, ϕ B H B, the composite system of system A and system B is described by the tensor product of Hilbert spaces, i.e., ψ A ϕ B H A H B. 8

19 Chapter 3 Single-Qubit and Two-Qubit Systems References: [Preskill]Chapter : Foundations I: states and ensembles; [Preskill] Chapter 4: Quantum entanglement; [Nielsen & Chuang] Chapter : Introduction and overview; [Nielsen & Chuang] Chapter : Introduction to quantum mechanics. 3. 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. gate: gate { one-bit gate NOT two-bit gate AND, OR,... NOT gate: a in mod. AND gate: a b a b, OR gate: a b a b in mod. 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 H : ( α β ) α 0 + β, with α + β, α, β C, (3..) 9

20 where 0 could mean spin up and for mean spin down, i.e., this is a two-level physical system, and the spin- system is a typical example. quantum gate: quantum gate single-qubit gate: ( α β ) SU() ( α β ), two-qubit gate: ( α β ) ( α β ) SU(4) ( α β ) ( α β ). 3. Pure state formalism of a qubit The Hilbert space H can be spanned by basis { 0, }, i.e., ψ α 0 + β ( α β ), ψ H, with α, β C and α + β. We may notice that α and β being complex numbers means four real numbers (four degrees of freedom). With the constraint α + β, 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 reduce the degrees of freedom to be two. Therefore, the state vector ψ can be described by two real numbers (θ, ϕ), for example ψ + (θ, ϕ) cos θ 0 + eiϕ sin θ ( cos θ e iϕ sin θ ), with 0 θ < π, and 0 ϕ < π. (3..) The two real variables (θ, ϕ) can actually determine a unit vector in the three-dimensional Euclidean R 3, namely ˆn (sin θ cos ϕ, sin θ sin ϕ, cos θ), as shown in Figure 3.. And we z θ ˆn (sin θ cos ϕ, sin θ sin ϕ, cos θ) ϕ y x Figure 3.: Bloch sphere shall also see that with ˆn pointing along different directions, namely different values of θ and ϕ, the state vectors ψ + (θ, ϕ) are different: () ˆn ê x (, 0, 0), θ π, ϕ 0, ψ +( π, 0) ( 0 + ); () ˆn ê y (0,, 0), θ π, ϕ π, ψ +( π, π ) ( 0 + i ); (3) ˆn ê z (0, 0, ), θ 0, ϕ [0, π), ψ + (0, ϕ) 0. The vector ˆn in three dimensional space is called Bloch vector, and the unit boundary of the vector set is called Bloch sphere, i.e., ˆn. 0

21 3.. Single-qubit gate SU() in the spin-/ case 3... Spin-/ operator and Pauli matrices The spin-/ operator can be expressed as J h σ or J σ, (3..) if we set the reduced Planck constant h to be one. And the σ is the so-called Pauli vector, defined as σ σ x ê x + σ y ê y + σ z ê z (3..3) where σ x, σ y and σ z are the Pauli matrices, which have the standard form σ x σ ( 0 0 ), σ x σ ( 0 i i 0 ), σ z σ 3 ( 0 0 ). The Pauli matrices satisfy the following properties, the anticommutative relation {σ i, σ j } δ ij ; (3..4) the commutative relation [σ i, σ j ] iε ijk σ k, with i, j, k {,, 3} and k {i, j}. (3..5) 3... Spinor representation of SU() group We can define a unitary operator D (θ, n) for the spin-/ system which is induced by the rotation in the Euclidean space R 3 along the direction n through an angle of θ: which is equivalent to D (θ, n) exp ( iθ n J) (3..6) D (θ, n) exp ( i θ n σ) cos θ i n σ sin θ. (3..7) D (θ, n) is a single-qubit gate. And it has some interesting features, for instance D (π, n), D (4π, n). (3..8) Though this would be meaningless if it only gives the global phase, there can be significant effect if the relative phase is changed because of this.

22 3.. Properties Thm 3... If we define σ n σ n, then σ n ψ + (θ, ϕ) ψ + (θ, ϕ), with n (sin θ cos ϕ, sin θ sin ϕ, cos θ). (3..9) Proof. Let s firstly express σ n with the Pauli matrices, σ n sin θ cos ϕσ + sin θ sin ϕσ + cos θσ 3 sin θ cos ϕ ( 0 i 0 ) + sin θ sin ϕ (0 ) + cos θ ( 0 i 0 0 ) cos θ ( sin θ cos ϕ + i sin θ sin ϕ sin θ cos ϕ i sin θ sin ϕ ) cos θ namely cos θ sin θ(cos ϕ i sin ϕ) ( ), sin θ(cos ϕ + i sin ϕ) cos θ σ n ( cos θ sin θe iϕ sin θe iϕ cos θ ). (3..0) Therefore, we can calculate σ n ψ + (θ, ϕ) in the following way There we get E.Q. (3..9) proved. Thm 3... cos θ sin θe iϕ σ n ψ + (θ, ϕ) ( sin θe iϕ cos θ ) ( cos θ e iϕ sin θ ) cos θ cos θ + sin θe iϕ e iϕ sin θ sin θe iϕ cos θ cos θeiϕ sin θ ( cos θ e iϕ sin θ ) ψ + (θ, ϕ). (3..) ψ + (θ, ϕ) σ m ψ + (θ, ϕ) n m, (3..) with n (sin θ cos ϕ, sin θ sin ϕ, cos θ) and m R 3, m. Proof. In analogy to E.Q.(3..0), we can get the expression for σ m: cos θ sin θ σ e iϕ m ( sin θ e iϕ cos θ ), with m (sin θ cos ϕ, sin θ sin ϕ, cos θ ). (3..3)

23 Therefore, we can get ψ + (θ, ϕ) σ m ψ + (θ, ϕ) (cos θ e iϕ sin θ ) ( cos θ sin θ e iϕ sin θ e iϕ cos θ ) ( cos θ e iϕ sin θ ) (cos θ e iϕ sin θ ) (cos θ cos θ + sin θ sin θ e i(ϕ ϕ ) cos θ sin θ e iϕ sin θ cos θ e iϕ ) cos θ ( cos θ cos θ + sin θ sin θ e i(ϕ ϕ ) ) +e iϕ sin θ ( cos θ sin θ e iϕ sin θ cos θ e iϕ ) ( cos θ θ sin ) cos θ + sin θ cos θ sin θ (e i(ϕ ϕ ) + e i(ϕ ϕ ) ) cos θ cos θ + sin θ sin θ cos(ϕ ϕ ) n m, (3..4) where we have used 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. (3..5) There, E.Q. (3..) is verified, too. Remark: For the expression in terms of density matrix, we have where ρ ψ + (θ, ϕ) ψ + (θ, ϕ). tr (ρ( σ m)) n m, (3..6) Thm As we shall know from the definition of ψ + (θ, ϕ) that then where 0 z ψ + (0, ϕ), (3..7) ψ + (θ, ϕ) D(ê z n) 0, with n (sin θ cos ϕ, sin θ sin ϕ, cosθ), (3..8) Proof. Firstly, we should realize that D(ê z n) ( cos θ e iϕ sin θ e iϕ sin θ cos θ ). (3..9) where we define D (ê z n) D (θ, n xy) exp ( i θ σ n xy), (3..0) n xy (cos ϕ, sin ϕ, 0), n xy ( cos(ϕ + π ), sin(ϕ + π ), 0) ( sin ϕ, cos ϕ, 0), (3..) 3

24 z x ϕ θ n (sin θ cos ϕ, sin θ sin ϕ, cos θ) n xy ( sin ϕ, cos ϕ, 0) ϕ y n xy (cos ϕ, sin ϕ, 0) Figure 3.: Rotation in the Euclidean space i.e., n xy is a unit vector that is orthogonal to both n and ê z, and they can form a righthand coordinate system, see Figure 3., which means that the corresponding rotation taken place in the Euclidean space should be a rotation around n xy through an angle θ. And, now it would be easy to show that E.Q. (3..8) is right. E.Q. (3..0) can be rewritten as D ( e z n) cos θ i σ n xy sin θ cos θ i sin θ ( sin ϕσ + cos ϕσ ) And we can verify E.Q. (3..8) directly, D Remarks: ( cos θ 0 0 cos θ ) + sin θ ( 0 cos ϕ + i sin ϕ ) cos ϕ + i sin ϕ 0 cos θ sin θ e iϕ sin θ eiϕ cos θ. (3..) ( e z n) 0 cos θ sin θ e iϕ sin θ eiϕ cos θ ( 0 ) cos θ sin θ eiϕ ψ +(θ, ϕ). (3..3) () As we have shown above If we define then, we could verify that Firstly, σ n ψ + (θ, ϕ) ψ + (θ, ϕ), ψ + (θ, ϕ) D ( e z n) 0. (3..4) ψ (θ, ϕ) D ( e z n), (3..5) σ n ψ (θ, ϕ) ψ (θ, ϕ). (3..6) ψ (θ, ϕ) cos θ sin θ e iϕ sin θ eiϕ cos θ (0 ) sin θ e iϕ cos θ. (3..7) 4

25 Then, cos θ sin θe iϕ σ n ψ (θ, ϕ) ( sin θe iϕ cos θ ) sin θ e iϕ cos θ sin θ e iϕ cos θ + cos θ sin θe iϕ sin θ e iϕ sin θe iϕ cos θ cos θ sin θ e iϕ cos θ ψ (θ, ϕ). (3..8) () The spinor representation of the SO(3) group, D(R), satisfies D(R) x σd (R) x σ, (3..9) in which the vector x is rotated under x R x, i.e., x j R ijx j. Example: D(R) D ( e z n) gives It implies that 3..3 Physical realization of qubit D(R)σ 3 D (R) σ n. (3..30) σ n n D(R)σ 3 D (R)D(R) e z Electron D(R)σ 3 0 n. (3..3) Photon Spin Mass 0.5Mev 0 Qubit Spin-state Photon-polarization Note: Though the two-level quantum system is equivalent with the -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. 3.3 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 Computation, arxiv: quant-ph/

26 Y.Z., Jinglong Pang, Space-Time Topology in Teleportation-Based Quantum Computation, arxiv: Y.Z., Kun Zhang, Bell Transform, Teleportation Operator and Teleportation-Based Quantum Computation, arxiv: Notation Pauli matrices are unitary matrices defined in spin- space: σ x X ( 0 0 ) ; σ y iy ( 0 i i 0 ) ; σ z Z ( 0 0 ). (3.3.) There we get three quantum gates: quantum gate X, quantum gate Z and quantum gate Y ZX. Def 3.3. (Bell states). The following four double-qubit states φ + ψ(0, 0) φ ψ(0, ) ψ + ψ(, 0) ( 00 + ), ( 00 ), ( ), (3.3.) ψ ψ(, ) ( 0 0 ), are the so called Bell states. Lemma All the four Bell states defined in E.Q. (3.3.) can be expressed as ψ(i, j) (I X i Z j ) ψ(0, 0), with i, j 0,. (3.3.3) Proof. With the definition of the Bell states (3.3.), we can check E.Q. (3.3.3) one by one for all the four cases of i, j 0,. () For the case of i j 0, E.Q (3.3.3) is absolutely right. () For the case of i 0, j, the right-hand-side (RHS) of E.Q. (3.3.3) should be RHS (I Z) ψ(0, 0) (I Z) ( 00 + ) ( 00 ) which means E.Q (3.3.3) is correct in this case. ψ(0, ), (3.3.4) 6

27 (3) For the case of i, j 0, we evaluate the right-hand-side (RHS) of E.Q. (3.3.3) RHS (I X) ψ(0, 0) (I X) ( 00 + ) ( ) ψ(, 0), (3.3.5) which also shows the legitimation of E.Q (3.3.3) in this circumstance. (4) For the case of i j, we can get the right-hand-side (RHS) of E.Q. (3.3.3) RHS (I XZ) ψ(0, 0) (I XZ) ( 00 + ) (I X) ( 00 ) ( 0 0 ) which says E.Q (3.3.3) for the last situation. ψ(, ), (3.3.6) Now, we can conclude that for all the four cases i, j 0,, E.Q (3.3.3) is valid. Remark: For four Bell states (3.3.), we have the following geometric representations, ψ(00) φ + ( 00 + ) (3.3.7) ψ(0) ψ + (I I X) ψ(00) ( ) X (3.3.8) ψ(0) φ (I I Z) ψ(00) ( 00 ) Z (3.3.9) ψ() ψ (I I XZ) ψ(00) ( 0 0 ) XZ (3.3.0) The vertical line denotes one Hilbert space H. Lemma The four Bell states can also be expressed as ψ(i, j) ( 0i + ( ) j ī ), (3.3.) where ī (i + ) mod and i 0,. We can show in the following that E.Q. (3.3.) is consistent with E.Q. (3.3.). 7

28 Proof. From E.Q. (3.3.3) we can get ψ(i, j) (I X i Z j )( 00 + ) ( 0 X i Z j 0 + X i Z j ) ( 0 X i 0 + ( ) j X i ) ( 0 i + ( ) j ī ), which is what exactly E.Q. (3.3.) shows. Remark: All the four Bell states defined in E.Q. (3.3.) are normalized: (i) ψ(0, 0) ψ(0, 0), since ( 00 + ) ( 00 + ) ( ) (ii) ψ(i, j) ψ(i, j), because ( ). ψ(i, j) ψ(i, j) ( ψ(0, 0) I Z j X i )(I X i Z j ψ(0, 0) ) ψ(0, 0) I Z j X i X i Z j ψ(0, 0) ψ(0, 0) ψ(0, 0). Lemma Let s M denotes an arbitrary SU() matrix, namely single-qubit gate, and M T is the transpose of M. Then with the diagrammatical representation (I M) ψ(0, 0) (M T I ) ψ(0, 0), (3.3.) M M T (3.3.3) 8

29 Proof. Firstly, evaluate the left-hand-side (LHS LHS) of E.Q. (3.3.) LHS (I M) ψ(0, 0) I M ii i0 i0 i M i ( 0 (M M 0 ) + (M M )) ((M M 0 ) 0 + (M M ) ) (M T M T ) M T I ( ), i.e., LHS (M T I ) ψ(00), (3.3.4) which happens to be equal to the right-hand-side of E.Q. (3.3.). In the derivation we have assumed that M has the matrix form M ( M 00 M 0 M 0 M ), namely M T ( M 00 M 0 M 0 M ). We may notice that Therefore, we can get the conclusion that X T X, Z T Z, (XZ XZ) T ZX. ψ + (I X) ψ(00) (X I ) ψ(00), φ (I Z) ψ(00) (Z I ) ψ(00), ψ (I XZ XZ) ψ(00) (ZX ZX I ) ψ(00). (3.3.5) (3.3.6) ψ(0) ψ + X X (3.3.7) ψ(0) φ Z Z (3.3.8) ψ() ψ XZ ZX (3.3.9) 9

30 3.3. Parity-bit (i) and Phase-bit (j) Parity-bit (i) i 0, two spins are aligned, denoted by φ ; i, two spins are anti-aligned, denoted by ψ. Phase-bit (j) j 0, superposition with +, i.e., with equal phase; j, superposition with, i.e., with opposite phase. Table 3.: Parity-bit (i) and Phase-bit (j) j i 0 0 φ + φ ψ + ψ Lemma Bell states are eigenstates of the two commutative operators:. parity-bit operator: Z Z,. phase-bit operator: X X, namely (Z Z) ψ(i, j) ( ) i ψ(i, j), (X X) ψ(i, j) ( ) j ψ(i, j). (3.3.0) Proof. The parity-bit operator and the phase-bit operator are commutative, because namely (X X) (Z Z) (XZ XZ) (XZ XZ) ( ZX ZX) ( ZX ZX) (ZX ZX) (ZX ZX) We may examine the two operators separately. (Z Z) (X X), (3.3.) [X X, Z Z] 0. (3.3.) 0

31 (a) Parity-bit operator Z Z. (Z Z) ψ(i, j) (Z Z) [ 0i + ( ) j ī ] [Z 0 Z i + ( ) j Z Z ī ] [ 0 ( ) i i + ( ) j ( ) ( ) i ī ] [( ) i 0 i + ( ) j ( ) i ī ] ( ) i [ 0 i + ( ) j ī ], i.e., (Z Z) ψ(i, j) ( ) i ψ(i, j). (3.3.3) In the derivation, we have used the facts that Z i ( ) i i, Z ī ( ) i ī. (3.3.4) (b) Phase-bit operator X X. (X X) ψ(i, j) (X X) ( 0i + ( ) j ī ) (X 0 X i + ( ) j X X ī ) ( ī + ( ) j 0 i ) ( ) j ( 0 i + ( ) j ī ), namely (X X) ψ(i, j) ( ) j ψ(i, j). (3.3.5) And we have used the facts in the following to go through the above derivation, { X i ī, X ī i. (3.3.6) With E.Q. (3.3.3) and E.Q. (3.3.5), we get E.Q. (3.3.0) proved Orthonormal basis of the two-qubit Hilbert space Thm Bell states form an orhonormal basis of two-qubit Hilber space, namely { φ +, ψ +, φ, ψ } is an orthonormal basis of two-qubit Hilbert space.

32 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. (3.3.3), we can derive that Because ψ(i, j) ψ(i, j ) ψ(0, 0) (I Z j X i )(I X i Z j ) ψ(0, 0) ( 00 + )(I Z j X i+i Z j )( 00 + ) kk (I Z j X i+i Z j ) ll k,l0 k l k Z j X i+i Z j l k,l0 δ kl k Z j X i+i Z j l k,l0 k Z j X i+i Z j k k0 tr(zj X i+i Z j ) tr(zj +j X i+i ). Z X I, and i, j, i, j {0, }, then we can obtain trz trx tr (ZX ZX) 0, ψ(i, j) ψ(i, j ) δ jj δ ii, (3.3.7) which means that the four Bell states are mutually 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 ψ(i j ) X X i Z j ψ(ij) Z Z j X i (3.3.8) Therefore, the orhonormal relation has the diagrammatical representation ψ(ij) ψ(i j ) Z Z j X i (3.3.9) X X i Z j From the diagrammatical rules, we would have the normalized trace of the singlequbit gates on the loop, ψ(ij) ψ(i j ) tr(zj X i X i Z j ) tr(zj +j X i+i ) δ jj δ ii. (3.3.30)

33 (b) The vector set consisted of the four Bell states is complete. By utilizing E.Q. (3.3.), we can get As we see that therefore i.e., Remark: ψ(i, j) ψ(i, j) i,j0 i,j0 ( 0i + ( ) j ī )( 0i + ( ) j ī ) ( 0i 0i + ( ) j ī 0i + ( ) j 0i ī + ī ī ). i,j0 ( ) j ī 0i 0, i,j0 ( ) j 0i ī 0, (3.3.3) i,j0 ψ(i, j) ψ(i, j) ( 0i 0i + ī ī ) i,j0 i,j0 i0 ( 0i 0i + i i ) ji ji i,j0 I 4, { ψ(i, j) i, j 0, } is the Bell basis of H H ; { i, j i, j 0, } is the product basis of H H How to distinguish (create) ψ(i, j). There are two different cases ψ(i, j) ψ(i, j) I 4. (3.3.3) i,j0 (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 ψ i,j. The reason is that local operation changes ψ(i, j), since [X I, Z Z] 0, (3.3.33) [Z I, X X] 0. 3

34 Remark: Quantum entanglement cannot be created by remote pairs by using local operation and classical communication. 4

35 Chapter 4 No-Cloning, Dense Coding, Teleportation and Cryptography Reference: [Preskill] Chapter 4: Quantum entanglement. [Nielsen & Chuang] Chapter : Introduction to quantum mechanics. [Nielsen & Chuang] Chapter : quantum information theory. 4. No-cloning theorem Def 4.. (Cloning Machine). The cloning machine is a unitary transformation U, which satisfies U( φ 0 ) φ φ (4..) for arbitrary state φ. It can be represented in the diagram shown in Figure 4.. Target object φ φ U Blank object 0 φ Figure 4.: Copy machine in quantum mechanics Thm (No-cloning theorem ). The cloning machine doesn t exist (in Quantum Mechanics). Proof. For simplicity, we deal with the two-dimensional Hilbert space. As the definition (4..) shows, we choose the blank object to be 0 and the target object to be be 0 or, and the copy machine satisfies the following equations: U , U 0. (4..) 5

36 For example, the most popular two-qubit quantum gate in quantum computation is the CNOT gate, which has the property, CNOT , CNOT 0. (4..3) For φ H, which has the form of we obtain i.e., φ a 0 + b, with a + b, (4..4) U φ 0 U(a 0 + b ) 0 au bu 0 a b, U φ 0 a b. (4..5) However, the copy machine (4..) tells us another thing: U φ 0 φ φ (a 0 + b ) (a 0 + b ) a b + ab 0 + ab 0. (4..6) In general, E.Q. (4..5) and E.Q. (4..6) are not consistent with one another. Therefore, the cloning machine is not available for φ H. Remarks: 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 (No-cloning theorem ). 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 While, being non-orthogonal means that Because the copy machine U satisfies E.Q. (4..), we find out φ ψ. (4..7) φ ψ 0. (4..8) ( φ 0 ) ( ψ 0 ) φ 0 U U ψ 0 ( φ φ ) ( ψ ψ ) φ ψ, 6

37 thus φ ψ φ ψ. (4..9) It s clear that the E.Q.(4..9) will either violate the relation (4..7) or (4..8). Therefore, 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 U( 0 ), (4..0) where U can be the CNOT gate in quantum computation. Thm (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, φ ψ, (4..) φ ψ 0. Assume that the unitary transformation U can distinguish φ and ψ, without disturbing them, namely U φ 0 φ e, (4..) U ψ 0 ψ f where e f, means φ can be distinguished from ψ. Because which is ( φ 0 )( ψ 0 ) ( φ 0 )U U( ψ 0 ) From E.Q.(4..3) we know that either or ( φ e )( ψ f ) φ ψ e f, φ ψ φ ψ e f. (4..3) φ ψ 0 (4..4) e f (4..5) must be true.therefore we are able to distinguish two orthogonal states without disturbing 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. 7

38 4. Dense coding Alice sends a qubit (st qubit) Alice φ + Bob (nd qubit) 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 Bob. Dense coding can be executed in the following manner step by step: Step : Experiment setup. Alice and Bob share a maximally entangled state, e.g. Alice φ + AB Bob which is the cup representation for the Bell state φ + defined in (3.3.7). Step : Local unitary transformation. Alice chooses one of the four unitary transformation {I, X, Z, ZX} and performs it on her qubit. ψ(0, 0) ( 00 + ) (4..) ψ(00) AB I I Alice Bob Alice Bob ψ(0, ) ( 00 ) I Z ψ(0, 0) Bob Z I ψ(0, 0) Alice (4..) ψ(0) AB Z Z Alice Bob Alice Bob ψ(, 0) ( ) I X ψ(0, 0) Bob X I ψ(0, 0) Alice (4..3) Note: The Holevo bound (Old Chapter 5.4., page 36, John Preskill s lecture notes) says that without entanglement at most a classical bit information can be transmitted via sending a qubit. 8

39 ψ(0) AB X X Alice Bob Alice Bob ψ(, ) ( 0 0 ) I XZ ψ(0, 0) Bob ZX I ψ(0, 0) Alice (4..4) ψ() AB ZX XZ 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. X i Z j X i Z j Alice Bob Bob Step 4: Bell measurements. Bob performs the Bell measurement: which gives the parity bit i, and (X X) ψ(ij) ( ) j ψ(ij), (4..5) (Z Z) ψ(ij) ( ) i ψ(ij) (4..6) from which gives the phase bit j. With (i, j), Bob 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 4., 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. 9

40 Table 4.: Dense coding Local unitary transf. Final state Two bits Alice Bob Bob I φ + ψ(0, 0) (0, 0) X ψ + ψ(, 0) (, 0) Z φ ψ(0, ) (0, ) ZX ψ ψ(, ) (, ) 4.3 Quantum teleportation Y.Z., Braid Group, Temperley Lieb Algebra, and Quantum Information and Computation, arxiv: quant-ph/ Y.Z., Jinglong Pang, Space-Time Topology in Teleportation-Based Quantum Computation, arxiv: Y.Z., Kun Zhang, Bell Transform, Teleportation Operator and Teleportation-Based Quantum Computation, arxiv: In some sense, Quantum Teleportation is a kind of inverse process of dense coding (see Table 4.). Alice sends two classical bits to Bob (st qubit) Alice φ + Bob (nd qubit) Bob gets one qubit from Alice Table 4.: Dense coding vs. Quantum teleportation Resource Send transmit Dense coding φ + qubit bits Quantum Teleportation bits qubit Task: Alice wants to send an unknown qubit to Bob, who is far away from her. Lemma ψ φ + ( φ+ ψ + (X I φ + ) X ψ + (Z I φ + ) Z ψ + (ZX I φ + ) XZ ψ, ) (4.3.) 30

41 which can be represented in the diagram formulism also, ψ φ + + X X + Z Z + ZX XZ. (4.3.) Proof. We can give an expression to unknown state ψ ψ a 0 + b, with a, b C and a + b. (4.3.3) From the definition of the four Bell states (3.3.), we can get ( φ + + φ ), ( ψ + + ψ ), ( ψ + ψ ), (4.3.4) ( φ + φ ). With these materials we can make the following derivation ψ φ + (a 0 + b )( 00 + ) [a 0 ( 00 + ) + b ( 00 + )] [a( ) + b( )] [a( φ+ + φ ) 0 + b( ψ + ψ ) 0 + a( ψ + + ψ ) + b( φ + φ ) ] [ φ+ (a 0 + b ) + φ (a 0 b ) + ψ + (a + b 0 ) + ψ (a b 0 )] [ φ+ ψ + (Z I φ + ) (Z ψ ) + (X I ) φ + (X ψ ) + (ZX I φ + )(XZ ψ )], which is equivalent to E.Q. (4.3.). Therefore, Lemma 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 ( N) 3

42 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 ) The Quantum Telecportation can be accomplished with the following steps: Step : 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: Alice & Bob: ϕ A A φ + B. Step : 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 4.3. Table 4.3: Alice s measurement result and the two-bit information post-measurement state Z Z X X two-bit φ + (0, 0) φ (0, ) ψ + (, 0) ψ (, ) 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 E 00 φ φ +, E 0 φ φ, (4.3.5) E 0 ψ + ψ +, E ψ ψ, which can be represented in a diagrammatic formalism shown as E 00, E 0 X X, E 0 Z Z ZX, E XZ. 3

43 We now utilize Lemma , which actually means ψ φ + ( φ+ ψ + φ (Z ψ ) + ψ + (X ψ ) + ψ )(XZ ψ )). (4.3.6) Therefore, after the Bell measurement carried out on the qubit of Alice and the unknown state, we obtain ( φ φ I )( ψ φ + ) φ Z ψ, ( ψ + ψ + I )( ψ φ + ) ψ+ X ψ, (4.3.7) ( φ + φ + I )( ψ φ + ) φ+ ψ, ( ψ ψ I )( ψ φ + ) ψ XZ ψ. E.Q.(4.3.7) can also be rewritten in the diagram language shown in the following context: ( φ + φ + I )( ψ φ + ) φ+ ψ Measurement Preparation A A B A A B (4.3.8) ( φ φ I )( ψ φ + ) φ Z ψ Measurement Preparation Z Z A A B Z Z A A B (4.3.9) ( ψ + ψ + I )( ψ φ + ) ψ+ X ψ Measurement Preparation X X A A B X X A A B (4.3.0) ( ψ ψ I )( ψ φ + ) ψ XZ ψ Measurement Preparation ZX XZ A A B ZX XZ A A B (4.3.) 33

44 Step 3: Classical communication between Alice and Bob. Alice informs her results (i, j) (two-bit of information as shown in Table 4.3) 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 4.4. Table 4.4: Quantum Teleportation Protocol Bell measurement Classical communication Unitary correction Alice two-bit Bob φ + (0, 0) I φ (0, ) Z ψ + (, 0) X ψ (, ) ZX Remarks: The state ψ, which Alice means to transmit to Bob, is unknown to Alice. No-cloning theorem 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: U( φ 0 ) φ φ ; (4.3.) Quantum teleportation: T ( φ 0 ) X φ. (4.3.3) Quantum teleportation has the interpretation of space-time topology. 4.4 The quantum teleportation using continuous variables Problem description : One complete orthonormal basis for the Hilbert space of two particles on the real line is the (separable) position eigenstate basis { q q }. Another is the entangled basis { Q, P }, where Q, P π dq e ip q q q + Q ; (4.4.) these are the simultaneous eigenstates of the relative position operator Q q q and the total momentum operator P p + p. (a) Verify that Q, P Q, P δ(q Q)δ(P P ); (4.4.) Originated from the exercise 4.3, revised Chapter 4 of John Preskill s online lecture notes. 34

45 (b) Since the states Q, p are a basis, we can expand a position eigenstate as q q dqdp Q, P Q, P ( q q ). (4.4.3) Evaluate the coefficients Q, P ( q q ). (c) Alice and Bob have prepared the entangled 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 C C 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? What information should 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? ) Background: 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 ) Notation: Define the U() phase operator as Ω dq qq Q 0, P 0 (4.4.4) U P e ip ˆq dq e ip q q q, (4.4.5) The unitary formalism shows as U P q e ip q q. (4.4.6) Define the translation operator as U P U P, q U P q U P q e ip q. (4.4.7) T Q e iˆp Q dq q + Q q, (4.4.8) The unitary formalism shows as T Q q q + Q. (4.4.9) T Q T Q, q T Q q T Q q + Q, (4.4.0) q dq q Q q dq δ q,q Q q q + Q. (4.4.) Note we have the relations U P T Q q U P q + Q e ip (q+q) q + Q, (4.4.) 35

46 T Q U P q T Q e ip q q e ip q q + Q, (4.4.3) therefore U P T Q e ip Q T Q U P. (4.4.4) The entangled basis is defined as Q, P (U P T Q ) Ω dqe ip q q, q + Q (4.4.5) with the cup configuration Q, P P Q (4.4.6) And the cap configuration expresses as Q, P P Q (4.4.7) The normalization relation in diagrammatical language: Q, P P Q Q, P δ(p P )δ(q Q ). P Q (4.4.8) Other diagrammatical rules: P P Q Q (4.4.9) P Q Q P (4.4.0) The product state has the diagrammatical representation: q, q q q The inner product of product state with the entangled state Q, P has the expression Q, P q, q P Q q q π e ip q δ(q (q q )). Note that Q stands for the relative position of two particles, namely Q q q. 3 ) Continuous teleportation: 36

47 i) State preparation where ψ C is the unknown state, expressed as ii) Bell measurement P Q ψ C Q, P AB (4.4.) ψ C dq q C q ψ C dqψ(q) q. (4.4.) P Q iii) Topological operation P Q C A B (4.4.3) P Q P Q P Q ψ C Q, P AB P Q Q P Q P ψ B iv) Unitary correction After Bell measurement, Bob will obtain the state ψ B T Q U P T Q U P ψ B. (4.4.4) Therefore, he is required to perform the unitary correction U U P T Q U P T Q e ip Q U P P T Q Q, (4.4.5) where the commutative relation (4.4.4) has been applied. 4.5 Quantum cryptography (information security) 4.5. Classical cryptography Alice and Bob want to communicate (transmit information) with each other.. To ensure information security, they choose to get an encryption key, e.g. K ( ).. Now, Alice wants to send Bob some information, e.g. A ( ). For information security consideration, she would encrypt her information before sending out, i.e., A + K ( 0 ). 37

48 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 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 step by step in the following: Step : State preparation. Alice and Bob share the Bell state. φ + ( z z + z z ) (( x + x ) ( x + x ) +( x x ) ( x x )) ( x x + x x ), i.e., φ + ( 00 + ) ( ++ + ). (4.5.) Step : Random local measurements by Alice. Alice makes purely random (Prob /) choices from {σ z, σ x } to conduct measurements on her qubit with the chosen observables (see Table 4.5). Table 4.5: Alice s purely random local measurements Alice random prob / Z Z X X Z X σ A z σ A z σ A x σ A x σ A z σ A x eigenstate z z x x z x eigenvalue 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 4.6). 38

49 Table 4.6: Bob s purely random local measurements Alice prob / X Z Z X X X σ B x σ B z σ B z σ B x σ B x σ B x eigenstate uncertain z uncertain x uncertain x eigenvalue 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 4.7. Table 4.7: The same observables that Alice and Bob share 4 6 measurement σ z σ x σ x eigenvalue + And then encode them as x, z + x, z 0. (4.5.) Finally, we can get the encryption key, i.e., K (0 ) (4.5.3) 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 BB84 quantum key distribution The BB84 Quantum key distribution was firstly presented in 984. Principle Non-orthogonal state (e.g. x, z ), cannot be distinguished without any changes, so Quantum infromation can be securely encoded in non-orthogonal states. To obtain the BB84 Quantum key distribution, we have the following steps: 39

50 Step: Alice randomly prepares one of the four states z, z, x, x, and labels with its observable Z, while with X. z, z x, x Step: 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. Step5: Develop the row key to a practical key. { x, z 0 x, z. 40

51 Chapter 5 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. Reference: [Preskill] Chapter 4: Quantum entanglement; James Clerk Maxwell Lorenzo Maccone, A simple proof of Bell s inequality, arxiv:.54v. 5. Einstein s quantum mechanics: local hidden variable theory (LHV) 5.. 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 5.. But, Bohr didn t agree with him, and argued that the Table 5.: QM and LHV Quantum theory ψ α 0 + β Hidden variable theory ( ψ, λ), λ: hidden variable 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 0th 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. 4

52 5.. What local theory? As we know that one of the two axioms of Special Relativity is that there is no faster-thanlight 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 + (3.3.), 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. 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 9.4.., we know that if two subsystems A and B share an entangled state in the composite system H A H B, 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 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. 5. Bell s inequality in the local hidden variable theory Let s consider the following experiment: () Alice has three coins, each with head or tail face. We can label the three coins, which Alice holds, with A, A and 3 A respectively. For each coin we assign a specific variable to describe the state of the coin, i.e., x for coin A, y for coin A and z for coin 3 A, and x, y, z {H, T}, with H for Head, and T for Tail. () 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). x,y,z {H,T} 4

53 The probability that the i-th coin and the j-th coin have the same value can be denoted as P same (i, j), e.g. P same (, ) Prob(H, H, H)+Prob(H, H, T)+Prob(T, T, H)+Prob(T, T, T). (5..) Now,we can define and evaluate it in the following way BI P same (, ) + P same (, 3) + P same (, 3), (5..) 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) + (Prob(H, H, H) + Prob(T, T, T)) (5..3) i.e., BI P same (, ) + P same (, 3) + P same (, 3), (5..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 5.. For another method of calculation BI, with the help of the diagram, we obtain BI P same (, ) + P same (, 3) + P same (, 3) (A + C) + (B + C) + (C + D) + C + (Prob(H, H, H) + Prob(T, T, T)), (5..5) where, through derivation, we have applied the relation A + B + C + D, 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 B, B, and 3 B. Alice and Bob obtain the same results when measuring their coins with the same label, i.e., when Alice measures the coin i A and Bob measures the coin i B, with i,, 3, then their results are always coincide. 43

54 A C B D 3 Figure 5.: Venn diagram of probabilities distribution Prob(x, y, z). The circles labeled by, and 3 represent the corresponding probabilities distribution of coin, 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 and share the same face value, whereas coin and 3, coin 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. 5.3 Bell s inequality in quantum mechanics () Experiment setup: Alice and Bob share the spin singlet state ψ, ψ ( 0 0 ). (5.3.) Here, we replaced a coin with a spin polarization direction. For example, we can choose the three orientations a i, i,, 3, for Alice s system, and b i, i,, 3, for Bob s system, with a i b i, and a i a j, bi b j, if i j. A spin pointing along the positive (negative) direction of an orientation corresponds to a coin with Head ( Tail ) face. () Mutually Exclusive Experiment (MEE). Because [ σ a i, σ a j ] 0, if i j, [ σ b i, σ b j ] 0, if i j, which means that σ a i and σ a j have no simultaneous eigenstates, neither σ a i and σ b i. As an result, there comes some differences between the quantum mechanics and the local hidden variable theory: 44

55 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 E A ( a) a a I E B ( b) I b b ( + a σ A ), ( + b σ B ), The density matrix of the composite system is expressed as with a b, and a, b R 3. (5.3.) ρ AB ψ AB ψ. (5.3.3) The probability, that Alice s spin is pointing along the direction a and Bob s spin along the orientation of b when we do the measurements, is calculated by Prob( a, b) AB ψ (E A ( a)e B ( b)) ψ AB 4 AB ψ ( + a σ A )( + b σ B ) ψ AB 4 AB ψ ( + a σ A + b σ B + ( a σ A )( b σ B ) ) ψ AB. (5.3.4) part I part II And we can calculate the two parts of expression separately. Firstly, for part I, where part I + AB ψ a σ A ψ AB + AB ψ b σ B ψ AB, AB ψ a σ A ψ AB ( AB 0 AB 0 ) a σ A ( 0 AB 0 AB ) ( A 0 a σ A 0 A + A a σ A A ) tr A( a σ A ) 0, (5.3.5) 45

56 and AB ψ b σ B ψ AB ( AB 0 AB 0 ) b σ B ( 0 AB 0 AB ) ( B b σ B B + B 0 b σ B 0 B tr B( b σ B ) 0, (5.3.6) because Pauli matrices are traceless. Therefore, part I. (5.3.7) Next, we come to part II, part II AB ψ ( a σ A )( b σ B ) ψ AB ( AB 0 AB 0 )( a σ A )( b σ B )( 0 AB 0 AB ) [ A 0 a σ A 0 A B b σ B B A 0 a σ A A B b σ B 0 B A a σ A 0 A B 0 b σ B B + A a σ A A B 0 b σ B 0 B ] ( a 3b 3 (a ia )(b + ib ) (a + ia )(b ib ) a 3 b 3 ) ( a 3b 3 a b a b a 3 b 3 ) a b. (5.3.8) 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 therefore the part II can be rewritten as ( σ A + σ B ) ψ AB 0, (5.3.9) part II AB ψ ( a σ A )( b σ B ) ψ AB AB ψ ( a σ A )( b σ A ) ψ AB AB ψ a i σ A i b j σ A j ψ AB AB ψ a i b j δ ij ψ AB AB ψ iε ijk a i b j σ A k ψ AB a b, (5.3.0) where the repeated indexes imply sum and the relation of the Pauli matrices is applied, σ i σ j δ ij I + iε ijk σ k. (5.3.) Hence, integrating the calculation of part I and part II, we get Prob( a, b) 4 4 a b 4 [ cos( a, b)]. (5.3.) 46

57 Following, we obtain the probability that Alice s spin points along a i and Bob s spin points along b j or Alice s spin points along a i and Bob s spin b j, namely measurement results are same, P ij same Prob( a, b) + Prob( a, b) [ cos( a i, b j )]. (5.3.3) There, we can see that with b j a i, E.Q.(5.3.3) should be P ij same, i.e., physical systems Alice and Bob are perfectly anticorrelated. Now, we can calculate the quantity BI in Quantum Mechanics view point, BI P same + P 3 same + P 3 same, with b i a i, i,, 3. (5.3.4) For the explanation of E.Q.(5.3.4), we may have some discussions. E.Q.(5.3.4) 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 (5..4) and relation (5.3.4) 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 a i, i,, 3 is to ensure that, when Alice and Bob measure the same axis, or same coin, their results are perfect correlated. Now, we can calculate out some special cases by specifically setting the orientations of a i, i,, 3. Case : a i with different i {,, 3} are mutually orthogonal, which is shown in Figure 5.a. Therefore, we can see Following, for BI, cos( a, b ) cos( a, b 3 ) cos( a, b 3 ) 0. (5.3.5) i.e., BI + + 3, BI. (5.3.6) In this case, the Bell inequality (5..4) is correct, and it means QM and LHV are consistent with each other. 47

58 Case : { a i, i,, 3} are placed in a counterclockwise way on a plane, as is shown in Figure 5.b. cos( a, b ) cos( a, b 3 ) cos( a, b 3 ), (5.3.7) namely BI , BI, (5.3.8) which violates the inequality (5..4), which implies QM and LHV disagrees with each other at this time. The Bell s inequality is violated in QM. b a 3 b a a b3 b a 0 a 0 0 a 3 b3 b (a) Mutually orthogonal setting (b) Planar setting Figure 5.: Orientation settings for a i and b i, i,, 3, where a i b i 5.4 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. () Experimental setup, see Table 5.. Table 5.: Experiment setup Alice ψ Bob observables observables A, B C, D A ±, B ± C ±, D ± () Local Hidden Variable theory Observables A, B, C, D can be assigned values simultaneously, because they are pre-determined or pre-existing. A + B 0 A B 0 } B A B A } A B A A + B A } { A B ±. A + B ±. Define the new observable, M M (A + B)C + (A B)D ±. (5.4.) 48

59 Since HV is unknown, we only have probability distribution. i.e., which is called CHSH inequality. (3) Quantum Mechanics We can set these four observable A, B, C, D as M ± M M, (5.4.) M AC AC + BC BC + AD AD BD BD, (5.4.3) A a σ A, B a σ A, C b σ B, D b σ B. (5.4.4) The relative positions of a to a and that of b to b are shown in Figure 5.3. a θ a b b θ (a) Alice (b) Bob Figure 5.3: Choice of the operators With the help of E.Q. (5.3.8), we can evaluate the left-hand-side of the CHSH inequality, M cos( a, b) cos( a, b) cos( a, b ) + cos( a, b ). (5.4.5) Now considering the special case depicted in Figure 5.4, We can derive D A π/4 π/4 C π/4 B Figure 5.4: Special setting for A, B, C, D Remarks: Therefore, AC AC AD AD BC BC BD M 4 It violates the CHSH inequaltiy (5.4.3). BD. (5.4.6). (5.4.7) The CHSH inequality is violated for all entangled pure states, see Chapter 4.3.4, John Preskill s lecture notes. 49

60 What is involved in our calculation is spin- system, but usually it is the photonpolarization 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. 5.5 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. Particle s trajectories are influenced with each other non-locally. QM True randomness Quantum non-locality No LHV theory { Randomness is intrinsic QM is complete Classical locality v c Perfect correlation 5.6 Hardy s theorem Problem description : Bob (in Boston) and Claire (in Chicago) share many identically prepared copies of the two-qubit state ψ ( x) 00 + x 0 + x 0. (5.6.) where x is a real number between 0 and /. They conduct many trials in which each measures his/her qubit in the basis { 0, }, and they learn that if Bob s outcomes is then Claire s is always 0, and if Claire s outcome is then Bob s is always 0. Bob and Claire conduct further experiments in which Bob measures in the basis { 0, } and they learn that if Bob s outcome is then Claire s is always 0, and if Claire s outcomes is then Bob s is always 0. Bob and Claire conduct further experiments in which Bob measures in the basis { 0, } 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 Originated from the exercise 4., revised Chapter 4 of John Preskill s online lecture notes. 50

61 measures in the basis { 0, } 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 wonder what will happen if they both measure 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 argues 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, } instead. So suppose 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, } instead, we can be certain that his outcome would have been, since experiment has shown that if Bob had obtain 0 then Claire could not have obtained ϕ. Similarly, if Claire had measured in the basis { 0, }, then she certainly would have obtained the outcome. We conclude that if Bob and Claire both measured in the basis { 0, }, both would have obtained the outcome. Bit this a contradiction, for experiment has shown that it is not possible for both Bob and Claire to obtain the outcome if they both measure in the basis { 0, }. We are therefore forced to conclude that if Bob and Claire both measure in the basis { ϕ, ϕ }, it is impossible for both to obtain the outcome ϕ. Though 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, }. (b) If Bob and Claire both measure in the basis { ϕ, ϕ }, what is the quantummechanical prediction for the probability P (x) that both obtain the outcome ϕ? (c) Find the maximal violation of Hardy s theorem: show that the maximal value of P (x) is P [(3 5)/] (5 5 )/.090. (d) Bob and Claire conduct an experiment that confirms the prediction of quantum mechanics. What was wrong with Albert s reasoning? ) 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. ) 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 u b and u c satisfy the experimental fact: u b u c 0, (5.6.) where u b, u c 0,. 5

62 The observables w b and w c satisfy the experimental facts: u b 0 w c 0; (5.6.3) u c 0 w b 0. (5.6.4) Conclusion: Therefore, namely u b u c 0 u b 0 or u c 0 w b 0 or w c 0. (5.6.5) Prob(w b 0 & w c 0 u b u c 0) LHV 0, (5.6.6) Prob(w b & w c u b u c 0) LHV 0. (5.6.7) In following, we will calculate it in quantum mechanics, and verify Prob(w b & w c u b u c 0) QM 0. (5.6.8) 3 ) In quantum mechanics, Bob and Charlie are correlated by entangled state ψ BC. Bob ψ BC Charlie ψ BC ϕ B 0 C + x 0 B C 0 B ϕ C + x B 0 C x 00 BC + x 0 BC + x 0 BC (5.6.9) where 0 < x <. The observables u b and u c (u b, u c 0, ) are corresponding to the measurements U b and U c defined as U b 0 B 0, U b B, (5.6.0) U c 0 C 0, U c C. (5.6.) Note that U b + U b, U c + U c. (5.6.) The value u b,c 0 or u b,c is associated with the measurement result U b,c or U b,c respectively. We have u bu c 0, because state ψ BC does not have the term BC. The observables w b and w c (w b, w c 0, ) are corresponding to the measurements W b and W c defined as W b ϕ B ϕ, W b ϕ B ϕ, (5.6.3) W c ϕ C ϕ, W c ϕ C ϕ, (5.6.4) where state ϕ is the normalized state ϕ defined in (5.6.9), and state ϕ is orthogonal to the state ϕ. Note that W b + W b, W c + W c. (5.6.5) The value w b,c 0 or w b,c is associated with the measurement result W b,c or W b,c respectively. 5

63 Bob U b ψ BC Charlie W c ϕ C 0 B u b 0 w c 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 u b 0 leads to w c 0. Bob W b ψ BC Charlie U c 0 C ϕ B w b 0 u c 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 u c 0 leads to w b 0. 4 ) The task is to calculate the probability From Prob(w b & w c u b u c 0) QM BC ϕ ϕ ψ BC. (5.6.6) we know the state ϕ can be expressed as 0 B 0 ψ AB 0 B ϕ C, (5.6.7) ϕ x 0 + x, (5.6.8) and the normalized form ϕ x ( x 0 + x. (5.6.9) Find the orthogonal state of ϕ, denoted as ϕ, { ϕ A 0 + B, ϕ B 0 A, (5.6.0) where A x x, B. (5.6.) x x Rewrite it into matrix formalism ( ϕ ϕ ) ( A B B A ) ( 0 ), (5.6.) 53

64 where the transformation matrix is a real orthogonal matrix, namely ( A B B A ) ( A B B A ) ( A + B 0 0 B + A ) ( 0 0 ). (5.6.3) Therefore, ( 0 ) ( A B B A ) ( ϕ ϕ ), (5.6.4) { 0 A ϕ + B ϕ, B ϕ A ϕ. (5.6.5) Consider the term ϕ A ϕ B solely in state ψ AB : 00 AB B ϕ A ϕ B ; (5.6.6) 0 AB AB ϕ A ϕ B ; (5.6.7) 0 AB AB ϕ A ϕ B. (5.6.8) Hence ψ AB ( xb xab) ϕ ϕ BC x x x ϕ ϕ BC. (5.6.9) Prob(w b & w c u b u c 0) QM BC ϕ ϕ ψ BC Find the maximal value of function p(x): p (x) x ( x) ( x) p(x). (5.6.30) x ( x) 3 (x 3x + ) 0. (5.6.3) And Due to 0 < x <, we have x ± 3 ± 5, x 0 0. (5.6.3) p max (x ) 5 5 The result is conflicted with local hidden variable theory (5.6.33) 5 ) Reason for conflict between quantum mechanics and local hidden variable theory: [U b, W b ] 0, [U c, W c ] 0. (5.6.34) Due to uncertainty relation, u b, w b and u c, w c can not be determined simultaneously. Note: [X, Z] 0 [X X, Z Z] 0. (5.6.35) 54

65 Table 5.3: A comparison between Bell s inequalities, Hardy s theorem and GHZ theorem Parties Formalism Correlation Statement Bell inequalities Inequality Statistical Hardy s theorem Equality Statistical LHV hypothesis is denied GHZ theorem 3 Equality Perfect 5.7 The GHZ theorem ) A comparison between Bell s inequalities, Hardy s theorem and GHZ theorem is shown in Table 5.3. ) 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., GHZ 3 ( ), (5.7.) as shown in the diagram formalism in Figure 5.5. Charlie(III) GHZ Alice(I) Bob(II) Figure 5.5: 3-qubit GHZ state. Step II: Chose four observables with the form of A σ () x σ y () σ y (3) ; (5.7.a) B σ () y σ x () σ y (3) ; (5.7.b) C σ y () σ y () σ x (3) ; (5.7.c) D σ x () σ x () σ x (3). (5.7.d) For these four observables, we may notice the following things: A B C D I 8 ; A, B, C, D are mutually commutative; ABCD I 8 ; D X X X 3, i.e. phase-bit operator, AD (σ x ) (σ y σ x ) (σ y σ x ) I Z Z, i.e. the second paritybit operator, CD (σ y σ x ) (σ y σ x ) (σ x ) Z Z I, i.e. the first parity-bit operator; 55

66 A GHZ 3 GHZ 3, (5.7.3a) B GHZ 3 GHZ 3, (5.7.3b) C GHZ 3 GHZ 3, D GHZ 3 GHZ 3, (5.7.3c) (5.7.3d) since AX ( iz X ) ( iz 3 X 3 ) X (Z X ) (Z 3 X 3 ), B( izx) X ( iz3x3) (ZX) X (Z3X3), C( iz X ) ( iz X ) X 3 (Z X ) (Z X ) X 3, DX X X 3. Step III: For LHV hypothesis, A, B, C, D can be exactly determined, i.e. Am () x m y () m y (3) ; (5.7.4a) Bm () y m x () m y (3) ; (5.7.4b) Cm y () m y () m x (3) ; (5.7.4c) Dm x () m x () m x (3) ; (5.7.4d) with Thus, m x (i), m (j) y ±, i, j,, 3. (5.7.5) ABCD (m () x m y () m x () m y () m y () m () x ). (5.7.6) Step IV: While for QM, we can get from E.Q.(5.7.3a) (5.7.3d) that ABCD GHZ GHZ, (5.7.7) i.e. ABCD. (5.7.8) Step V: Experiment support E.Q.(5.7.8), which means LHV is denied. Remark: In QM, [σ x (i), σ y (i) ] 0, with i,, 3, thus m (i) x cannot be determined simultaneously. and m (i) y 56

67 Part II Quantum Computing and Quantum Algorithm 57

68 Chapter 6 Classical Circuit and Quantum Circuit 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. Reference: Nielsen & Chuang [Preskill] New Chapter 5: Classical circuit and quantum circuit; [Nielsen & Chuang] Chapter 3: Introduction to computer science; [Nielsen & Chuang] Chapter 4: Quantum circuits. 6. Classical circuit Def 6... Classical circuit, a circuit model of classical computation, is a finite sequence of elementary gates applied to a finite string of input bits. 58

69 6.. Universal gate set 6... Elementary logical gates () NOT gate: from which we shall see that NOT x x x, (6..) NOT NOT Id, (6..) i.e. NOT gate is a reversible gate. The truth table of the NOT gate is shown in Table 6.. Table 6.: Truth table of NOT gate. input output 0 0 () AND gate: AND (x, y) (x y) mod x y. (6..3) The AND gate is irreversible. Its truth table is Table 6.. Table 6.: Truth table of AND gate. input output (3) OR gate: OR (x, y) x + y xy x y. (6..4) As we can see that OR gate is irreversible. The corresponding truth table is Table 6.3. Table 6.3: Truth table of OR gate. input output

70 6... Universal gate set Def 6... 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: general function F {0, } n {0, } n typical function f i {0, } n {0, } F (f, f,, f n ), i.e., the general function can be represented by typical functions. Examples: Thm: {NOT, AND} is an universal gate set; Thm: {NOT, OR} is an universal gate set; Thm3: {NAND, COPY} is an universal gate set; Remark: and Since NAND NOT AND, (6..5) COPY x (x, x). (6..6) NOT(x) x x AND(x) NOT AND(x, x), (6..7) thus NOT(x) NAND(x, x) NAND COPY(x). (6..8) Thm4: {NOR, COPY} is an universal gate set. 6. Reversible classical computation 6.. Irreversible computation NOR NOT OR. (6..9) Classical computation in terms of irreversible gates, such as AND, OR, NAND, NOR gates, is irreversible. Thm 6... (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. 60

71 Table 6.4: Irreversible and reversible computation Computer Gates Erasure of information Irreversible Irreversible gates Irreducible expenditure Reversible Reversible gates No expenditure 6.. Classical reversible gate One-bit reversible gates: { NOT(x) 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). (6..) With three XOR gate, we can form a swap gate, which is defined as And the construction should be And, we can verify that, SWAP (x, y) (y, x). (6..) SWAP XOR XOR XOR. (6..3) XOR XOR XOR (x, y) XOR XOR (x, x y) XOR (x x y, x y) (x x y, x y x x y) (y, x). Similarly, the quantum analogy should be SWAP CNOT CNOT CNOT, (6..4) with SWAP defined as SWAP ψ φ φ ψ. (6..5) And in quantum circuit model, it has the diagrammatical expression ψ φ φ ψ Thm 6... One-bit gates and two-bit gates cannot suffice universal classical reversible computation. 6

72 6..3 Three-bit Toffoli gate Def 6.. (Three-bit Toffoli gate). Toffoli gate θ (3) (x, y, z) (x, y, z xy). (6..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 controlledcontrolled NOT gate with two control-bit and one target bit, besides it is a nonlinear gate. Thm 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. ) Toffli gate with z : here we get the NAND gate; ) Toffli gate with x, z 0: the COPY gate for this time; 3 ) Toffli gate with x : and we get the XOR gate. θ (3) (x, y, ) (x, y, xy) (x, y, NAND(x, y)), (6..7) θ (3) (, y, 0) (, y, 0 + y) (, y, y) (, COPY(y)), (6..8) θ (3) (, y, z) (, y, z y) (, XOR(y, z)), (6..9) 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: [θ (3) ] (x, y, z) θ (3) (x, y, z xy) (x, y, z xy xy) (x, y, z), 6

73 namely [θ (3) ] Id. (6..0) Remark: To construct given classical gate, one may need the exponential number of θ (3) gates, which is terrible in practice Three-bit Fredkin gate Def 6.. (Three-bit Fredkin gate). The three bit Fredkin gate is defined as Fredkin gate (x, y, z) (x, xz + xy, xy + xz) (6..) This definition shows that the three-bit Fredkin gate is nonlinear, too. Thm 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. ) Fredkin gate with z 0: here we get the AND gate; ) Fredkin gate with y 0, z : Fredkin(x, y, 0) (x, xy, xy) (x, xy, AND(x, y)), (6..) Fredkin(x, 0, ) (x, x, x) and we obtain here the COPY gate and NOT gate; 3 ) Fredkin gate with x : and we get the SWAP gate. (COPY(x), NOT(x)), (6..3) Fredkin(, y, z) (, z, y) (, SWAP(y, z)), (6..4) 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 63

74 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 i.e., Fredkin (x, y, z) Fredkin(x, xz + xy, xy + xz) (x, x(xy + xz) + x(xz + xy), x(xz + xy) + x(xy + xz)) (x, x y + x xz + xxz + x y, x z + x xy + xxy + x z) (x, xy xy, xz xz) (x, (x + x)y, (x + x)z) (x, y, z), Fredkin Id. (6..5) Hence, the three-bit Fredkin gate with constant bits is universal for classical reversible computation. 6.3 The construction of an n-bit Toffoli gate using the 3-bit Toffoli gate Def 6.3. (n-bit Toffoli gate). θ (n) (x, x,, x n, y) (x, x,, x n, y x x x n ) (6.3.) which can also be shown in the diagram formalism, x x x x x n x n x n which has n control bits and one target bit. y x x x n Thm With only one bit of scratch space, performing θ (n) gate needs at least number of n 3 + n of θ (3) gates. Remark: With more bit of scratch space, the number of θ (3) needed to perform θ (n) can be reduced in polynomial scale. Thm 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 n 5. Proof. The proof is done by induction. When n 4, we need scratch bit and (n 5) 4 3 number of θ (3) gates. θ (3) θ (3) x x x x 0 x x x x 0 x 3 θ (3) x 3 y y x x x 3 64

75 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 (n 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+) gate, we have θ (3) θ (3) x x x x 0 0 x 3 x x 4 x 4 θ (n) 0 0 x 5 x 5 x n y x n y x x x 3 x n Therefore, the number of scratch space needed to construct θ (n+) gate is (n 3) + (n + ) 3. And the number of θ (3) gates is n 5 + (n + ) 5. Thm 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. Proof. When n 4, we need scratch bit and (4n ) 4 4 number of θ (3) gates. x x After the first θ (4) gate, we have After the second θ (3) gate, we obtain θ (3) x x s θ (4) s x 3 x 3 y y x x x 3 y y (x x s )x 3 y x x x 3 s x 3. (6.3.) y y x x x 3. (6.3.3) In the case of construction θ (n) gate, we can find we need one θ (n) gate and one θ (n ) 65

76 gate, shown as x x x x s s x 3 x 3 s s x 4 θ (n) x 4 s 3 θ (n ) s 3 x 5 x 5 x n After the first θ (n) gate, we have y x n y x x x 3 x n y y ((((x x s )x 3 s )x 4 s 3 )x 5 )x n y x x x 3 x n (((s x 3 s )x 4 s 3 )x 5 )x n y x x x 3 x n z. (6.3.4) Note the redundant term z can be rewritten as Therefore, one more θ (n ) gate can wipe it out, z (((x x s )x 3 s )x 4 )x n. (6.3.5) y y z y x x x 3 x n. (6.3.6) We used totally n 3 scratch bits and (n 5) + ((n ) + 5) 4n number of θ (3) gates here. 6.4 Quantum circuit model A Quantum Circuit model is composed of a series of single-qubit-gates SU()/U() and two-qubit-gates SU(4)/U(4). Diagrammatic representation as shown in Figure 6.. x U y (a) single-qubit gate x y x U y (b) two-qubit gate x y x y U x n y n (c) n-qubit gate Figure 6.: Quantum Circuit Model of single-qubit, two-qubit,and n-qubit gates 66

77 6.4. Definition of quantum circuit Def 6.4. (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 6.5. Table 6.5: Classical Circuit and Quantum Circuit Note Computation Classical Computation Quantum Computation Object Operation Reversibility Complexity n-bit logic gate: {0, } n AND, OR, NOT,etc. n-qubit H n quantum gate: U( n ) group irreversible (most cases) reversibe reversible P vs. NP BPP vs. BQP 6.4. One-qubit gates The one qubit gates are U() or SU() operators. And we shall see that U() operator is equivalent to SU() operator up to a phase coefficient. There are some examples of the one-qubit gate: Hadamard gate H, Phase gate S, π 8 gate T : Pauli X gate: Pauli Z gate: Phase-shift gate R θ : H ( ) ; (6.4.) S ( 0 0 i ) ; (6.4.) T ( 0 0 e iπ/4) eiπ/8 ( e iπ/8 0 0 eiπ/8) ; (6.4.3) X σ x ( 0 0 ) ; (6.4.4) Z σ z ( 0 0 ) ; (6.4.5) R θ ( 0 0 e iθ ) ; (6.4.6) 67

78 Thm Any U() group element can be expressed as U e iα D z (β)d y (γ)d z (δ), (6.4.7) where D z (β), D y (γ), and D z (δ) are rotation gates defined as D y (γ) e iσ y γ/ ( cos γ sin γ sin γ cos γ ), (6.4.8b) D z (β) e iσ z β/ ( e iβ/ 0 0 eiβ/), (6.4.8a) D z (δ) e iσ z δ/ ( e iδ/ 0 0 eiδ/). (6.4.8c) Proof. Now, we are going to prove the theorem step by step. Step. Any U() group element U should satisfy the unitary condition Let s assume that From E.Q. (6.4.9) we can get UU U U I. (6.4.9) U ( a b c d ), with a, b, c, d C. (6.4.0) det U det (UU ), namely det U e iα, with 0 α < π and α R. (6.4.) Step. With the E.Q. (6.4.) in hand, we can denote U as U e iα U, (6.4.) with and Thus det U, and U U U U I, (6.4.3) U ( a b ) SU(), with a, b, c, d C. (6.4.4) c d (a b) ( a b )aa + bb, (c d) ( c d )cc + dd, (a b) ( c d )ac + bd 0, (6.4.5a) (6.4.5b) (6.4.5c) det ( a b ) ad bc. (6.4.5d) c d Thus, we can conclude that 68

79 E.Q.(6.4.5a) and E.Q.(6.4.5b) 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.(6.4.5c) can spell two constraint on U. It seems that E.Q.(6.4.5d) can also put two constraint on U. But, we can find out that the constraint (6.4.5d) can be derived from the former three equations (6.4.5a) (6.4.5c). Therefore, the E.Q.(6.4.5a) (6.4.5d) would set totally five constraints on the parameters of U. Consequently, there are only degrees of freedom left for the SU() group element U. Step 3. From the constraint E.Q.(6.4.5c) we can derive namely ac + bd 0 ac bd, a c + b d0 a c b d, a c b d. (6.4.6) Combine E.Q.(6.4.5a) and E.Q.(6.4.5b) with E.Q(6.4.6), and we obtain a ( d ) ( a ) d, i.e., Therefore Thus, a d. a d, b c, a + b. a d, b c, a + b, 0ac + bd, ad bc. (6.4.7) (6.4.8a) (6.4.8b) (6.4.8c) (6.4.8d) (6.4.8e) Hence, we can denote a, b, c, d in the form of af a cos γ, b f b sin γ, cf c sin γ, df d cos γ. (6.4.9a) (6.4.9b) (6.4.9c) (6.4.9d) 69

80 Step 4. Then, we have f a f b f c f d, f a f c f b f d, f a f d f b f c, (6.4.0a) (6.4.0b) (6.4.0c) namely f a f b f c f d, f c f b, f d f a. (6.4.a) (6.4.b) (6.4.c) Therefore, we can assume that f a e i(β+δ)/, (6.4.a) fbe i(β δ)/, (6.4.b) f c e i(β δ)/, f d e i(β+δ)/, (6.4.c) (6.4.d) i.e., U e iα ( e i(β+δ)/ cos γ e i(β δ)/ sin γ e i(β δ)/ sin γ e i(β+δ)/ cos γ ) e iα ( e iβ 0 (e iδ/ 0 e iβ/) cos γ e iδ/ sin γ e iδ/ sin γ e iδ/ cos γ ) e iα ( e iβ 0 γ 0 e iβ/) (cos sin γ sin γ cos γ ) ( e iδ/ 0 0 e δ/) e iα D z (β) D y (γ) D z (δ). (6.4.3) Notes: Relations between rotations: D y ( π ) D x (α)d y ( π )D z (α) (6.4.4) D z ( π ) D x (α)d z ( π )D y (α) (6.4.5) Examples: HD x (π)d y ( π ) Ph (π ), Hadamard gate, (6.4.6) NOTD x (π)ph ( π ), NOT gate, (6.4.7) with These can be verified easily. Ph(θ) e iθ I. (6.4.8) 70

81 Hadamard gate: D x (π)cos ( π ) i sin (π ) σ x iσ x, D y ( π )cos (π 4 ) i sin (π 4 ) σ y (I iσ y ), from which we can get the left-hand-side of E.Q. (6.4.6) D x (π)d y ( π ) Ph (π ) σ x (I iσ y ) (σ x iσ x σ y ) (σ x + σ z ) H. NOT gate X: D x (π)ph ( π ) (cos π i sin π σ x ) i 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() 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 Hadamard gate: H Def 6.4. (Hadamard gate). is named as Hadamard gate. H H (X + Z) ( ), (6.4.9) We can easily verify the following relations by utilizing the properties of the Pauli matrices, HXH Z; (6.4.30) HZH X; (6.4.3) HY H Y ; (6.4.3) H H; (6.4.33) H I. (6.4.34) As we can see that Hadamard gate H is not only a unitary transformation but also selfadjoint. Proof. From the definition (6.4.9) of the Hadamard gate H, and the properties of the Pauli matrices, we can derive the properties (6.4.30) (6.4.34) of the Hadamard gate. 7

82 () For E.Q. (6.4.30), we can evaluate its left-hand-side, HXH (X + Z)X (X + Z) (X + Z)X(X + Z) (XXX + ZXX + XXZ + ZXZ) (X + Z + Z XZZ) (X + Z + Z X) Z, which is equivalent to the corresponding right-hand-side of E.Q. (6.4.30). () As for E.Q. (6.4.3), we can get in the following manner HZH (X + Z)Z (X + Z) (X + Z)Z(X + Z) (XZX + ZZX + XZZ + ZZZ) ( ZXX + X + X + Z) ( Z + X + X + Z) X. (3) We do the same trick as in the former two cases to derive E.Q. (6.4.3), HY H (X + Z)Y (X + Z) (X + Z)Y (X + Z) (XY X + ZY X + XY Z + ZY Z Z) ( Y i + i Y ) Y. (4) Because the Pauli matrices X and Z are self-adjoint, therefore should be the Hadamard operator H, H (X + Z) (X + Z ) (X + Z) H. 7

83 (5) Still from the definition (6.4.9) of the Hadamard gate H, we can calculate the square of the Hadamard operator H, H (X + Z) (X + Z + {X, Z}) (I + I + 0) I. Therefore, we have shown that H gate is a self-adjoint unitary matrix. In the view point of rotation operator, we have H (σ x + σ z ) ( e x + e z ) σ, namely H n σ, with n (ê x + ê y ) R 3. (6.4.35) Thus, ih exp ( i θ n σ), θπ which we can put in another way ih D( n, π), (6.4.36) namely a unitary transformation induced by the 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: and { { H 0 +, H, H + 0, H. (6.4.37) (6.4.38) E.Q. (6.4.37) can be easily derived by utilizing the definition (6.4.9) of the Hadamard gate H as shown in the following H 0 (X + Z) 0 ( + 0 ) +, and H (X + Z) ( 0 ). 73

84 While E.Q. (6.4.38) can be deduced by applying H on both sides of E.Q. (6.4.37) and utilizing E.Q. (6.4.34). Furthermore, We can reformulate E.Q. (6.4.37) into one index form, expressed as H i (( ) i i + ī ), (6.4.39) where i 0,. On the other hand we can also get also as a consequence of E.Q. (6.4.37) H( 0 ±i ) H y +i y, H y i y, (H 0 ±ih ) (6.4.40) ( + ±i ) ( 0 + ±i 0 i ) ((±i) 0 + ( i) ) ±i i ( 0 + ±i ) ±i ( 0 i ) Controlled two-qubit gates and controlled three-qubit gates Def (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 (Controlled U gate). Let s denote the Controlled U gate as CU, then which means CU c t c U c t, (6.4.4) if c 0, 0 t 0 t, if c, t U t, where c is the control qubit, and t is the target qubit. This can also be represented in the diagrammatical formalism: CU One example of the Controlled U gate is the CNOT gate, i.e., U X. 74

85 CNOT gate Def (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, (6.4.4) where we call a as the controlled qubit, b as the target qubit. The CNOT gate is usually represented with the diagram Remarks: () With a 0, then With a, then CNOT ab a a b b a. CNOT 0 b 0 b. (6.4.43) CNOT b b. (6.4.44) It is the reason why the CNOT gate is called controlled not gate, namely if target bit is, we do the NOT operation. () The controlled-u gate is defined as CU c t c U c t. (6.4.45) CU See more details and properties of the controlled-u gate in following section (3) From the definition of the CNOT gate, we can verify that since U (CNOT) I 4, (6.4.46) with a, b 0,. (CNOT) a, b CNOT a, b a a, (b a) a a, b, (4) The CNOT gate can be expressed as by which we can get the matrix formalism, CNOT 0 0 I + X. (6.4.47) CNOT ( ) ( 0 0 ) + (0 0 0 ) (0 0 ) namely CNOT. (6.4.48) 75

86 (5) CNOT gate is the quantum analogy of the classical gate XOR (the Exclusive OR gate): XOR (a, b) a b, irreversible, bits bit CNOT ( a, b ) qubits ( a, a b ), reversible. qubits (6) CNOT SU(4). Since we have derived the matrix formalism, we can see and On the other hand, we know that (CNOT) I 4, thus det CNOT, (6.4.49) (CNOT) CNOT. (6.4.50) CNOT(CNOT) CNOT(CNOT) (CNOT) I 4. Therefore, we can infer that CNOT SU(4). Lemma The CNOT gate has the following properties (X X)CNOTCNOT(X I ), (I X)CNOTCNOT(I X), (6.4.5a) (X I )CNOTCNOT(X X), (6.4.5b) (6.4.5c) (Z Z)CNOTCNOT(I Z), (Z I )CNOTCNOT(Z I ). (6.4.5d) (I Z)CNOTCNOT(Z Z), (6.4.5e) (6.4.5f) Proof. Now, we are going to verify E.Q.(6.4.5a) E.Q.(6.4.5f), one by one, from the expression (6.4.47) of CNOT gate. (a) (X X)CNOT CNOT(X I ) (b) (X I )CNOT CNOT(X X) (X X)CNOT (X X)( 0 0 I + X) X 0 0 X + X X X XI X I I ( X I )(X I ) CNOT(X I ). (6.4.5) (X I )CNOT (X I )( 0 0 I + X) X 0 0 I + X X X XX XX X I X ( X I )(X X) CNOT(X X). (6.4.53) 76

87 (c) (I X)CNOT CNOT(I X) (d) (Z Z)CNOT CNOT(I Z) (e) (I Z)CNOT CNOT(Z Z) (f) (Z I )CNOT CNOT(Z I ) (I X)CNOT (I X)( 0 0 I + X) 0 0 X + X 0 0 I X + XX ( 0 0 I + X)(I X) CNOT(I X). (6.4.54) (Z Z)CNOT (Z Z)( 0 0 I + X) Z 0 0 Z + Z ZX 0 0 Z ( XZ XZ) 0 0 I Z + XZ ( 0 0 I + X)(I Z) CNOT(I Z). (6.4.55) (I Z)CNOT (I Z)( 0 0 I + X) 0 0 Z + ZX 0 0 Z I Z + Z XZ ( 0 0 I + X)(Z Z) CNOT(Z Z). (6.4.56) (Z I )CNOT (Z I )( 0 0 I + X) Z 0 0 I + Z X 0 0 Z I + Z XI ( 0 0 I + X)(Z I ) There we get all the six relations (6.4.5a) (6.4.5f) verified. CNOT(Z I ). (6.4.57) On the other hand, we can present these properties of the CNOT gate, namely the six relations (6.4.5a) (6.4.5f), in the form of quantum circuit diagram, which is shown in Figure 6.. Lemma The Hadmard gate and CNOT gate have the following connection with CNOT and CNOT defined as (H H)CNOT (H H) CNOT, (6.4.58) { CNOT a, b a, a b, CNOT a, b a b, b. (6.4.59) 77

88 Proof. The prove is very simple. We can utilize the properties of the Hadmard gate as expressed in E.Q.(6.4.30) (6.4.34) in the following way (H H)CNOT (H H) (H H)( 0 0 I + X)(H H) H 0 0 H H + H H HXH HXH + + I + Z + + ( ) + ( 0 0 ) ( ) ( + + ) I X CNOT. (6.4.60) In this derivation, we have also used the following facts CNOT 0 0 I + X, CNOT I X, and I 0 0 +, I + + +, Z 0 0, X + +. And we can also represent the relation (6.4.58) between the Hadmard gate and CNOT gate in the diagram formalism, see Figure 6.3: X X X (a) (X X)CNOT CNOT(X I ) X X X (b) (X I )CNOT CNOT(X X) X X Z Z Z (c) (I X)CNOT CNOT(I X) (d) (Z Z)CNOT CNOT(I Z) Z Z Z Z Z (e) (I Z)CNOT CNOT(Z Z) (f) (Z I )CNOT CNOT(Z I ) Figure 6.: Properties of the CNOT gate (6.4.5a) (6.4.5f). 78

89 H H H H Figure 6.3: (H H)CNOT (H H) CNOT 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 Analogy to the classical reversible gate, the Quantum Toffoli gate is defined as where x, y, z 0, and it has the diagram formalism, θ (3) x, y, z x, y, z xy, (6.4.6) x x y y z z xy. Remark: Quantum Toffoli gate with the Hadamard gate and phase gate is the universal quantum gate set, namely these three gates can perform universal quantum computation. Def The Quantum Fredkin gate is defined as where x, y, z 0, and it has the diagram formalism, θ (3) x, y, z x, xz xy, xy xz, (6.4.6) x x y xz xy z xy xz Quantum circuit model of Bell states Thm 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 ψ(i, j) CNOT(H I ) j, i, with i, j 0,, (6.4.63) which can also be expressed in the Quantum circuit model phase-bit j H parity-bit i Here we present two types of proofs. ψ(i, j). Proof. The first type of proof. From the index formulation of the Hadamard gate E.Q. (6.4.39), (H I ) j i (( ) j j i + j i ). (6.4.64) 79

90 Following the CNOT gate, we get CNOT(H I ) j i (( ) j j i j + j i j ). (6.4.65) And considering the case j 0 and j, j 0 j ( 0 i + ī ) ψ(i, 0) ( 0 i ī ) ψ(i, ) (6.4.66) Conclusively, we have CNOT(H I ) j i ψ(i, j). (6.4.67) Proof. The second type of proof. To prove E.Q.(6.4.63), we may make use of the definition (3.3.) of the Bell states and E.Q. (3.3.3). We are going to reach our destination by two main steps. Step : We will prove that ψ(0, 0) CNOT(H I ) 00. As we can show that CNOT(H I ) 00 CNOT(H 0 0 ) CNOT( + 0 ) CNOT( ) ( 00 + ) ψ(0, 0). (6.4.68) Step : With the equation ψ(i, j) (I X i Z j ) ψ(0, 0), we can get ψ(i, j) (I X i Z j ) ψ(0, 0) (I X i Z j )(CNOT(H I ) 00 ) (I X) i (I Z) j CNOT(H I 00 ) (I X) i CNOT(Z Z) j (H I 00 ) CNOT(I X) i (Z Z) j (H I ) 00, in the last two jumps, we have used the relations (6.4.5e) and (6.4.5c), which we can verified in the following context: and (I X) i CNOT (I X) i CNOT(I X) CNOT(I X) i, (I Z) j CNOT (I Z) i CNOT(Z Z) CNOT(Z Z) j. 80

91 And we should notice that Thus, we can infer that (I X) i (Z j Z) j Z j X i Z j ( ) j Z j X i Z j X ( ) ij Z j Z j X i ( ) ij (Z j Z j )(I X i ). (6.4.69) ψ(i, j) ( ) ij CNOT(Z j Z j )(I X i )(H I ) 00 On the other hand, if we ve noticed that then Hence, ( ) ij CNOT(Z j Z j )(H X i ) 00 ( ) ij CNOT(Z j Z j )(H I )(I X i ) 00 ( ) ij CNOT(Z j Z j )(H I ) 0i. HZH X H } ZH HX, (6.4.70) I Z j H Z j HX ψ(i, j) ( ) ij CNOT(Z j H Z j ) 0i which is equivalent to E.Q. (6.4.63). HX j. (6.4.7) ( ) ij CNOT(HX j Z j ) 0i ( ) ij CNOT(H I )(X j Z j ) 0i ( ) ij CNOT(H I )(X j 0 Z j i ) ( ) ij CNOT(H I )( j ( ) ij i ) CNOT(H I ) ji, Quantum circuit model of GHZ states Bell states Most important and popular state Two-qubit maximally entangled Bell inequality GHZ states Most important and popular state Multi-qubit maximally entangled GHZ theorem 8

92 Thm GHZ n ( 0x x 3 x n + ( ) x x x 3 x n ) CNOT n CNOT (n ) CNOT H x x x n, (6.4.7) where CNOT ij is the CNOT gate with the i-th qubit as the control qubit and j-th qubit as the target qubit. And this can be represented in the diagram formulism, shown in Figure 6.4. x H x x 3 x n GHZ Figure 6.4: GHZ state generated by the CNOT and H gates. e.g.. Three qubit GHZ state. GHZ 3 ( ). (6.4.73) 0 H 0 GHZ 3 0 t 0 t t t 3 Figure 6.5: GHZ 3 state generated by the CNOT and H gates. t ; t ( ); t t 3 ( ); ( ). e.g.. Four qubit GHZ state. GHZ 4 ( ). (6.4.74) 8

93 H 0 GHZ 4 t 0 t t t 3 t 4 Figure 6.6: GHZ 4 state generated by the CNOT and H gates. t 0 0 ; t ( 00 0 ); t t 3 t 4 ( 00 + ); ( ); ( ). Let s now consider the complete set of observables defining the GHZ state. The phase-bit operator X X X n : (X X X n ) CNOT n CNOT (n ) CNOT H x x x n CNOT n (X X X n I ) CNOT (n ) CNOT H x x x n i.e., CNOT n CNOT (n ) (X X n I I ) CNOT (n ) CNOT H x x x n CNOT n CNOT (n ) CNOT (X I I )H x x x n n CNOT n CNOT (n ) CNOT H (Z I I ) x x x n, n (X X X n )CNOT n CNOT (n ) CNOT H x x x n ( ) x CNOT n CNOT (n ) CNOT H x x x n, (6.4.75) where we have utilized the relation (6.4.5a), namely (X X ) CNOT CNOT (X I ), and the property HXH Z H } XH HZ. I 83

94 The first parity-bit operator Z Z I I : n (Z Z I I )CNOT n CNOT (n ) CNOT H x x x n n CNOT n CNOT (n ) CNOT 3 (Z Z I I )CNOT H x x x n n CNOT n CNOT (n ) CNOT (I Z I I )H x x x n n CNOT n CNOT (n ) CNOT H (I Z I I ) x x x n n ( ) x CNOT n CNOT (n ) CNOT H x x x n, (6.4.76) where we have used the relations (6.4.5d) and (6.4.5f), namely (Z Z)CNOT CNOT (I Z), (Z I )CNOT CNOT (Z I ). The second parity-bit I Z Z 3 I I : n 3 (I Z Z 3 I I )CNOT n CNOT (n ) CNOT H x x x n n 3 CNOT n CNOT 4 (I Z Z 3 I I )CNOT 3 CNOT H x x x n n 3 CNOT n CNOT (n ) CNOT 3 (Z Z Z 3 I I )CNOT H x x x n n 3 CNOT n CNOT (n ) CNOT (I Z Z 3 I I )H x x x n n 3 CNOT n CNOT (n ) CNOT H (I Z Z 3 I I ) x x x n n 3 ( ) x +x 3 CNOT n CNOT (n ) CNOT H x x x n, (6.4.77) to derive this result we have to employ the relations (6.4.5d) and (6.4.5f), i.e., (Z Z)CNOT CNOT (I Z), (I Z)CNOT CNOT (Z Z). 84

95 The third parity-bit I I Z 3 Z 4 I I : n 4 (I I Z 3 Z 4 I I )CNOT n CNOT (n ) CNOT H x x x n n 4 CNOT n CNOT 5 (I I Z 3 Z 4 I I )CNOT 4 CNOT 3 CNOT H x x x n n 4 CNOT n CNOT 4 (Z I Z 3 Z 4 I I )CNOT 3 CNOT H x x x n n 4 CNOT n CNOT 3 (I I Z 3 Z 4 I I )CNOT H x x x n n 4 CNOT n CNOT H (I I Z 3 Z 4 I I ) x x x n n 4 ( ) x 3+x 4 CNOT n CNOT H x x x n. (6.4.78) The (n )-th parity-bit I I Z n Z n : n ( I I Z n Z n )CNOT n CNOT (n ) CNOT H x x x n n CNOT n (Z I I Z n Z n )CNOT (n ) CNOT H x x x n n 3 CNOT n CNOT (n ) ( I I Z n Z n )CNOT (n ) CNOT H x x x n n CNOT n CNOT H ( I I Z n Z n ) x x x n n ( ) x n +x n CNOT n CNOT H x x x n. (6.4.79) 6.5 Universal quantum computation 6.5. Quantum universal gate set The definition of the Universal Quantum Gate set is analog to the Universal Classical gate set. Def 6.5. (Universal Quantum Gate set). A set of quantum gates is universal if any unitary operator U SU() (or U U(), 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()}. 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. 85

96 Def 6.5. (Finite Approximately Universal Quantum Gate set). A finite gate set is approximately universal, if any unitary operator in SU( n ) can be approximately expressed as a product of elementary gates in this set to arbitrary precisions. Examples: ) {H, S, θ (3) }, H being the Hadamard gate, S representing the Phase gate, and θ (3) for the three-bit quantum Toffoli gate, i.e., H ( ), S ( 0 0 i ), θ(3) x, y, z x, y, z xy. 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. ) {H, T, CNOT} with T as the π 8 gate and CNOT denoted the CNOT gate,namely, Remark: S T. T e i π e i π ( 0 e i π ), CNOT x, y x, x y. 8 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 Universal quantum gate set of two-qubit gates Def (Generic two-qubit gates). The two-qubit gate with eigenvalues e iθ, e iθ, e iθ 3 and e iθ 4, where θ i π and θ i θ j are irrational numbers, is defined as Generic two-qubit gate. To completely understand this, we have to learn the Number Theory in mathematics. Thm {Any Generic two-qubit gate} is an universal quantum gate set. Thm (Barenco s gate). Barenco s gate (995) is defined as Controlled Ph ( π 4 ) D x ( θ ), (6.5.) where θ π is an irrational number. It can be represented in the diagram formalism as U with U Ph ( π 4 ) D x ( θ ) e i π cos θ 4 ( 4 i sin θ 4 i sin θ 4 cos θ ). (6.5.) 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. 86

97 Thm {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 CNOT CNOT 3 CNOT, (6.5.3) and with the following Lemmas, reformulate Barenco s gate as an expression in terms of the CNOT gate and single-qubit gates. Lemma Any single-qubit gate U, can be described in terms of Euler angles, namely U e iα D z (β)d y (γ)d z (δ), with α, β, γ, δ R. (6.5.4) Lemma Any single-qubit gate U can be formulated as U Ph (α)axbxc AXBXC, with ABC I, A, B, C SU() and α R. (6.5.5) Lemma Any controlled two-qubit gate can be decomposed in the following way U R α C B A (6.5.6) with R α ( 0 0 e iα ). Proof. We are now going to prove the three lemmas 6.5.., and Lemma : The lemma is proved in subsection 6.4., section 6.4, chapter 6. Lemma : The lemma can be proved through lemma Because A, B, C are all elements of the SU() group, therefore, from lemma 6.5.., we can construct A, B, C in the form of: AD z (φ)d y (ϕ)d z (ψ), (6.5.7a) BD z (φ )D y (ϕ )D z (ψ ), CD z (φ )D y (ϕ )D z (ψ ), (6.5.7b) (6.5.7c) where we have introduced 9 parameters. The expressions (6.5.7a) (6.5.7c) ensures that A, B and C all belong to SU(), therefore ABC SU(). Since we can parameterize an arbitrary SU() operator with only two independent real variables, the expressions (6.5.7a) (6.5.7c) have two constraints. And the relation ABC I (6.5.8) gives rise to another constraint on the parameters that we have introduced in the expressions (6.5.7a) (6.5.7c). Therefore, we actually have 6 independent variables in the expressions (6.5.7a) (6.5.7c). We do not need that much of independent 87

98 variables, since U U() can be determined with three independent real variables. We may set As we shall see that AD z (β)d y ( γ ), (6.5.9a) BD y ( γ ) D z ( β + δ ), (6.5.9b) CD z ( β + δ ). (6.5.9c) ABC D z (β)d y ( γ ) D y ( γ ) D z ( β + δ ) D z ( β + δ ) D z (β)d y (0) D z ( β) I, and AXBXC D z (β)d y ( γ ) XD y ( γ ) D z ( β + δ ) XD z ( β + δ ) D z (β)d y ( γ ) D y ( γ ) D z ( β + δ ) D z ( β + δ ) D z (β)d y (γ) D z (δ), (6.5.0) from which we can see that theorem can be derived from theorem Lemma 3: The lemma can be proved by utilizing the lemma The controlled-u gate can be expressed as CU 0 0 I + U ( I U ) (6.5.) By inserting E.Q.(6.5.5) into E.Q.(6.5.), and we can get CU ( ABC e iα AXBXC ) ( I e iα I ) ( A A ) (I X ) (B B ) (I X ) (C C ) C Ph(α) ( A ) CNOT (B A ) CNOT (C B C ), (6.5.) where C Ph(α) is the controlled-phase gate, C Ph(α) 0 0 I + e iα I ( I e iα I ) ( 0 0 e iα ) I R α I, (6.5.3) 88

99 namely Ph (α) On the other hand, we can also obtain R α. (6.5.4) ( B B )I B B ; (6.5.5b) ( A A )I A A ; (6.5.5a) ( C C )I C C. (6.5.5c) On the other hand we know that CU U. (6.5.6) By substituting E.Q. (6.5.3), (6.5.5a) (6.5.5c) and (6.5.6) in E.Q. (6.5.), we can acquire E.Q. (6.5.6) Deutsch s gate is a universal quantum gate Thm (Deutsch s gate). Deutsch gate (989) is a universal quantum gate, where Deutch s gate x y z x y R xy z, (6.5.7) R Ph ( π ) D x (θ) ie iσ x θ, (6.5.8) with θ π being an irrational real number. And it has the diagrammatical formalism, Deutch s gate This is the first universal quantum gate, and is a three-qubit gate. Note that R Ph ( π ) D x (θ) [Ph ( π 4 ) D x ( θ )] ( U ), 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, R R U U U t 0 t t t 3 t 4 t 5 (6.5.9) 89

100 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. (6.5.9): () ψ(t 0 ) x y z ; () ψ(t ) x y U y z ; (3) ψ(t ) x x y U y z ; (4) ψ(t 3 ) x x y (U ) x y U y z x x y U x y U y z ; (5) ψ(t 4 ) x x y x U x y U y z x y U x y U y z ; (6) ψ(t 5 ) x y U x U x y U y z x y U x+y x y z x y U x+y (x+y xy) z x y U xy z x y R xy z. Note: the calculation here is binary, namely where x, y {0, }. x y (x + y) mode x + y xy, (6.5.0) 90

101 Chapter 7 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. Reference: [Preskill] Chapter 6: Quantum computation; [Nielsen & Chuang] Chapter 6: Quantum search algorithms 7. Classical and quantum algorithm Def 7.. (Algorithm). An algorithm is. a well-defined procedure;. with finite descriptions; Nielsen & Chuang 9

102 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 7.: Table 7.: Examples of Quantum Algorithm Speed up Non-exponential speed up Relativized exponential speed up Exponential speed up Examples Grover s search algorithm (Oracle model) Simon s algorithm (Oracle model) Shor s factoring algorithm 7. Oracle model Oracle model is the black-box model. Def 7.. (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: () 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 y U f x y f(x). 9

103 () Quantum black box is a quantum gate U f computing f(x): x U f x y y f(x), where x is called the register qubit, and y is called the ancilla qubit. Note that quantum gate U f satisfies the unitary (reversible) condition given by The Query Complexity: U f U f U f U f I d. (7..) In the oracle model, we only count the minimal number of the black box to characterize 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 7.3 Deutsch s algorithm Time (U f ) year; Time (other gates) sec. (7..) The Deutsch s algorithm is the first Quantum Algorithm, presented in Definitions Def 7.3. (Constant function and Balanced function). Let s define a function f: Then, the function f is f x {0, } y f(x) {0, }. (7.3.) Constant function, if f(0)f(); (7.3.) { Balanced function, if f(0) f(). (7.3.3) In the case that f is constant, i.e., f(0) f(), we can acquire f(0) f() 0, or f(0) f(). (7.3.4) Therefore, with the binary addition. f(0) f() 0 (7.3.5) In the case that f is balanced, i.e., f(0) f(), we can know Similarly, f(0) 0, f(), or f(0), f() 0. (7.3.6) f(0) f() (7.3.7) which suggests that the number of x which satisfies f(x) 0 and x that f(x) is same, namely #{x f(x) 0, x {0, } } #{x f(x), x {0, } }. (7.3.8) 93

104 Question: Assume that the U f is a black box of computing f(x), what is the minimum number of performing U f to judge whether f(x) is a constant or balanced function? 7.3. Classical algorithm The minimum number of performing U f to know about whether f(x) is constant or balanced is two. For x x y U f y f(x), we can construct the following classical circuit to verify whether f(x) is constant or balanced, 0 0 U f 0 f(0) (7.3.9) U f 0 f(0) f(), where obviously the f(0) and f() have to be calculated indepently Deutsch s problem Deutsch s problem is firstly presented in 989, 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 U f x y y f(x). (7.3.0) Promise: f(x) is constant or balanced. Question: Determine f(x) constant or balanced. Answer: time of using U f. Note: In quantum mechanics, f(0) and f() can be computed simultaneously, because of quantum superposition principle Phase kick-back Lemma With the quantum black box U f defined as with the diagrammatical representation U f x y x y f(x), (7.3.) x we can prove the following algebraic formula, U f x y y f(x), (7.3.) U f x 0 ( ) f(x) 0 x, (7.3.3) 94

105 i.e., the quantum circuit model given by or equivalently x 0 x 0 U f U f x ( ) f(x) 0. ( ) f(x) x 0. (7.3.4) (7.3.5) Note: Why is the above quantum circuit model called the phase kick-back? The global phase ( ) f(x) can be kicked back from the target qubit to the control quibt, so that the target qubit 0 seems unchanged from the input to the output. This lemma can be proved in the following manner. Proof. U f x 0 U f x 0 U f x x f(x) x f(x) x f(x) x f(x) x f(x) f(x) if f(x) 0, x ( 0 ); if f(x), ( ) f(x) x There we get E.Q. (7.3.3) verified. Example: If f(x) x, then U x CNOT, since x ( 0 ) ( ) x ( 0 ); 0. (7.3.6) U x x y x y f(x) x y x CNOT x y. (7.3.7) And we can see that CNOT x 0 x x x x ( ) x x so that in such the circumstance we have CNOT x ( ) x x. 0, (7.3.8) Note: The phase kick-back is the key technique in the performance of such quantum algorithms as Deutsch s algoithm, Deutsch-Jozsa a algorithm, Grover s search algorithm, and so on. 95

106 7.3.5 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 0 H U f H t t t 3 t 4 (7.3.9) where we study it in the following five steps:. Input: state preparation. ψ(t ) 0 0,. ψ(t ) (H I ) ψ(t ) 0 + 0, We have utilized the relation that 0 + H 0 0 H H i ( ) i j j (7.3.0) j0 where i j is AND operation. 3. ψ(t 3 ) U f ψ(t ) i0 U f i 0 i0 ( )f(i) i 0, 4. ψ(t 4 ) (H I ) ψ(t 3 ) i0 ( )f(i) (H I ) i 0 i,j0 ( )f(i)+i j j Output: quantum measurement. And the icon stands for the measurement with respect to the standard basis { 0, }. If we perform the measurement 0 0 I, there we can get the state after measurement for the bipartite system as, And we can see that, j 0 ψ(t 4 ) i0 ( ) f(i) 0 0. (7.3.) 96

107 () if f is a constant function, i.e., f(0) f(), the probability that the first qubit is in the state 0 is, since ( 0 0 I ) ψ(t 4 ) ( ) f(0) 0 0, (7.3.) ( I ) ψ(t 4 ) i0 ( ) f(i)+i i0 0 ( ) f(0) ( ) i 0 0; (7.3.3) () if f is a balanced function, i.e., f(0) f(), the probability that the first qubit is in the state 0 is 0, due to ( 0 0 I ) ψ(t 4 ) i0 ( ) f(i) (7.3.4) Note: What is underlying Deutsch s algorithm? The f(0) and f() are calculated simultaneously so that f(0) f() is calculated using the black box U f once. This is called quantum parallelism due to quantum interference or quantum superposition principle. 7.4 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 Constant and balanced function in n-qubit Def 7.4. (Constant and Balanced function (n-qubit)). Let s define the function f to be Then function f is f x {0, } n f(x) {0, }. (7.4.) Constant function, if f(x) c, x {0, } n, where c is a constant in {0, }; Balanced function, if the number of elements in the set {x f(x) 0, x {0, } n } is the same as that in {x f(x), x {0, } n }, i.e., the numbers both equal to n. #{x f(x) 0, x {0, } n } #{x f(x), x {0, } n } n. (7.4.) 7.4. 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: 97

108 If we choose to use the classical deterministic algorithm, there we should make at least n + queries in the worst case. If we use the classical random algorithm, there are at least order() queries. If we choose Deutsh-Jozsa s algorithm, one query will suffice, which is due to the application of the quantum superposition principle Notation and lemma State: Product : with Summation: { x x x x n, (7.4.3a) y y y y n. (7.4.3b) x y x y x y x n y n, (7.4.4) x i y i x i AND y i. (7.4.5) y {0,} n y 0 y 0 y n0. (7.4.6) Lemma H (n) x ( H H H ) x x x n n n/ ( ) x y y. (7.4.7) y {0,} n Proof. H (n) x ( H H H ) x x x n n H x H x H x n ( )n y 0 ( )n ( )n ( ) x y y y 0 y 0 y 0 y 0 y n0 y n0 y 0 ( ) x y y y n0 ( ) xn yn y n ( ) x y +x y + +x n y n y y y n ( ) x y x y x n y n y y y n n/ y {0,} n ( ) x y y, (7.4.8) where we have utilized E.Q. (7.3.0). Therefore, we have verified E.Q. (7.4.7). Note: H (n) 0 n/ x {0,} n x. (7.4.9) 98

109 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 (Phase kick-back for n-qubit). Let s define U f as with and U f a y a y f(a) (7.4.0) f a {0, } n f(a) {0, }, y {0, }. 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, 0 U f x ( ) f(x) 0 x, (7.4.) with x {0, } n. This can also be presented in the diagram formalism: Proof. x x x x U f x n x n 0 ( ) f(x) 0. 0 U f x f(x) + f(x) x x f(x) f(x) (7.4.) ( ) f(x) x 0. (7.4.3) We have made use of the same technique exploited to prove lemma , here Deutsch-Jozsa s algorithm Deutsch-Jozsa s algorithm can be expressed in quantum circuit model: 0 H H where 0 H H U f 0 n H H 0 t t t 3 t 4 ( ) f(x) 0, (7.4.4) 99

110 () Input: state preparation. ψ(t ) 0 n 0, () ψ(t ) [(H n ) I ] ψ(t ) n/ x {0,} n x 0, (3) The phase kick-back technique. ψ(t 3 ) U f ψ(t ) U n/ f x {0,} n x 0 n/ x {0,} n( ) f(x) x 0, (4) ψ(t 4 ) [(H n ) I ] ψ(t 3 ) [(H n ) I n/ ] x {0,} n( ) f(x) x 0 n x,y {0,} n( )f(x)+x y y 0. (5) Output: quantum measurement. If the post-measurement state of the first n-qubit is 0 n, then the corresponding state of the n + -qubit system should be Then, we can infer that if f is a constant function, i.e., ψ(t 4 ) n ( ) f(x) 0 n 0. (7.4.5) x {0,} n f(x) f(x ), x, x {0, } n, then the probability to find out that the first n-qubit is in the state 0 n is, since ( 0 n 0 I ) ψ(t 4 ) n ( ) f(0) 0 n 0 x {0,} n ( ) f(0) 0 n 0, (7.4.6) and with ( n I ) ψ(t 4 ) n ( ) f(x)+x n 0 x {0,} n n ( ) f(0)+x n 0 x {0,} n ( ) x n 0 x {0,} 0, (7.4.7) ( ) f(0) n ( ) x ( ) x + +x n. 00

111 if f is a balanced function, i.e., # {x f(x) 0, x {0, } n } # {x f(x), x {0, } n }, then the probability to find out that the first n-qubit is in the state n n is 0, since ( 0 n 0 I ) ψ(t 4 ) n ( ) f(x) 0 n 0 x {0,} n n ( ) f(x) 0 n 0 x {0,} 0. (7.4.8) Remark: Quantum superposition principle makes it possible that computing f(x) for all x {0, } n simultaneously, which is impossible in classical computation. 7.5 Bernstein-Vazirani s algorithm Firstly, let s introduce a lemma. Lemma ( ) (a+y) x n δ a,y, a, y {0, } n. (7.5.) x {0,} n This lemma can proved in the following way. Proof. ( ) (a+y) x x {0,} n x 0 ( ) (a +y ) x ( ) (a +y ) x x 0 x n0 ( ) (an+yn) xn which is equivalent to E.Q. (7.5.). Let s assume that there is a function f a defined as n δ a,y δ a,y δ an,yn n δ a,y, (7.5.) f a x {0, } n f a (x) {0, }. (7.5.3) And we also know some facts about the function (black box) that f a (x) a x a x a x a n x n. (7.5.4) Question: How can we determine a, which is an n-bit string? Solution With classical algorithm, we need totally n queries, i.e., n equations, to get a. 0

112 Solution With Quantum Algorithm, only one query is sufficient to obtain a. For the Quantum Algorithm, the corresponding Quantum Circuit model should be 0 H H 0 H H U fa 0 n H H 0 t t t 3 t 4 ( ) f(x) 0, (7.5.5) where () ψ(t ) 0 n 0, () ψ(t ) [(H n ) I ] ψ(t ) n/ x {0,} n x 0, (3) ψ(t 3 ) U fa ψ(t ) n/ U fa x {0,} n x 0 n/ x {0,} n( ) a x x 0, (4) ψ(t 4 ) [(H n ) I ] ψ(t 3 ) n/ [(H n ) I ] x {0,} n( ) a x x 0 n x,y {0,} n( )x (a+y) y 0 y {0,} n δ a,y y 0 a 0. With the quantum measurement on the n register qubits, we can get a, and equivalently we obtain a (a, a,, a n ). 7.6 Simon s algorithm Simon s Algorithm has no phase kickback; is the -st algorithm that shows Quantum Algorithm can have exponential speed up for apparently hard problems. 7.7 Grover s algorithm Let s now consider a search problem, the unstructured database search, with N 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. 0

113 7.7. Overview of the problem Let s assume that there is a black-box for computing an unknown function f ω defined as f ω x {0, } n f ω (x) {0, }. (7.7.) Question: How can we find a unique marked term x ω {0, } n such that f ω (x) {, if x ω; (7.7.a) 0, if x ω. (7.7.b) Solution With classical deterministic algorithm, we need N n steps, in the order of N, to find ω, namely checking the entire database item by item. Solution With classical random algorithm, we need N n steps, in the order of N, to find ω. 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 0 0 items Grover s algorithm Let s denote that the size of the database Z n as N #Zn n. In Quantum Mechanics, it is possible to encode the entire database into one state, denoted as state s given by s x N x {0,} n ω + N N x {0,} x ω x N ω + N N ω, (7.7.3) (7.7.4) where ω x. (7.7.5) N x ω Rewrite the state s into special type of two-dimensional Hilbert space H spanned by { ω, ω }, namely, H Span{ ω, ω }, where As we can show in Figure 7., where ω ω 0, ω ω. (7.7.6) θ arcsin N. (7.7.7) Therefore, we can also rewrite state s in the form of s sin θ ω + cos θ ω. (7.7.8) 03

114 ω s θ ω Figure 7.: State s as a state vector in the two-dimensional Hilbert space H Span{ ω, ω }. 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 H Span{ ω, ω }. The hint is that a geometric rotation can be decomposed as a product of two reflection operations. Now, we will show the way to find ω from s in the following steps. The first reflection operator U ω is defined as And we can get that And we can show that U ω I ω ω. (7.7.9) U ω ω (I ω ω ) ω ω ω ω, (7.7.0) U ω ω (I ω ω ) ω ω. (7.7.) U ω (I ω ω ) I 4 ω ω + ( ω ω ) Thus, with U ω acting on the state s, we can get I. (7.7.) s U ω s U ω (sin θ ω + cos θ ω ) sin θ ω + cos θ ω, (7.7.3) which can be shown in Figure 7., and the geometric implication of operator U ω is reflecting the state s around the ω axis. The second reflection operator U s is defined as U s s s I. (7.7.4) 04

115 ω s θ θ ω Figure 7.: The first reflection operator U ω in the geometrical picture. The state s is reflected around the axis ω. s As we can see that U s s ( s s I ) s s s s, (7.7.5) U s s ( s s I ) s s, (7.7.6) where s s 0 and s s. (7.7.7) With the second reflection operator acting on the first reflected state, we can get s U s s U s U ω s ( s s I ) ( sin θ ω + cos θ ω ) sin θ s s ω + cos θ s s ω + sin θ ω cos θ ω sin θ s + cos θ s + sin θ ω cos θ ω cos θ s + sin θ ω cos θ ω ( cos θ + ) sin θ ω + ( cos θ ) cos θ ω (cos θ + cos θ) sin θ ω + (cos θ sin θ) cos θ ω (cos θ sin θ + cos θ sin θ) ω + (cos θ cos θ sin θ sin θ) ω sin 3θ ω + cos 3θ ω. (7.7.8) And this can be shown in Figure 7.3, 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 R grov U s U ω. (7.7.9) Therefore, we can attain s R grov s. (7.7.0) 05

116 s ω 3θ s θ θ ω Figure 7.3: The second reflection from the state s to the s with the second reflection operator U s. s As we can see that the R grov rotates the original state s by the angle θ. We can define further that R n grov R grov R grov R grov. (7.7.) n Therefore, we can infer that R n grov s sin (n + )θ ω + cos (n + )θ ω. (7.7.) This is the so-called Grover iteration. As we shall see that, in the Grover problem, the angle θ is very small, since sin θ, with n. (7.7.3) N n Suppose that with T -iteration we can get the marked ω, therefore (T + )θ π sin θ N T π 4θ, θ, N namely Nπ T [ + O(N / )] N. (7.7.4) Example: N 4 more explicitly, it can be rewritten as f ω x {0, } f ω (x) {0, }, (7.7.5) (0, 0) f ω (0, 0), (0, ) f ω (0, ), (, 0) f ω (, 0), (, ) f ω (, ). (7.7.6) 06

117 Suppose that 0 is the marked item we are looking for in the database Z, i.e., f ω (, 0), f ω (0, 0) f(0, ) f(, ) 0, (7.7.7) and the initial state encoded the database can be set as s sin θ ω + cos θ ω, (7.7.8) where θ π, and ω 0. (7.7.9) 6 With one Grover s rotation R grov, we can get the initial state rotated counterclockwise through the angle θ π 3, which will lead to the angle between second reflected state and the state ω as θ + θ 3θ π, (7.7.30) which implies that the marked state 0 is found through one R grov operation. Check the result in algebraic approach s ; s U ω s ; s U s U ω s sin ( π 6 + π 6 ) 0 + cos (π 6 + π 6 ) 0 ; 0 (7.7.3) This can be shown in the Figure 7.4. ω s π 3 s π 6 π 6 ω s Figure 7.4: One Grover s Rotation with N 4 Note: The other diagrammatical approach to quantum search algorithm is shown below. s ( ) (7.7.3) 07

118 s U ω s ( ) (7.7.33) s U s U ω s (7.7.34) The needle is represented for the state, and the direction of the needle is represented for the relative sign of the state Quantum circuit model of Grover s algorithm Initial state is prepared in the way as in quantum circuit model, which can be shown as s H n 0 n, (7.7.35) 0 H s 0 H 0 H. (7.7.36) As we know that U ω I ω ω, (7.7.37) we can define the unitary gate U fω as U fω x y x y f ω (x), (7.7.38) with x being a n-qubit state, and y being a single qubit state, i.e., n qubits x x U fω single qubit y y f ω (x). (7.7.39) Therefore, we can get the phase kick-back U fω x 0 ( ) fω(x) 0 x. (7.7.40) Lemma U fω x 0 0 (U ω I ) x, (7.7.4) with the function f ω defined as f ω (x) {, if x ω, (7.7.4a) 0, if x ω. (7.7.4b) 08

119 This lemma can be represented in the form of Quantum Circuit diagram 0 0 U ω s H n U fω (7.7.43) 0 n 0 As we shall see that with ω as an unknown state, U fω 0. is actually an oracle. Proof. In the case x ω. U fω ( ω 0 ) ( ) f(ω) 0 ω 0 ω 0 U ω ω ; (7.7.44) In the case x ω. U fω ( x ω 0 ) ( ) f(x ω) 0 x ω 0 x ω 0 U ω x ω. (7.7.45) Note: How to perform U ω has to use of the black box U f, which suggests that performing U ω takes the main part of the total performing time. 3 The second reflection operator U s s s I, (7.7.46) is at hand, because its realization does not involve the black-box and only includes elementary quantum gates. And we can derive that U s H n ( 0 n 0 I n ) H n H n X n ( n I n ) X n H n H n X n (I (n ) H) θ (n) (I (n ) H) X n H n, (7.7.47) 09

120 where θ (n) is the n-qubit Toffoli gate, which can be written as θ (n) (I (n ) (n ) ) I + ( (n ) ) X. (7.7.48) Therefore, we can verify the E.Q.(7.7.47) in the following manner, (I (n ) H) θ (n) (I (n ) H) (I (n ) H) [ (I (n ) (n ) ) I + ( (n ) ) X] (I (n ) H) (I (n ) (n ) ) I + ( (n ) ) Z I n + (n ) ( ) I n n. (7.7.49) Now, we can make use of the above components to construct the quantum circuit model for Grover s algorithm. Step The quantum circuit should be 0 0 H n R R Rgrov R R Rgrov R R R grov 0 n (7.7.50) 0 0. Note that R R Rgrov (U s I )U fω. Step Suppose that the result we get in Step is state ω, then we can use the oracle U fω to check if it is the state ω, i.e., ω 0 U fω 0. (7.7.5) With the measurement, we can do the check. If we find out the result is wrong, then we should start from Step again. If it is the rightful state, namely, the measurement results are, then the work is done. 0

121 Part III Density Matrix and Quantum Entanglement

122 Chapter 8 Quantum Mechanics (II): Density Matrix References: [Preskill] Chapter : Foundations I: states and ensembles; [Nielsen & Chuang] Chapter : Introduction to quantum mechanics. 8. 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, with the density operator defined as ψ P n ψ a k ψ P n a k a k ψ a k a k ψ ψ P n a k tr( ψ ψ P n ) tr(ρp n ), (8..) ρ ψ ψ, (8..) and all a k { a j A a j a j a j, a j H } being normalized. Therefore, A ψ A ψ a n ψ P n ψ n a n tr(ρp n ) n tr(ρa). (8..3) We can claim that the probability to obtain the outcome a n when measuring A is actually Prob(a n ) tr(ρp n ). (8..4)

123 In the physical sense, for the quantum closed system with the initial state ψ, ψ c n n, n where { n } is an orthonormal basis for the Hilbert space H. The probability to get the post-measurement states n is Prob( n ) n ψ ψ ψ. (8..5) Question: What is the post-measurement state of an open system? ρ p n n n, n with p n and p n 0. (8..6) n The ρ is the so-called density matrix, which will be defined later. And the density operator ρ is equivalent to state ensemble { n, p n }, in the Quantum Mechanical sense. Remark. In the description of the density matrix, the ambiguity of the global phase factor with non-physical meaning is removed, ψ e iα ψ, (8..7) ρ ψ ψ, ρ α e iα ψ ψ e iα ρ. (8..8) 8.. State ensemble formalism of density matrix Def 8... The state ensemble is a set of state ψ i each attached with a probability p i, namely { ψ i, p i i,..., n}, with p i [0, ] and n i p i. (8..9) It means that the physical system has the probability of p i to be prepared in the state ψ i. For any observable A, the expectation value of A is with ρ n i p i ψ i ψ i, A n i dim H k dim H k dim H k dim H k p i ψ i A ψ i n i n i tr(aρ), A n i p i ψ i k k A ψ i k A ψ i p i ψ i k n k A ( ψ i p i ψ i ) k i k Aρ k p i ψ i A ψ i tr(aρ), (8..0) where { k k,..., dim H } is an orthonormal basis for the Hilbert space H. 3

124 Def 8... The density matrix for the system described by the state ensemble, as shown in E.Q. (8..9), is defined as ρ n i p i ψ i ψ i, (8..) if n, this represents a pure state, and the system is completely specified; if n, this represents a mixed state, and the system is partially specified. Thm 8... A sufficient and necessary condition to tell whether the density operator ρ represents a pure state or a mixed state is { pure state ρ ρ, trρ ; mixed state ρ ρ, trρ <. (8..) 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 λ i lying along the diagonal position of the matrix: There we can also get ρ: ρ dim H i λ i λ i λ i. (8..3) ρ dim H i,j dim H i,j dim H i λ i λ j λ i λ i λ j λ j λ i λ j λ i λ j δ ij λ i λ i λ i. (8..4) trρ dim H i dim H λ i ( i λ i ). (8..5) Therefore, the sufficient and necessary condition that ρ is equal to ρ should be λ i λ i, with i,..., dim H. (8..6) With ρ being self-adjoint and non-negative with unit trace, namely dim H i λ i, with λ i R and λ i 0, i,..., dim H, (8..7) it then can be inferred that E.Q. (8..6) is equivalent to that there should be only one λ j being while all other λ i with i j are vanishing. And it also says that the same thing as E.Q. (8..), since with only one eigenvalue for the density operator, it is sufficient and necessary to get the pure state. Thm 8... The density operator ρ should be subject to the following time-evolution equation i h ρ(t) [H, ρ(t)]. (8..8) t 4

125 Notice: We can see that E.Q. (8..8) 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. (8..9). Every state vector ψ i within the state ensemble should follow the Shrödinger equation or we can get the state ensemble at time t with i h t ψ i(t) H ψ i (t, { ψ i (t), p i i,..., n}, (8..9) ψ i (t) U (t) ψ i, (8..0) where U (t) is defined in E.Q. (..7). Therefore, the density operator at time t should be ρ(t) n i Now, we can derive the time-evolution equation for ρ: p i ψ i (t) ψ i (t). (8..) i h n t ρ(t) p i (i h t ψ i(t) ψ i (t) + ψ i (t) i h t ψ i(t) ) i n i [ρ, H] which is equivalent to E.Q. (8..8). p i (H ψ i (t) ψ i (t) + ψ i (t) ψ i (t) ( H)) 8.. Operator formalism of density matrix Now, we come to discuss the properties of the density matrix. Def 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, (8..) trρ 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 8... 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 (8..), for it not only represents the mixed state case, but also the pure state case. 5

126 Self-adjoint since p i R for i,..., n. Non-negative n ρ ( φ ρ φ i n p i ψ i ψ i ) (8..3) p i ψ i ψ i (8..4) i n i p i ψ i ψ i (8..5) ρ, (8..6) n p i φ ψ i ψ i φ i n i p i ψ i φ, (8..7) where φ is an arbitrary vector in Hilbert space H. Given p i 0 for all i,..., n, it s obvious that φ ρ φ is non-negative, namely ρ is non-negative. Unit trace n trρ tr ( p i ψ i ψ i ) (8..8) n i p i tr ( ψ i ψ i ) (8..9) i n i p i (8..30), (8..3) because p i with i,..., n would add up to and all ψ i are assumed to be normalized 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, ρ AB ψ AB ψ. (8..3) That represents a pure state, as we consider the composite physical system as a whole. 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 H AB of the composite system C as M A I B. Therefore, the expectation value of the observable M A is then M A AB ψ M A I B ψ AB tr((m A I B )ρ AB ) tr A (M A ρ A ), (8..33) 6

127 where And we have used the facts that { ρ A tr B ρ AB. (8..34) tro AB tr A (tr B O AB ), tr (M A O AB ) tr A (M A tr B O AB ), (8..35) which can be proved simply by expressing all the operators in the form of kets and bras. The reduced density matrix ρ A can describe the subsystem A, given the density operator of the compound physical system is ρ AB, in the following sense. If O A is an operator corresponding to an arbitrary observable defined in the subsystem A, namely O A B(H A ), then we can express it in the total Hilbert space H AB H A H B as Therefore, the expectation value of the observable O A should be O A I B. (8..36) O A tr AB (O A ρ AB ) tr A [tr B (O A ρ AB )] tr A (O A tr B ρ AB ) On the other hand, it really makes a density operator because self-adjoint: tr A (O A ρ A ). (8..37) ρ A (tr Bρ AB ) tr B ρ AB tr B ρ AB ρ A ; (8..38) unit trace: tr A ρ A tr A (tr B ρ AB ) nonnegative: namely tr AB ρ AB ; (8..39) A ψ ρ A ψ A tr A (ρ A ψ A ψ ) tr AB [ρ AB ( ψ A ψ I B )] A ψ ρ A ψ A 0, ψ A H A (8..40) since the probability for the compound system ρ AB found in state ψ A ψ I B should be nonnegative. E.g. Reduced density matrix ρ A of the general bipartite system where the partial trace ρ A tr B ρ AB µ ψ AB i a iµ i A µ B, (8..4) µ µ ρ AB µ B µ i a iµ a iµ i A j, (8..4) j tr B ( µ B ν ) ν µ B δ µν. (8..43) 7

128 8. Mixed state formalism of a qubit Thm Density matrix in the two-dimension Hilbert space H, can be represented as ρ( p) (I + p σ), (8..) with p R 3 satisfying p. And we give the vector p the name of the polarization vector. Proof. Firstly, we would prove that any density matrix in the two-dimensional Hilbert space H can be expressed in the form as E.Q. (8..). Secondly, we would show that with the constraint p, E.Q. (8..) is a density matrix. a) Any density matrix in the -dimensional Hilbert space H can be represented in the form of E.Q. (8..). The most general -dimensional matrix would have the form ρ ( a b c d ), (8..) which has 4 8 degrees of freedom. But to make a density matrix for a physical system, the matrix ρ must satisfy three constraints. (i) (ii) Self-adjoint: A density matrix must be self-adjoint, i.e., ρ ρ, which means a ( a c a b d ) ( a b c d ) d d b, (8..3) c c b which actually makes four constraints for ρ, that would reduce the degrees of freedom from eight to four. Unit trace: The trace of a density matrix must be equal to one, i.e., trρ a + d, (8..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 H can actually be viewed as a vector space B(H ). The three Pauli matrices plus the identity matrix can be the vector basis for B(H ), 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. (8..). 8

129 b) Now, we set to prove that iff p, E.Q.(8..) would be a density matrix. From E.Q.(8..) we get ρ ( + p 3 p ip p + ip p 3 ) with p, p, p 3 R. (8..5) This definition implies apparently that ρ is self-adjoint. And we can immediately get the trace and the determinant of ρ: trρ (( + p 3) + ( p 3 )), (8..6) det ρ 4 [( + p 3)( p 3 ) (p ip )(p + ip )] 4 ( p ). (8..7) Let s denote the two eigenvalues of ρ with λ and λ, then trρ λ + λ, det ρ λ λ 4 ( p ), from which we can infer that ρ being nonnegative is equivalent to λ 0 λ 0 } p. (8..8) Therefore, ρ defined in E.Q. (8..) is a density matrix iff p. c) We can now claim that, any density operator for two-level system attached with Hilbert space H, can be written as E.Q. (8..). 8.. 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. (8..), Amid, we can get where we have used the fact that σ n tr[( σ n)ρ] tr [ ( σ n)( + p σ)] {tr( σ n) + tr[( σ n)( p σ)]}. (8..9) tr( σ n) tr(n σ + n σ + n 3 σ 3 ) n trσ + n trσ + n 3 trσ 3 0, (8..0) trσ i 0, with i,, 3. (8..) 9

130 And, we can also get tr[( σ n)( p σ)] tr 3 i,j n i p j σ i σ j 3 n i p j tr(σ i σ j ) i,j 3 i,j In the above derivation we have used the relation which can be derived in the following manner n i p j δ ij n p. (8..) tr(σ i σ j ) δ ij, (8..3) trσ i tri i j tr(σ i σ j ) tr(i 3 k ε ijkσ k ) i 3 k ε ijktrσ k 0 } tr(σ i σ j ) δ ij, i, j,, 3. Now, by substituting E.Q. (8..0) and E.Q. (8..) into E.Q. (8..9), we can conclude that σ n n p, which is equivalent to Thm 8... with n, p R 3 and n. tr(( σ n)ρ( p)) n p, (8..4) With this theorem, we can understand why we call p the polarization vector : (i) σ n tr(( σ n)ρ( p)) 0, for p n; (ii) σ n tr(( σ n)ρ( p)), for p n; (iii) σ n tr(( σ n)ρ( p)), for p n; (iv) σ n tr(( σ n)ρ( p)) 0, for p 0, n R 3. We can see that p 0 implies no polarization orientation, which shows no information. 8.. Pure state and mixed state in two-dimensional Hilbert space H Thm 8... If the state in the two-dimensional Hilber space H, which means that the state has the density matrix defined in E.Q. (8..). Then the state of the physical system is a pure state, if p is on the Bloch sphere, p, { a mixed state, if p is inside the Bloch sphere, p <. 0

131 We can show that ρ ( p) 4 ( + p σ + ( p σ) ) 4 ( + p σ + p ), (8..5) because ( p σ) 3 i 3 i p i σ i + p i i,j i>j p i p j {σ i, σ j } p. (8..6) Therefore, { ρ ( p) ρ( p), if p is on the Bloch sphere, p ; ρ ( p) ρ( p), if p is inside the Bloch sphere, p <. (8..7) Thus ρ ( p) ρ( p) or p < is sufficient to claim that ρ( p) is a mixed state. On the other hand, we can get some examples of case p : p e z, ρ( e z ) ( ) 0 0 z z ; p e x, ρ( e x ) ( ) ( 0 + )( 0 + ) + + x x ; p e y, ρ( e y ) ( i i ) ( 0 + i )( 0 i ) y y. In general, if p, we can express p as p (sin θ cos ϕ, sin θ sin ϕ, cos θ), ρ( p) ( + 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 θe iϕ sin θe iϕ cos θ ) cos θ sin θ cos θ e iϕ sin θ cos θ eiϕ sin θ ( cos θ sin θ eiϕ) (cos θ sin θ e iϕ ) ψ + (θ, ϕ) ψ + (θ, ϕ) p p. (8..8) Therefore, p is a sufficient condition for that ρ( p) is a pure state. Given that any state for a two-level physical system, such as spin-/ systems, which can be described by ρ( p) as defined in E.Q. (8..), we can conclude that now: p is a sufficient and necessary condition for that ρ( p) is a pure state; p < is a sufficient and necessary condition for that ρ( p) is a mixed state.

132 Pure state vs. Mixed state in -dimensional Hilbert space pure state ψ ( 0 + ) x σ x x x ρ ψ ψ x x ρ ( ) mixed state ρ ( ) Measurement M x x x, N x x x M x, N x 0 M x, N x 8.3 Convexity of density matrix Thm (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 H. As we can show that by using the concept of the Bloch ball, in H, the most general density matrix can be expressed as ρ( p) (I + p σ), with p. (8.3.) If p, then ρ( p) represents a pure state. p also means that the state can be represented as one point on the Bloch sphere. So, the pure states is the external points of the Bloch ball, which is a convext set. If p <, then ρ( p) represents a mixed state. p < also means that the state can be represented as one point inside the Bloch ball. For any vector p with the constraint p, we can express it in the form of p n + λ( n n ), (8.3.) with 0 λ and n n, n, n R 3. Therefore, we can get the density matrix ρ( p) expressed as ρ( p) {I + [ n + λ( n n )] σ} i.e., with [λ(i + n σ) + ( λ)(i + n σ)] λ (I + n σ) + λ (I + n σ), ρ( p) λρ( n ) + ( λ)ρ( n ), (8.3.3) ρ( n ) (I + n σ), ρ( n ) (I + n σ).

133 Therefore, we can also get ρ( p) {λ, ψ + ( n ) ; ( λ), ψ + ( n ) }. (8.3.4) Remark: The same mixed state density matrix may have different interpretations of state ensembles. We can make an easy example to demonstrate this, for instance ρ I. (8.3.5) 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 /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. 8.4 Two-qubit system and its subsystem Two-qubit system and its subsystem entire system subsystem two-qubit H H one-qubit H density matrix reduced density matrix pure state (maybe) mixed state (exactly known) (partially known or completely unkonwn) 8.4. 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), ψ AB ( 00 + ) AB ( 0 A 0 B + A B ). (8.4.) And the corresponding density matrix should be ρ AB ψ AB ψ ( 00 AB AB + AB 00 + AB ). (8.4.) We can check that ρ AB ρ AB, ρ AB ψ AB ψ ψ AB ψ ψ AB ψ ρ AB, (8.4.3) 3

134 amid we have use the fact that AB ψ ψ AB ( A 0 B 0 + A A B )( 0 A 0 B + A B ) ( A 0 0 A B 0 0 B + A 0 A B 0 B + A 0 A B 0 B + A A B B ) ( ). 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, ρ A tr B ρ AB. (8.4.4) From above definition, we can get ρ A B i ρ AB i B i0 ( 0 A 0 + A ) I. (8.4.5) For an arbitrary orientation, n (sin θ cos ϕ, sin θ sin ϕ, cos θ), n σ A tr A [( n σ A )ρ A ] tr A( n σ A ) 0. (8.4.6) Therefore, in this case the polarization vector for the subsystem A should be a null vector in R 3, 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. Def 8.4. (Maximally entanglement). If a bipartite ρ AB is a pure state with ρ A I A, then ρ AB is called maximally entangled Maximally entangled two-qubit pure states Thm All Bell states are maximally entangled. Proof. This is transparent from E.Q. (3.3.), ψ(i, j) AB [ 0i AB + ( ) j ī AB ] 4

135 that ρ A tr B ψ(i, j) AB ψ(i, j) B i ψ(i, j) AB ψ(i, j) i B + B ī ψ(i, j) AB ψ(i, j) ī B 0 A 0 + A I A, (8.4.7) and ρ B tr A ψ(i, j) AB ψ(i, j) B 0 ψ(i, j) AB ψ(i, j) 0 B + B ψ(i, j) AB ψ(i, j) B i A i + ī A ī I B. (8.4.8) Therefore, we can conclude now that every Bell state ψ(i, j) is maximally entangled. Remark: () Local unitary transformations, e.g. I X i Z j, preserve the entangling property; () With ρ A ρ B I, 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 density matrices { ρ A n A n, ρ B m B m, with n m. There is no information can be hidden in the subsystem in this case. We may say that the more entangled a state is, the more hidden of the quantum information is 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 +. φ + AB 0 E separable ψ ABE entangled 00 AB e 00 E + 0 AB e 0 E + 0 AB e 0 E + AB e E. (8.4.9) But, Alice and Bob can avoid it by Bell measurements X A X B and Z A Z B and informing each other their measurements results. 5

136 After measuring Z A Z B, they get the result of, then ψ ABE entangled ψ ABE entangled 00 AB e 00 E + AB e E (8.4.0) After measurement of X A X B, the result is, then the final state has to be ψ ABE entangled ψ ABE separable ( 00 AB + AB ) e E. (8.4.) in which the state of Eve e E has been decoupled with φ + AB. 6

137 Chapter 9 Schmidt Decomposition, Purification and GHJW Theorem References: [Preskill] Chapter : Foundations I: states and ensembles; [Nielsen & Chuang] Chapter : Introduction to quantum mechanics. 9. 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 ρ AB ψ AB ψ. And the density matrix of the subsystem A should be ρ A tr B ρ AB tr A ( ψ AB ψ ). Before further talking about the measurement and quantum operation on the subsystem of the composite system, in this chapter, we would introduce some relations between subsystem and composite system. And Figure 9. illustrates the plan for this chapter. Schmidt decomposition Quantum entanglement Purification GHJW theorem Figure 9.: Plan for chapter Schmidt decomposition and quantum entanglement 9.. Schmidt decomposition Thm 9... (Schmidt decomposition). In a bipartite system H A H B, a bipartite pure state ψ AB can be expressed in the form of ψ AB pi i A i B, i with p i 0, p i, (9..) i 7

138 where { i A } and { i B } are orthonormal basis for H A and H B respectively. This is the so-called Schmidt decomposition. Example: Bell state (EPR state) ψ AB ( 00 + ) AB ii AB i0 i A i B, (9..) i0 with p 0 p. We can also get ρ A I A, ρ B I B. (9..3) Remark: For ψ AB i pi i A i B, we may have dim H A dim H B. But ρ A and ρ B would have the same diagonalized formalism, i.e., ρ A p i i A i, i ρ B p i i B i. (9..4) i Def 9... Schmidt coefficient and Schmidt number:. The non-vanishing p i in E.Q. (9..) are called Schmidt coefficients.. The number of Schmidt coefficients is called Schmidt number. 9.. Quantum entanglement Def 9... Separable and Entangled bipartite pure states: if the Schmidt number is, ψ AB is called separable, which means that the state vector of the composite physical system can be expressed as which is the so-called product state; ψ AB φ A χ B, (9..5) if the Schmidt number is greater than, then the state ψ AB is called entangled state; if the reduced density matrices for the subsystems are all identity matrices, namely ρ A I A, ρ B I B, (9..6) then ψ AB is called maximally entangled. (We have actually mentioned this at the end of the last chapter (Ref. page 4).) Note: We would discuss about the entanglement definition for bipartite mixed state and multipartite mixed state later. 8

139 9..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 ψ AB b iµ i A µ B, (9..7) i,µ with { i A } and { µ B } being orthonormal basis for H A and H B respectively, is to get both the reduced density matrices ρ A and ρ B diagonalized. And therefore the corresponding density matrix of the composite physical system should be ρ AB ψ AB ψ Now, we can go on step by step. b iµ b jν i A j µ B ν. (9..8) i,j,µ,ν Step : Compute ρ A by the taking the partial trace of ρ AB over H B : ρ A tr B ρ AB µ B µ ρ AB µ B i,µ,j,ν,µ b iµ b jν( i A j ) ( µ µ B ν µ ) b iµ b jν( i A j ) B ν µ B i,µ,j,ν b iµ b jµ i A j. (9..9) i,µ,j Step : Diagonalize ρ A by choosing another orthonormal basis { i A }, all vectors in which are all eigenvectors of ρ A, such that ρ A p i i A i, i with p i and p i 0, (9..0) i since ρ A 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. Step 3: With the basis { i A } for H A and the basis { µ B } for H B, we can reformulate the state vector for the composite system, ψ AB a iµ i A µ B. (9..) i,µ Now, we can choose another basis for H B to simplify the expression of ψ AB : ĩ B A i ψ AB a jµa i j A µ B a iµ µ B, (9..) j,µ µ with which we can rewrite the state vector ψ AB again, ψ AB i A ĩ B. (9..3) i 9

140 Correspondingly, the density matrix ρ AB should take the form of ρ AB i A j ĩ B j. (9..4) i,j Step 4: We can prove that the newly chosen basis { ĩ B } is orthogonal and we also could normalize it. To accomplish those, again we can compute ρ A through partial trace calculation of ρ AB over H B : ρ A B µ ρ AB µ B µ ( i A j ) ( B µ ĩ B j µ B ) i,j,µ ( i A j ) ( B j ĩ B ). (9..5) i,j On the other hand, we know from E.Q. (9..0) that ρ A p i δ ij i A j. (9..6) i,j By comparing E.Q. (9..5) with E.Q. (9..6), we can get that which means that the basis { ĩ B } is actually orthogonal. Now we can normalize { ĩ B } by defining B j ĩ B p i δ ij, (9..7) i B pi ĩ B, (9..8) where p i > 0, which means that we are discussing in the subspaces of H A and H B which ensures p i > 0. Now, with the basis { i } for H A and { i } for H B we can rewrite the state vector for the third time ψ AB pi i A i B, i which is exactly the same as E.Q. (9..). 9.3 Example for the Schmidt decomposition As an example of the Schmidt decomposition, here we present the solution of the exercise., Chapter of John Preskill s online lecture notes. Problem: For the two-qubit state Φ A ( B + (a) Compute ρ A tr B ( Φ Φ ) and ρ B tr A ( Φ Φ ). 3 B) + 3 A ( B + B) (9.3.) 30

141 (b) Find the Schmidt decomposition of Φ. Notation: () 0,. () matrix: N where i is the row index and j is the column index. (3) 4 4 matrix: M i N ij j, (9.3.) i,j0 ij M ij,kl kl, (9.3.3) i,j,k,l0 where i, j are the row indexes and k, l are the column indexes. (4) Product basis in Hilbert space H H : Rewrite the state Ψ AB as ; ; ; (9.3.4) Φ AB 0 A ( 3 0 B + B) + A ( ( 00 AB AB AB + AB ) B + B) a ij ij AB. (9.3.5) 8 i,j0 And in the matrix expression, we have Φ AB (9.3.6) Note: Ψ AB is symmetric under the exchange of system A and system B, which implies ρ A ρ B. 3

142 ρ AB Ψ AB Ψ ij AB kl a ij a kl i,j,k,l0 ij AB kl ρ ij,kl i,j,k,l AB AB AB AB 3 3 (9.3.7) AB AB AB AB AB AB AB AB 0 + And in the matrix expression, we have (a) 8 0 AB (9.3.8) AB (9.3.9) 3 8 AB AB. (9.3.0) (9.3.) ρ AB ( 3 3 ) (9.3.) 3 3 ρ A tr B ρ AB B m ρ AB m B m0 m0 i,j,k,l0 m0 i,k0 B m ij AB ρ ij,kl kl m B i A k ρ im,km ( ρ 00,00 + ρ 0,0 ρ 00,0 + ρ 0, ) ρ 0,00 + ρ,0 ρ 0,0 + ρ, 8 ( ) X, (9.3.3) 4 3

143 where X is the Pauli matrix, X σ x ( 0 0 ). (9.3.4) (b) Eigenstate of Pauli gate Z. ρ B tr A ρ AB A m ρ AB m A m0 m0 i,j,k,l0 m0 i,k0 A m ij AB ρ ij,kl kl m A j B l ρ mj,ml ( ρ 00,00 + ρ 0,0 ρ 00,0 + ρ 0, ) ρ 0,00 + ρ,0 ρ 0,0 + ρ, 8 ( ) X ρ A. (9.3.5) Z 0 0, Z. (9.3.6) Eigenstate of Pauli gate X. X ± ± ±. (9.3.7) ± ( 0 ± ); + +, + + 0; (9.3.8) I Find the eigenstate of reduced density matrix ρ A and ρ B. Therefore ρ A,B ± λ ± ±, λ ± ± 3 4, (9.3.9) ρ A,B λ + + A,B + + λ A,B. (9.3.0) Ψ AB + A + Ψ AB + A Ψ AB + A ϕ B + A ϕ B, (9.3.) where 6 + ϕ B A + Ψ AB + B λ + + B, 4 6 ϕ B A Ψ AB B λ B. 4 Thus, the Schmidt decomposition of state Ψ AB shows as (9.3.) Ψ AB λ + + A + B λ A B. (9.3.3) 33

144 9.4 The purification theorem and GHJW theorem 9.4. Purification Thm (Purification). For any given mixed state ρ A for system A, there is a bipartite pure state Φ AB such that Then Φ AB is called the purification of ρ A. For example, the density matrix ρ A tr B ( Φ AB Φ ). (9.4.) ρ A I A, as we have shown in E.Q. (8.4.5), has the purification ψ AB ( 00 AB + AB ). Proof. The key point is the Schmidt decomposition. Suppose that ρ A p i ϕ i A ϕ i, i with p i, and p i > 0, (9.4.) i where { ϕ i } is an orthonormal basis of H A. Now, we introduce another system B and construct the following state for the composite physical system H A H B, Φ AB i pi ϕ i A α i B, with p i, (9.4.3) i amid, { α i B } is the orhonormal basis of H B. And we can show that Φ AB is a pure state, which is self-convincing, and it is also normalized since AB Φ Φ AB pj p i ( A ϕ j B α j )( ϕ i A α i B ) i,j pi p ja ϕ j ϕ i A B α j α i B i,j pi p j δ ij δ ij i,j p i i. We can also verify that Φ AB defined in E.Q. (9.4.3) is truly purification of the state ρ A as defined in E.Q. (9.4.). As we can see that tr B ρ AB B α i ρ AB α i B i i,j,k i,j,k i,j,k pj p kb α i ( ϕ j A α j B )( A ϕ k B α k ) α i B pj p k ( ϕ j A ϕ k ) ( B α i α j B α k α i B ) pj p k ( ϕ j A ϕ k )δ ij δ ki p i ϕ i A ϕ i i ρ A, 34

145 i.e., Therefore, Φ AB is a purification of ρ A. There are some things that we should remark: ρ A tr B ρ AB. (9.4.4) ρ A 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 ρ A, 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 tr A (M A ρ A ). 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 The GHJW theorem Thm (GHJW). Consider many ensembles that realising ρ A. There is a Hilbert space H B and a bipartite pure state Φ AB, by which any one of these ensembles can be realized measuring a suitable observable of B. Proof. Firstly, we would show that two arbitrary purification of ρ A, e.g. Φ and Φ, are connected with a local unitary transformation on the subsystem B. Let s denote that Φ A i pi ϕ i A α i B, Φ A i qi ψ i A β i B, (9.4.5) where {(p i, ϕ i A )} and {(q i, ψ i A )} are two state ensemble interpretations of ρ A and { α i B } and { β i B } are two orthonormal basis for H B. Assume that ρ A has the following eigenvectors and the associated eigenvalues ρ A k A λ k k A, with λ k, and λ k 0. (9.4.6) k Since Φ AB and Φ AB are two different purification of ρ A, it s apparent that tr A Φ AB tr A Φ AB ρ A. (9.4.7) With E.Q. (9.4.7) in mind, we can see that the respective Schmidt decompositions for Φ AB and Φ AB are Φ AB k λk k k B, Φ AB k λk k k B. (9.4.8) Therefore, Φ AB and Φ AB are connected with the local unitary translation Φ AB (I A U B ) Φ AB, (9.4.9) 35

146 with Substitute E.Q. (9.4.5) into E.Q. (9.4.9), and we can get that namely where U B k B k. (9.4.0) k Φ AB qi ψ i A U B β i B, i Φ AB qi ψ i A γ i B, (9.4.) i γ i B U B β i B. (9.4.) Surely, { γ i B } makes an orthonormal basis for H B, too. Secondly, let s consider two observables M B and N B defined in the subsystem B. And we consider that case the two vector basis { α i B } and { γ i B } respectively are eigenvectors of M B and N B, namely { M B α i B M i α i B, N B γ j B N j γ j B, with i, j,..., dim H B. Therefore, we can conclude that: After the measurement of M B, there would be a probability of p i for the subsystem B in the state α i B, namely the subsystem A in the state ϕ i A. Hence, measuring M B in the subsystem B would prepare the subsystem A in the mixed state ρ A with the state ensemble {(p i, ϕ i A )}. After the measurement of N B, there would be a probability of q j for the subsystem B in the state γ j B, namely the subsystem A in the state of ψ i A. Thus, the measurement of N B in the subsystem B would prepare the subsystem A in the mixed state ρ A with the state ensemble {(p i, ψ i A )}. Therefore, we get the GHJW theorem. 9.5 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 ψ AB ( 00 AB + AB ). (9.5.) 36

147 We can easily get the reduced density matrices for Alice and Bob respectively, i.e., ρ A tr B ρ AB B i ρ AB i B i0 B i ( j A j B )( A j B j ) i B i,j,j 0 ( j A B i j B )( A j B j i B ) i,j,j 0 j A j δ ij δ j i i,j,j 0 i0 i A i, ρ A I. (9.5.) Reasoning in the same manner, we can get Now, let s consider the following processes: The first process: ρ B I. (9.5.3) 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 transfer between Alice and Bob. In this case, we can get ρ A I, 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 ρ A 0 A 0, which is a pure state. 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 {(, 0 ), (, )}. (9.5.4) Bob measures his qubit along the x-axis, and 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 {(, + ), (, )}. (9.5.5) 37

148 The two state ensembles described by (9.5.4) and (9.5.5) 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 ρ A I. 38

149 Chapter 0 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?. References: [Preskill] Chapter 4: Quantum entanglement; Nielsen & Chuang [Nielsen & Chuang] Chapter : Quantum information theory. 0. Bipartite mixed state entanglement 0.. Separability 0... 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, (0..) which is equivalent to say that the bipartite pure state has the Schmidt number as is separable. 39

150 0... Bipartite mixed state A bipartite mixed state ρ AB is separable if it admits the ensemble description given by where ρ AB p ij ρ A,i ρ B,j, (0..) i,j p ij, p ij 0 i,j and {ρ A,i i} are density operators for subsystem A, while {ρ B,j j} for subsystem B. Example: This is a separable mixed state ρ AB p i α i α i q j β j β j (0..3) i j where p ij p i q j and p i, i q j, i with p i 0, i, with q j 0, j. 0.. Quantum bipartite entanglement A bipartite state is entangled if it is not a separable state. Here is one separable state And another separable state shows as ρ i 4 A i j B j 4 i0 i,j0 j0 ij AB ij And for the state ρ 3, is expressed as ρ 4 I I. (0..4) 4 ( 00 AB AB AB 0 + AB ). (0..5) ρ 3 4 ( φ+ φ + + φ φ + ψ + ψ + + ψ ψ ), (0..6) which is hard to determine whether it is separable or entangled. However, as we shall see that ρ, ρ and ρ 3 are actually same, since both four product states { ij i, j 0, } and four Bell states { ψ(ij) i, j 0, } can span the Hilbert space H H, namely ρ ρ ρ 3. (0..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. 40

151 0..3 Positive-partial transpose (PPT) criterion for quantum bipartite separability Thm If ρ AB is separable, namely be of the form (0..), then the partial transpose of ρ AB, defined as (ρ AB ) PT (I T )ρ AB p ij ρ A,i (ρ B,j ) T, (0..8) i,j is still non-negative, where T denotes the transpose of matrix, namely T ( i j ) j i. (0..9) Proof. As we shall know that {ρ B,j j} are density operators of the subsystem B. Therefore, for an arbitrary ρ B,j, the eigenvectors can span the local Hilbert space H B. If we can diagonalize ρ B,j with the basis being the complete set of independent orthonormal eigenvectors of ρ B,j. After ρ B,j being diagonalized, we can get that its transpose is diagonalized, too. It would obvious that, that the transpose would preserve the eigenvalues of ρ B,j. In another worlds, (ρ B,j ) T should also be nonnegative, namely (ρ B,j ) T 0, j. (0..0) On the other hand, the fact that {ρ A,i i} being the density operators of the subsystem A, ensures that they are all nonnegative. Therefore, (ρ AB ) PT is nonnegative. Corollary If (ρ AB ) PT is not positive, then ρ AB is an entangled state. This is a direct corollary of the theorem Example: the Werner state and the PPT criterion Werner State, firstly presented in 989 by Reinhard F. Werner, is defined as ρ(λ) 4 ( λ)i 4 + λ φ + φ + (0..) with λ being a real number. We can show that if ρ(λ) with definition (0..) is a density matrix, iff Unit trace. tr [ρ(λ)] tr [ 4 ( λ)i 4 + λ φ + φ + ] 4 ( λ) tr (I 4) + λtr ( φ + φ + ) ( λ) + λ. 4

152 Self-adjoint. [ρ(λ)] [ 4 ( λ) I 4 + λ φ + φ + ] [ 4 ( λ) I 4] + (λ φ + φ + ) 4 ( λ) I 4 + λ ( φ + φ + ) 4 ( λ)i 4 + λ φ + φ + ρ(λ), 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, } are four eigenstates of ρ(λ). ρ(λ) ψ(0, 0) ρ(λ) φ + and with i 0 and j 0, we can get ρ(λ) ψ(i, j) ρ(λ) ψ(i, j) [ 4 ( λ)i 4 + λ ψ ψ ] φ + 4 ( + 3λ) φ+. (0..) ( 4 ( λ)i 4 + λ ψ(0, 0) ψ(0, 0) ) ψ(i, j) ( λ) ψ(i, j), (0..3) 4 as we know that ψ(i, j), with (i, j 0, ), are mutually independent. Therefore, we ve gotten the complete set of mutually independent eigenvectors of ρ(λ). Therefore, for the constrain of nonnegative: ρ(λ) 0 4 ( + 3λ) 0 4 ( λ) 0 3 λ. (0..4) When λ 3 ρ ( 3 ) 3 (I 4 φ + φ + ) 3 ( φ φ + ψ + ψ + + ψ ψ ), therefore ρ ( 3 ) φ+ 3 (I 4 φ + φ + ) φ + 0 φ +, (0..5) 4

153 and ρ ( 3 ) ψ(i, j) 3 (I 4 ψ(0, 0) ψ(0, 0) ) ψ(i, j) ψ(i, j) (0..6) 3 with i 0 and j 0. When λ which is a pure state with the relations and with i 0 and j 0. ρ() φ + φ +, ρ () φ + ( φ + φ + ) φ + φ +. (0..7) ρ () ψ(i, j) ( φ + φ + ) ψ(i, j) 0 ψ(i, j) (0..8) Calculate the partial transpose of Werner state ρ(λ). [ρ(λ)] PT (I T )ρ(λ) from which we can see that (I T ) [ 4 ( λ)i 4 + λ φ + φ + ] 4 ( λ) (I T )I 4 + λ(i T ) φ + φ + 4 ( λ)i 4 + λ(i T ) ( ) 4 ( λ)i 4 + λ ( ), (0..9) [ρ(λ)] PT φ + PT 00 + [ρ(λ)] [ 4 ( λ) + λ] φ+ 4 ( + λ) φ+, (0..0) [ρ(λ)] PT φ PT 00 [ρ(λ)] [ 4 ( λ) + λ] φ 4 ( + λ) φ, (0..) [ρ(λ)] PT ψ + PT [ρ(λ)] [ 4 ( λ) + λ] ψ+ 4 ( + λ) ψ+, (0..) [ρ(λ)] PT ψ PT 0 0 [ρ(λ)] [ 4 ( λ) λ] ψ 4 ( 3λ) ψ. (0..3) 43

154 In another viewpoint, we can see ( T ) φ + φ + ( T ) ii jj i,j0 ( T )( i j i j ) i,j0 i j T ( i j ) i,j0 i j j i i,j0 ij ji i,j0 SWAP (0..4) where the Swap gate is defined in (6..4) and has the matrix expression SWAP ij ji i,j Note that the SWAP gate is identical with its inverse, namely which implies the characteristic equation. (0..5) SWAP SWAP I 4, (0..6) SWAP I 4 0. (0..7) Thus, we can see the eigenvalue of the SWAP gate is ±. In two electrons system, i.e., two spin- system, triplet states are eigenstate of SWAP gate with eigenvalue, expressed as φ + φ + SWAP φ φ ψ + ψ + For singlet state, it is the eigenstate of SWAP gate with eigenvalue, Therefore, rewrite the SWAP gate as. (0..8) SWAP ψ ψ. (0..9) SWAP φ + φ + + φ φ + ψ + ψ + ψ ψ I 4 ψ ψ. (0..30) And the partial transpose of Werner state ρ(λ) can be calculated as [ρ(λ)] PT (I T )ρ(λ) (I T ) [ 4 ( λ)i 4 + λ φ + φ + ] 4 ( λ)i 4 + λ SWAP 4 ( λ)i 4 + λi 4 λ ψ ψ 4 ( + λ)i 4 λ ψ ψ. (0..3) 44

155 And the eigenvalues can be obtained by the way [ρ(λ)] PT ψ ( 4 ( + λ) λ) ψ 3λ ψ, (0..3) 4 [ρ(λ)] PT φ+ φ ψ + 4 ( + λ) φ + φ ψ +. (0..33) Therefore, ρ(λ) has the eigenvalues { 4 ( + λ), 4 ( 3λ)}, and the eigenstates { φ +, φ, ψ +, ψ 4 (+λ), triplet state 4 ( 3λ), singlet state }. (0..34) For the partial transpose criterion, we have the constrain for the parameter λ [ρ(λ)] PT 0 4 ( + λ) 0 λ 4 ( 3λ) 0 From inequality (0..4) and inequality (0..35), we can infer that That means ρ(λ) 0 3 λ, [ρ(λ)] PT 0 λ 3. ρ(λ) is separable 3 λ 3, ρ(λ) is entangled 3 < λ. 3. (0..35) 0..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 σ A a and B σ A a in subsystem Alice, and C σ B b and D σ B b in subsystem Bob. Firstly, we can calculate (σ A a) (σ B b) tr AB [(σ A a) (σ B b) ρ(λ)] Therefore, we can get tr AB [(σ A a) (σ B b) ( 4 ( λ)i 4 + λ φ + φ + )] λ 4 tr A (σ A a) tr B (σ B b) + λ φ + (σ A a) (σ B b) φ + λ φ + (σ A a) (σ B b) φ + λ a b. (0..36) AC AC + BC BC + AD AD BD BD λ a b + a b + a b a b λ ( a + a ) b + ( a a ) b. (0..37) 45

156 Since ( a + a ) and ( a a ) are mutually orthogonal,therefore To ensure the CHSH inequality being correct, we only need 0 ( a + a ) b + ( a a ) b. (0..38) AC AC + BC BC + AD AD BD BD (0..39) AC AC + BC BC + AD AD BD BD max, (0..40) i.e., λ, λ. (0..4) Combine above constrain of parameter λ with the inequality (0..4) which ensures ρ(λ) nonnegative namely a density operator, and we can get 3 λ. (0..4) As we shall see that with λ [ /3, ], the operator ρ(λ) is ensured to be a possible density operator, and if λ (/3, ], the Werner state ρ(λ) is entangled; if λ [ /3, / ],the Werner state ρ(λ) obeys the CHSH inequality; if λ [/3, / ], 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 contradicted with the local hidden variable theorem, namely the entanglement can exist in the local hidden variable theory in some case. 0. Multi-partite entanglement 0.. Definition Def 0... We define the multi-partite separability and entanglement in the following. The N-partite state is fully separable iff Pure state: ψ A A A n ψ A ψ A ψ An ; (0..) Mixed state: ρ A A A n p i ρ i A ρ i A ρ i A n. (0..) i A fully separable tripartite is shown in a diagram form in Figure 0., where the dashed lines represent separable states and A i denote the i-th particle, with i,, 3. 46

157 A A A 3 Figure 0.: Fully separable tripartite A A A A A 3 A A 3 A A 3 (a) (b) Figure 0.: Full entangled tripartite (c) N-partite fully entangled iff it s not fully separable. We can also represent this in a diagram. We still take the tripartite system as an example, as shown in Figure 0. are the three cases of the fully entangled tripartite state, where the solid lines represents the entangled states. 3 N-partite genuinely entangled iff all bipartite partitions are entangled. We can take the tripartite for an example, as shown in Figure 0.3. A A A 3 Figure 0.3: Genuinely entangled tripartite 0.. The GHZ state Def 0.. (N-qubit GHZ states). The state defined as GHZ± N ( x ± x ), (0..3) is the so called N-qubit GHZ state, where x is an N-bit string of 0 and and x + x ( 0 0 N ) binary. From this definition we can see that n : ± ( 0 ± ), which are eigenstates of X with eigenvalues of ±; 47

158 n : ψ(0, 0) ( 00 + ) ψ(0, ) ( 00 ) ψ(, 0) ( ) Bell states { φ ±, ψ ± }, ψ(, ) ( 0 0 ) all of which are eigenvectors of the parity-bit operator Z Z and that of the phase-bit operator X X. 3 n 3: Φ ( ), Φ 8 ( 000 ), Φ ( ), Φ 7 ( 00 0 ), Φ 3 ( ), Φ 6 ( 00 0 ), Φ 4 ( ), Φ 5 ( 0 00 ), with the phase-bit operator X X X and st parity-bit operator: Z Z I, nd parity-bit operator: I Z Z. 4 n-qubit: GHZ n ( 00 0 n n observables + ), with n observables n phase-bit X X X, n st parity-bit Z Z I I, n nd parity-bit I Z Z I I, n 3 (n )-th parity-bit I I I Z Z; n (X X X n ) GHZ n (X X X n ) [ 0x x 3 x n + ( ) x x x 3 x n ] [ x x 3 x n + ( ) x 0x x 3 x n ] ( ) x [ 0x x 3 x n + ( ) x x x 3 x n ] ( ) x GHZ n. 48

159 [Z Z I I ] GHZ n n [Z Z I I ] ( 0x x 3 x n + ( ) x x x 3 x n ) n [( ) x 0x x 3 x n + ( ) x ( )( ) x+ x x 3 x n ] ( ) x [ 0x x 3 x n + ( ) x x x 3 x n ] ( ) x GHZ n. (I Z Z 3 I I ) GHZ n n 3 (I Z Z 3 I I ) [ 0x x 3 x n + ( ) x x x 3 x n ] n 3 [( ) x +x 3 0x x 3 x n + ( ) x ( ) x+ ( ) x3+ x x 3 x n ] ( ) x +x 3 [ 0x x 3 x n + ( ) x x x 3 x n ] ( ) x +x 3 GHZ n. (I I Z 3 Z 4 I I ) GHZ n n 4 (I I Z 3 Z 4 I I ) [ 0x x 3 x n + ( ) x x x 3 x n ] n 4 [( ) x 3+x 4 0x x 3 x n + ( ) x ( ) x3+ ( ) x4+ x x 3 x n ] ( ) x 3+x 4 [ 0x x 3 x n + ( ) x x x 3 x n ] ( ) x 3+x 4 GHZ n. ( I I Z n Z n ) GHZ n n ( I I Z n Z n ) [ 0x x 3 x n + ( ) x x x 3 x n ] n [( ) x n +x n 0x x 3 x n + ( ) x ( ) xn + ( ) xn+ x x 3 x n ] ( ) x n +x n [ 0x x 3 x n + ( ) x x x 3 x n ] ( ) x n +x n GHZ n. 49

160 We can then conclude that with the input qubit x x x n, we can get the output qubit GHZ n which is the common eigenvector of the phase-bit operator, the st parity-bit operator, the nd parity-bit operator, and the (n )-th parity bit operator, with the corresponding eigenvalues (x, x, x + x 3,, x n + x n ). 5 n large number, GHZ n ( live + dead ) (cat state), which means the macroscopically distinguished states used in Schrödinger s cat experiment Properties of GHZ states For the GHZ states GHZ±, each qubit is maximally entangled with the other n qubit, e.g. ρ A tr A A 3 A n GHZ± n GHZ± tr A A 3 A n ( 0y n ± ȳ n )( n 0y ± n ȳ ), with y being a (n )-bit string of 0 and, thus ρ A y A A 3 A n y ( 0y n ± ȳ n )( n 0y ± n ȳ ) y A A 3 A n y ( 0 δ y y± δ y ȳ)( 0 δ yy ± δȳy ) ( ), i.e. ρ A I. (0..4) Cutting one-qubit gives rise to separable (n )-qubit mixed state with rank. namely ρ A A n tr A GHZ± n GHZ± A i ( 0y n + ȳ n )( n 0y + n y ) i A i0 ρ A A n ( y n y + ȳ n ȳ ), (0..5) from which it would be obvious that ρ A A n can be rewritten in the form as illustrated in E.Q. (0..), namely ρ A A n is separable. Note that we can apply the PPT criterion in this case since ρ P T A A n (Id T )ρ A A n ρ A A n 0 (0..6) which is the necessary condition for quantum separability. 3 n-qubit GHZ states GHZ± n form an orthonormal basis of n-qubit Hilber space, i.e. 4 Given a n-qubit GHZ n, other ( n ) GHZ± n states can be generated via LOCC, e.g. for the case of n, the other Bell states can be get from φ +, as shown in Figure

161 Table 0.: GHZ± n makes an orthonormal basis for n-qubit Hilbert space. n GHZ states (orthnormal basis) n-qubit Hilbert space n ± H n ψ(i, j), i, j 0, H H n 3 Φ i, i,,, 8 H H H n GHZ± n H H H n φ + I 4 I Z I X I XZ φ + φ ψ + ψ Figure 0.4: Bell states generated from φ + 5

162 Part IV Quantum Open System and Quantum Error Correction Codes 5

163 Chapter 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. References: Nielsen & Chuang [Preskill] Chapter : Foundations II: measurement and evolution; [Nielsen & Chuang] Chapter 8: Quantum noise and quantum operation. 53

164 . Introduction.. Why we talk about quantum open system Reason Quantum Computer is an open system. Quantum computer would interact with the environment (noise). The observers (control) would also interact with the Quantum Computer. Reason Quantum Computer is a many-qubit system. And it consists of a set of subsystems which interacting with each other... Closed system and open system Table.: 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 Shcrödinger Equation Quantum operation/superoperator Pure State vs. Mixed State for Qubit and Quantum Gate Qubit cos θ Pure state: ψ sin θ eiϕ Mixed state: ρ ( + p σ) (Density matrix) Quantum Gate SU() Superoperator: ρ ρ. Projective measurement The following three point are equivalent: Arbitrary observable O can be expressed as where O a n Π n, (..) n Π n n n (..) with a n being the eigenvalue of O associated with the eigenvector n. 54

165 The mean value of the observable should be with O ψ O ψ ψ a n Π n ψ n a n ψ n n ψ n a n P n, (..3) n P n ψ n (..4) In the orthonormal basis { n }, observables can be determined by the basis. The probability, that the post-measurement is n, should be The post-measurement state should be Prob(n) n ψ ψ n ψ n n ψ ψ Π n ψ tr(ρπ n ). (..5) n ψ Prob(n), (..6) or That is for the pure state. ρ ψ ψ Π n ψ ψ Π n Prob(n) Π n ρπ n Prob(n). (..7) 3 Complete set of orthogonal projection {Π n }. And we know that {Π n } should satisfy Π n Π m δ mn, orthonormal; Π n Π n, Hermition; Π n Id, completeness. n Now, if we consider a mixed state described by the density matrix, then (..8) which defines the ensemble. ρ ρ Prob(n) Π n ρπ n n Prob(n) n ρ {Prob(n), Π n ρπ n Prob(n) } Example: A single-qubit density matrix is defined as Π n ρπ n (..9) ρ( p) ( + p σ), with p R3, and 0 p. (..0) 55

166 The projectors are defined as, Π ( + n σ) n n, (..a) Π ( n σ) n n. (..b) It would be obvious that Π + Π Id, Π i Π j δ ij Id, Π i Π i. (..) The probabilities that the post-measurement state should be n, and that for n, are P tr(ρπ ) ( + p n), (..3a) P tr(ρπ ) ( p n) (..3b).3 General measurement theory 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 projectors is defined as The measurement operators Therefore, namely Π n n n. (.3.) M n Π n. (.3.) M nm n Π n (.3.3) n M nm n Π n n n n n Id. (.3.4) State pure state ψ ; mixed state ρ. 3 Measurement statistics pure state: Prob(m) ψ M mm m ψ ; mixed state: Prob(m) tr (M mm m ρ); 56

167 4 total probability: pure state: m Prob(m) ψ M mm m ψ m ψ M mm m ψ m ψ ψ ; (.3.5) mixed state: 5 post-measurement state: m Prob(m) tr (M mm m ρ) m tr ( M mm m ρ) m trρ. (.3.6) pure state: ψ M m ψ Prob(m) ; mixed state: ρ M m ρm m Prob(m)..4 Definition of POVM POVM means Positive Operator-Valued Measurement. There are two things that we care most at the present time: measurement probability; the post measurement state. The post measurement states may be destroyed in experiments immediately. Def.4.. General Measurement with negligible post-measurement state, with F a M am a, (.4.) F a M a (M a ) M am a F a, (.4.) F a 0, (.4.3) 57

168 and F a Id. (.4.4) a {F a a} is the set of the so-called positive operator-valued measurements, and {M a a} is the set of the general measurements. The probability is determined by F a, and M a is connected with the post measurement state, which we have no interest. Examples: () For the pure state ψ ψ, the probability should be P a ψ F a ψ 0 P a a F a Id. (.4.5) a () A single qubit with density matrix ρ ( p) as defined in E.Q. (..0). Let s denote POVM with {F, F, F 3 }, which are defined as F 3 n n 3 ( + n σ), F 3 n n 3 ( + n σ), (.4.6a) (.4.6b) F 3 3 n 3 n 3 3 ( + n 3 σ), (.4.6c) where n, n and n 3 are three coplanar unit vectors the angle between any pair of them is π 3 in R3, as shown in Figure.. n π 3 π 3 π 3 n n 3 Figure.: n, n and n 3. And we can get that 3 i F i Id; F a F b 4 9 n a n b n a n b 0, with a, b,, 3. We can verify these properties one by one. (.4.7a) F a F a, with a,, 3; (.4.7b) (.4.7c) 58

169 a) 3 i F i 3 ( + n σ) + 3 ( + n σ) + 3 ( + n 3 σ) 3 [3 + ( n + n + n 3 ) σ] (3 + 0) 3 Id. (.4.8) b) F a F a ( 3 ) n a n a n a n a ( 3 ) n a n a 3 F a. (.4.9) c) F a F b 9 ( + n a σ) ( + n b σ) 9 [ + ( n a + n b ) σ + ( n a σ) ( n b σ)] 9 [ + ( n a + n b ) σ + n a n b ], (.4.0) therefore since F a F b vanishes iff F a F b 0, (.4.) n b n a. (.4.).5 More on POVM.5. Dimensional analysis For a Hilbert space H, the number of projectors should be while the number of POVM should satisfy # {Π n } dim H, (.5.) # {F n } dim H. (.5.).5. POVM on subsystem can be viewed as projective measurements on the entire system 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 H A and {F A a } being the POVM on the H A. There is a larger Hilbert space H A with H A H A, such that {F A a } are the projective measurements on H A. 59

170 .5.3 Tensor product realization of POVM Thm Given one dimension non-negative POVM {F A a } on H A, there exist H B such that Prob(a) tr AB [E AB a (ρ A ρ B )] tr A (F A a ρ A ), (.5.3) where {E AB a } are the projective measurements on H A H B. We would construct {F A a } from {EAB a }, namely P a A ρ A F A a A ρ A ρ B E a B, (.5.4) where the circles mean traces. Let s assume that { i A } { j A } is an orthonormal basis of the Hilbert space H A, and { µ B } { ν B } is an orthonormal basis of the Hilbert space H B. Then { iµ AB } { jν AB } is an orthonormal basis of the Hilbert space H A H B Operator formula of F a and ρ B I A ρ B, and tr B ρ B tr AB (I A ρ B ), (.5.5) tr A (F A a ρ A ) tr AB [E AB a (ρ A ρ B )] tr A {tr B [E AB a (I A ρ B )] ρ A } (.5.6) Therefor, F A a tr B [E AB a (I A ρ B )]. (.5.7).5.3. Matrix formalism of F a tr A (F a ρ A ) j F a ρ A j j j F a i i ρ A j, (.5.8) i,j Therefor, we can get that tr AB [E AB a (ρ A ρ B )] j F A a i µ,ν j,ν i,µ jν E AB a iµ iµ ρ A ρ B jν jν E AB a iµ i ρ A j µ ρ B ν (.5.9) j,ν i,µ jν E AB a E AB a iµ µ ρ B ν. (.5.0) 60

171 The properties of F A a ) Hermitian. With E.Q.(.5.7) and E.Q.(.5.0), we can infer from (E AB a ) E a and ρ B ρ B, that F a F a. (.5.) ) Positivity, i.e. F A a 0. We can get the diagonal form of ρ B, i.e. ρ B P µ µ B µ, µ with P µ, and 0 P µ. (.5.) µ Combine this with E.Q.(.5.7), and we can get A ψ F A a ψ A A ψ tr B [E AB a (I A ρ B )] ψ A µ,ν A ψ B ν E AB a µ B ψ A B µ ρ B ν B A ψ B ν E AB a µ B ψ A P µ δ µν µ,ν A ψ B µ E AB a µ B ψ A P µ. (.5.3) µ Thus, since A ψ F A a ψ A E AB a 0, and P µ 0. 0, (.5.4) 3) Completeness. a F A a tr B [E AB a (I A ρ B )] a tr B [ E AB a (I A ρ B )] a tr B (I A ρ B ) I A tr B ρ B I A. (.5.5) Example We are going to talking about an example that has been discussed at the beginning of section.4. The POVM are {F A a a,, 3}, and the explicit expression should be F A a 3 n a n a, with a,, 3. (.5.6) And the vectors { n a a,, 3.} are shown in Figure., from which we can conclude that n + n + n 3 0. (.5.7) 6

172 Now, we introduce an auxiliary Hilbert space H B and the density matrix ρ B on H B. The density matrix of the subsystem B is ρ B 0 B 0. (.5.8) Which makes the density matrix of the compound Hilbert space H A H B be We can also define that Φ a AB 3 n a A 0 B + ρ AB ρ A ρ B ρ A 0 B 0. (.5.9) 3 0 A B, with a,, 3. (.5.0) It can easily verify that { Φ a AB a,, 3.} makes an orthonormal basis for the compound Hilbert space H A H B. AB Φ a Φ a AB 3 A n a B A 0 B 3 n a A 0 B A B 3 A n a n a A + 3, where we have the power to set A n a n a A, if a a ;, if a a. (.5.) If we define further that then we can get Φ 0 AB A B, (.5.) Φ a Φ a δ aa, a, a {0,,, 3}. (.5.3) With the orthonormal basis { Φ a AB a 0,,, 3.}, we can construct the projectors in the following manner namely E AB a Φ a AB Φ a 3 n a A 0 B A B 3 A n a B n a A 0 B A n a B A B A 0 B 3 n a A 0 B A 0 B A B A n a B 0, E AB 3 A 0 B a 3 n a A n a 0 B A 0 B + 3 n a A 0 0 B A n a B 0, (.5.4) 6

173 for a,, 3. And E AB 0 on the other hand, is defined as Therefore, for a,, 3, and Thus, for a,, 3, and E AB E AB 0 Φ 0 AB Φ 0 A B. (.5.5) a (I A ρ B ) E AB a (I A 0 B 0 ) 3 n a A n a 0 B A n a B 0, (.5.6) E AB a (I A ρ B ) 0. (.5.7) F A a tr B [E AB a (I A ρ B )] 3 n a A n a, (.5.8) Therefore, the relation.5.7 is verified in this special case..5.4 Direct-sum realization of POVM F A 0 0. (.5.9).5.5 POVM as quantum operation (superoperator).6 Quantum operation (superoperator).6. Definition of the superoperator Question: What s the evolution equation of density matrix? As we know, for the pure state, ψ, ρ ψ ψ, ψ satisfies the Shrödinger Equation, which means it would evolve under unitary transformation, i.e. ψ ρ ψ ψ U U ψ, (.6.) U U ψ ψ U UρU. (.6.) As for the mixed state described by ρ, let s assume that ρ S ρ S(ρ), (.6.3) where S is the so-called superoperator. We are now going to explore some properties of the superoperator S. As we shall see that both ρ and ρ are density matrices. Therefore, S is a mapping between density matrices. We can assume that the superoperator is a linear mapping on density matrices, i.e. S(λ ρ + λ ρ ) λ S(ρ ) + λ S(ρ ), with λ, λ C; (.6.4) 63

174 trace-preserving mapping (TP), namely tr [S(ρ)] tr(ρ) ; (.6.5) positive mapping (P), i.e. S(ρ) 0. (.6.6) Therefore, S should be a linear positive trace-preserving operator..6. The wonderful theorem The following are equivalent. ) S(ρ) is a CPTP mapping (Complete-Positive-trace-preserving). S(ρ) 0, (S Id)ρ AB 0, ) The Kraus representation theorem. Positive; Complete Positive. (.6.7) Thm.6.. (Kraus representation). A CPTP mapping S(ρ) has the operator sum representation, S(ρ) M µ ρm µ, (.6.8) µ with the Kraus operators M µ satisfying 3) The Stinespring representation or in the diagram representation M µm µ Id. (.6.9) µ S(ρ A ) tr B [U AB (ρ A 0 B 0 )U AB ], (.6.0) A ρ A U AB S(ρ A ) B 0 B 0. (.6.) The symbol stands for the recycle bin, which means that the corresponding output qubit is within our interest. Math tool: Function Analysis. 4) The Choi-Jamiolkowski representation 64

175 .6.3 The CPTP mapping S(ρ A ) is positive, and (S I B )ρ AB is also positive. Remarks: Separable system ρ AB ρ A ρ B. Entangled state. ρ A S S(ρ A ) 0 ρ B ρ B. ρ AB S S(ρ A ) 0 ρ B. The CPTP map is designed for the entangled system in some degree. (.6.) (.6.3) Difference between CPTP mapping and PTP mapping. For example, transpose is PTP mapping, but not CPTP mapping. We can see that T ρ 0 ρ T 0, (.6.4) i.e. transpose is PTP mapping. On the other hand, we can consider a density matrix ρ AB, ρ AB N ii AB jj. (.6.5) i,j If we denote the transpose with T, then T I B means the partial transpose acting one subsystem A. As we know that ρ AB (T I B )ρ AB N (T I B ) ii AB jj i,j N ji AB ij i,j SWAP N (.6.6) (SWAP) Id (SWAP Id)(SWAP + Id) 0, (.6.7) thus SWAP has eigenvalues and. That means SWAP is not positive, namely ρ AB (T I B )ρ AB is not positive. Therefore, T I B is not CPTP mapping..6.4 The Kraus representation with Examples: ρ S(ρ) M µ ρm µ, (.6.8) µ M µ M µ Id. (.6.9) µ 65

176 (i) Unitary evolution of closed system, ρ UρU, (.6.0) with It can be easily verified that M U, M U ; M 0, M 0. (.6.) M M + M M Id. (.6.) (ii) Projective measurement. Π n n n, Π nπ n Id. (.6.3) n ρ ψ ψ {Π n} (iii) POVM with one-dimensional operators {F a }. ρ S(ρ) Π n ρπ n. (.6.4) n ρ ρ a F a ρ F a, with F a 0, and F a. (.6.5) a.6.5 The Stinespring representation A ρ A U AB S(ρ A ) B 0 B 0. (.6.6) The Stinespring representation is actually equivalent to Kraus representation. ρ AB ρ A 0 B 0 ρ AB U AB ρ AB U AB, (.6.7) i.e. ρ A tr B ρ AB tr B (U AB ρ AB U AB ) We can denote that B µ U AB ρ AB U AB µ B µ B µ U AB (ρ A 0 B 0 )U AB µ B µ B µ U AB 0 B (ρ A ) B 0 U AB µ B. (.6.8) µ M µ B µ U AB 0 B, M µ B 0 U AB µ B, (.6.9) then µ M µm µ B 0 U AB µ B µ U AB 0 B µ B 0 U AB U AB 0 B I A. (.6.30) 66

177 .6.6 Remarks on quantum operation.7 Quantum channel Example for Quantum Operation: ρ S ρ S(ρ), (.7.) where the superoperator S is a channel..7. The bit-flip channel The Quantum Bit-flip: 0 X, X 0, with X σx. (.7.) This means Quantum error-model: p p 0 0 p 0, p, (.7.3) where p is the error-probability, 0 p, namely Therefore We can then get the Kraus operators And we can verify that { S( 0 0 ) ( p) p ; (.7.4a) S( ) p ( p). (.7.4b) ρ ρ S(ρ) ( p)ρ + pxρx (.7.5) M 0 pi p ( 0 0 ) ; M px p ( 0 0 ). (.7.6) M 0 M 0 + M M ( p)(i ) + px ) Unitary Representation (Stine Spring representation). ( p)i + pi I. (.7.7) A ψ A U AB S ( ψ A ψ ) B 0 B, (.7.8) where U AB ( ψ A 0 B ) (M 0 I + im X) ψ A 0 B p ψ A 0 B + i px ψ A B (.7.9) 67