Basic Notation and Background

Transcription

1 Department of Mathematics, The College of William and Mary, Williamsburg, Virginia, USA; Department of Mathematics, Taiyuan University of Technology, Taiyuan, Shanxi, P.R. of China.

2 Hilbert spaces The mathematical platform of quantum mechanics/computing is Hilbert space (complete inner product space) V. We mainly focus on finite dimensional complex inner product space C n, the set of n 1 column vectors. Let C n be the dual vector space of C n consisting of 1 n row vectors

3 Hilbert spaces The mathematical platform of quantum mechanics/computing is Hilbert space (complete inner product space) V. We mainly focus on finite dimensional complex inner product space C n, the set of n 1 column vectors. Let C n be the dual vector space of C n consisting of 1 n row vectors In physics, we use the bra and ket vector notation (Dirac notation). Let Then is the dual vector. x = (x 1,..., x n) t = x 1. x n x = ( x 1,..., x n) C n C n.

4 The norm (length) of x is { n 1/2 x = x x 1/2 = {( x 1,..., x n)(x 1,..., x n) t } 1/2 = x j }. j=1 (a) For x C n, we can construct its transpose x t, the conjugate x, the conjugate transpose x (instead of x ). (b) The inner product of x, y C n is x y = n j=1 x j y j. The vectors are orthogonal if their inner product is zero. (c) Given a set of vectors S in C n, we can determine whether it is a linearly independent set, a generating set, an orthonormal set, a basis, or an orthonormal basis.

5 Matrices Linear maps (transformations/functions) on finite dimensional vector spaces can be identified with matrices, namely, A : C n C n so that x A x. Operations and Properties Let M n be the set (vector space/algebra) of n n matrices. (a) One can perform A + B, AB and µa for A, B M n and µ C. (b) One can compute the eigenvalues and eigenvectors of A M n.

6 Let { e 1,..., e n } be the standard basis for C n. Then A ij = e i A e j and A = A ij e i e j. The trace of A is defined by Tr(A) = n j=1 Ajj. Exercise 1) If A is m n and B is n m, then Tr(AB) = Tr(BA). 2) If R is an n n matrix, and ψ C n, then ψ R ψ = Tr(R ψ ψ ). i,j

7 Gram-Schmidt process and orthogonal projectors If S is linearly independent set in C n, one can apply the Gram-Schmidt process to S to get an orthonormal set. If e k is a unit vector, then the projection of a vector v in the direction of e k is v P k v, where P = e k e k is the projection operator. The vector v P k v is orthogonal to e k. If { e 1,..., e n } is an orthonormal basis and P k = e k e k for k = 1,..., n, then (i) P 2 k = P k, (ii) P jp k = 0 for j k, (iii) n k=1 P k = I n.

8 More notation, definitions and examples Let A M n. One can compute its transpose A t, the conjugate A and the conjugate transpose A. The matrix is Hermitian if A = A ; it is skew-hermitian if A = A ; it is normal if AA = A A; it is unitary if A = A 1. If A is real and A t = A 1, the A is a real orthogonal matrix. In quantum information science, the following Pauli matrices are useful: ( ) ( ) ( ) i 1 0 σ 0 = I 2, σ x =, σ 1 0 y =, σ i 0 z =. 0 1 Exercises 1) The Pauli matrices are trace zero Hermitian unitary matrices. 2) Let {i, j, k} = {x, y, z}, then σ iσ j = iγ ijσ k = σ jσ i, and [σ i, σ j] = 2iγ ij, where γ ij = 1 for (i, j) = (x, y), (y, z), (z, x).

9 Some useful facts Theorem (Schur Triangularization Lemma) Every matrix in M n is unitarily similar to a matrix in upper or lower triangular form. Sketch of proof. Solve det(λi A) = 0 to get A x = λ x with x x = 1. Construct U 1 with x as the first column. Then U AU = By induction, U 2 A22U2 is in upper triangular form. ( ) 1 0 Let U = U 1. Then U AU is in triangular form. 0 U 2 ( ) λ A12. 0 A 22

10 Some useful facts Theorem (Schur Triangularization Lemma) Every matrix in M n is unitarily similar to a matrix in upper or lower triangular form. Sketch of proof. Solve det(λi A) = 0 to get A x = λ x with x x = 1. Construct U 1 with x as the first column. Then U AU = By induction, U 2 A22U2 is in upper triangular form. ( ) 1 0 Let U = U 1. Then U AU is in triangular form. 0 U 2 ( ) λ A12. 0 A 22 If AA = A A, then (U AU)(U A U) = (U A U)(U AU). Comparing the (1, 1), (2, 2),..., (n, n) entries on both sides, we see that U AU is in diagonal form.

11 Theorem (Spectral Theorem) If A M n is normal, then there is a unitary U such that UAU = diag (λ 1,..., λ n); hence if U has columns u 1,..., u n then A = λ 1 u 1 u λ n u n u n. (a) For any positive integer m, A m = λ m 1 u 1 u λ m n u n u n. The formula holds for negative integers m as well if A is invertible. (b) If f(z) is an analytic function, then f(a) = n f(λ j) u j u j. j=1 (c) In particular, if f(z) = e z, then f(a) = n j=1 eλ j u j u j.

12 Theorem (Singular Value Decomposition) For every m n matrix A, there are unitary U M m and V M n so that U AV = D such that the (j, j) entries of D is s j for 1 j min{m, n}, where s 2 1 s 2 2 are the eigenvalues of A A. Sketch of Proof. Construct unitary V so that V A AV is in diagonal form with diagonal entries s 2 1,..., s 2 n. Note that the columns of AV are orthogonal vectors with lengths s 1,..., s n. Let the first k = min{m, n} columns of AV be s 1 u 1,..., s k u k, and let U M m be unitary so that the first k columns are u 1,..., u k. Then U AV has the asserted form.

13 Theorem Every unitary matrix U M n is a product of no more than n(n 1)/2 tridiagonal unitary matrices, each of them differs from I n by a 2 2 principal submatrix. Sketch of proof on white board! Remark Using the Gray code labeling of 2 m 2 m, all the tridiagonal unitary matrices involve a change of two basic vectors with binary labels differ in one position.

14 Theorem Every unitary matrix U M n is a product of no more than n(n 1)/2 tridiagonal unitary matrices, each of them differs from I n by a 2 2 principal submatrix. Sketch of proof on white board! Remark Using the Gray code labeling of 2 m 2 m, all the tridiagonal unitary matrices involve a change of two basic vectors with binary labels differ in one position. Definition Given two real vectors x = (x 1,..., x n) and y = (y 1,..., y n), we say that x is majorized by y if n j=1 xj = n yj and the sum of the k j=1 largest entries of x is not larger than that of y for k = 1,..., n 1. Theorem The vector of diagonal entries (d 1,..., d n) of a Hermitian matrix in M n is majorized by the vector of its eigenvalues (λ 1,..., λ n).

15 Tensor products Let A = (a ij) and B be two rectangular matrices or vectors. Then their tensor product (Kronecker product) is the matrix A B = (a ijb). The following equalities hold: (A B)(C D) = (AC) (BD), A (B+C) = A B+A C, (A B) = A B. Note that A, B, C can be ψ 1, ψ 2, ψ 3 vectors. If A M m and B M n are invertible, then (A B) 1 = A 1 B 1. Every T on M mn can be written as T = cjaj Bj with cj C so that j=1 for any vectors u C m, v C n, ( ) T u v = c ja j B j u v = c ja j u B j v. j j

16 Remark We often use the abbreviation: Example Denote by 0 = x 1 x n = x 1 x n = x 1 x n. ( ) 1 and 1 = 0 ( ) ( ) Let H = (1) H 0 = ( )/ 2, H 1 ) = ( 0 1 )/ 2, and H 2 = I 2. (2) Label the rows and columns of A M 2 n by (x 1 x n) with x j {0, 1}. Then n {}}{ H n = H H = 2 n/2 (( 1) x y ), where x y is the inner product of (x 1,..., x n) and (y 1,..., y n). (3) We have H n 0 0 = 2 n/2 x, where the summation ranges x through all x {0, 1} n.

17 Exercise Let A M m and B M n. Prove the following. 1) If UAU and V BV are in upper triangular form, then (U V )(A B)(U V ) is in upper triangular form. 2) det(a B) = det(a) n det(b) m. 3) A B has eigenvalue λ iµ j corresponding to the eigenvector λ i µ j for 1 i m and 1 j n, where A M m has eigenvectors λ 1,..., λ m corresponding to the eigenvalues λ 1,..., λ m, and B M n has eigenvectors µ 1,..., µ n corresponding to the eigenvalues µ 1,..., µ n.

18 Quantum Mechanics I will use the Schrödinger cat and some Youtube clips to tell you the strange world of quantum mechanics, and explain the mathematical model behind that. Here is the experiment: A cat is put in a metal box. A device may release poisonous gas to kill the cat with a certain probability. If we open the box to see (measure) the outcome, we will either a alive cat or a dead cat. In classical world, the cat is either alive or dead in the box once the device is activated. In quantum world, the cat in the box is in superposition condition so that it is alive and dead with a certain probability. However, once we see (measure) the outcome. We can only see the alive cat or dead cat.

19 The Copenhagen interpretation A1 A state x is a unit vector in a Hilbert space H (usually C n ). Linear combinations (superposition) of the physical states are allowed in the state space. A2 Every physical quantity (observable) corresponds to a Hermitian operator (matrix) A. Suppose a state x = c 1 u 1 + c 2 u 2 such that A u i = a i u i for i {1, 2}. Then applying a measurement of x corresponding to A will cause a wave function collapse to u 1 or u 2 with probability of c 1 2 and c 2 2, respectively. Here c 1, c 2 are called the probability amplitude of the state x. A3 The time dependence of a state is governed by the Schrödinger equation ih x t = H(t) x, where h is the Planck constant, and H is a Hermitian operator (matrix) corresponding to the energy of the system known as the Hamiltonian.

20 Remarks 1. The phase of the state does not matter, i.e., x and e iα x represents the same states. 2. In the finite dimensional case, if the state and the observable are represented by n n n x = c j u j C n and A = λ j u j u j = λ jp j, j=1 j=1 j=1 then the projective measurement of the state results in n x A x = λ j c j 2 P i x and becomes c i with a probability of c i In the Schrödinger equation, if H(t) does not depend on t, then j=1 Otherwise, x(t) = exp x(t) = e iht/h x(0). (1) ( i h t 0 H(s)ds ) x(0). (2)

21 Measurements In connection to (A2), quantum measurements are described by a set of measurement operators {M m : 1 m r} such that r j=1 M r M r = I. For each outcome m, construct a measurement operator so that the probability of obtaining outcome m in the state x is computed by p(m) = x M mm m x = x P m x and the state immediately after the measurement is m = Mm x p(m). If there are many copies of a state x, we can let M = mp m. Then the expected value of M is Exp x (M) = M = mp(m) = m x P m x = x M x. m m The variance (square of standard deviation) is (M M ) 2 = x M 2 x x M x 2.

22 The uncertainty principle Let Exp x (A) = x A x = µ and Var x(a) = Exp x ((A µi) 2 ) = x (A µi) 2 x = (A µi) x 2. Theorem For any observables A and B and for any state x, we have Var x(a)var x(b) 1 x [A, B] x, 4 where [A, B] = AB BA is the commutator of A and B. Proof. Let C = A µ AI and D = B µ BI. Then AB BA = CD DC and x (CD DC) x 2 + x (CD+DC) 2 = 4 x CD) x 2 4 x C 2 x x D 2 x.

23 Remark There is no quantum measurement that can distinguish non-orthogonal states ψ 1, ψ 2, reliably. Suppose there is such a measurement. Let E 1 = M j Mj such that ψ1 E1 ψ1 = 1, and E 2 = M k M k such that ψ 2 E 2 ψ 1 = 1. Then E j ψ j ψ j is positive semidefinite for j = 1, 2. Since I = Ei and Ej ψj ψj for j {1, 2}, we have i which is a contradiction. 1 = ψ 1 I ψ 1 ψ 1 (E 1 + E 2) ψ 1 ψ 1 ( ψ 1 ψ 1 ) ψ 1 + ψ 1 ( ψ 2 ψ 2 ) ψ 1 > 1,

24 Positive Operator-Valued Measure (POVM) POVM is a set of positive operators E j = M j Mj corresponding to the measuring operators M j so that Ej = In. j The measurement(s) would allow Bob to identify correctly the state he receives or gets no information at all. Note that a special case of POVM is the projective measurement {P 1,..., P r}, say, arising from an observable A = r j=1 λjpj. Example Alice sends Bob ψ 1 = 0 or ψ 2 = ( )/ 2. Consider the POVM {E 1, E 2, E 3} with E 1 = α 1 1, E 2 = β( 0 1 )( 0 1 ), and E 3 = I E 1 E 2. If Bob gets ψ 1, there is zero probability to get E 1. Thus, getting E 1 measurement means that the received state is ψ 2. Similarly, getting E 2 measurement means that the received state is ψ 1. Getting E 3 measurement yields no information.

25 The evolution Example Recall that the differential equation y = ay has solution y = e at y 0. If H = n j=1 λjpj is Hermitian, then eiωht = n j=1 eiωλ j t P j. If H = h ωσ x/2, then ( ) ψ(t) = e iωtσx/2 cos ωt/2 i sin ωt/2 ψ(0) = ψ(0). i sin ωt/2 cos ωt/2 If ψ(0) = ψ(t) = eiωt/2 2 ( 1 1 ( ) 1, then ψ(t) = 0 ). ( ) cos ωt/2. If ψ(0) = 1 i sin ωt/2 2 ( ) 1, then 1 General solution Let n = (n x, n y, n z) and H = h ωn σ/2 = h ω(n xσ x + n yσ y + n zσ z)/2. Then U(t) = exp( iht/h ) = cos(ω/2)ti + i(n σ) sin(ω/2)t.

26 Multipartite system, tensor product and entangled states A system may have two components described by two Hilbert spaces H 1 and H 2. Then the bipartite system is represented by H = H 1 H 2. A general state in H has the form x = c ij e 1,i e 2,j i,j with c ij 2 = 1, i,j where {e r,1, e r,2,... } is an orthonormal basis for H r with r {1, 2}. A state of the form x = x 1 x 2 is a separable state or a tensor product state. Otherwise, it is an entangled state.

27 Schmidt decomposition Proposition Every state x in H 1 H 2 admits a Schmidt decomposition x = r sj u j v j, j=1 where s j > 0 are the Schmidt coefficients satisfying r sj = 1, r is the j=1 Schmidt number of x, { u 1,..., u r } is an orthonormal set of H 1 and { v 1,..., v r } is an orthonormal set of H 2. Remark In matrix theory, the Schmidt decomposition is just the singular value decomposition if one identify C m C n with the space of m n matrices. The following has a wide research interest in different branches of study. Open problem Extend the Schmidt decomposition to H 1 H k for k 3.

28 Copying information Theorem (No cloning) There is no unitary ψ 0 C n and U M n 2 such that U ψψ 0 = ψψ for a pair of non-orthogonal states ψ 1, ψ 2 C n. Suppose U ψ 1 s = ψ 1 ψ 1 and U ψ 2 s = ψ 2 ψ 2 for some ψ 1, ψ 2 with α = ψ 1 ψ 2 (0, 1). Then α 2 = ψ 1ψ 1 ψ 2ψ 2 = sψ 1U (Uψ 2s) == sψ 1 ψ 2s = α, which is a contradiction. Remark There is unitary U M n 2 such that U e j e 1 = e je j for j = 1,..., n, where { e 1,..., e n } is an orthonormal set. So, classical information can be copied!

29 Manipulation of multiple qubit states Suppose two people, Alice and Bob, each possess one of the (maximally) entangled state, which is known as a Bell state: ψ 0 = ( )/ 2. Each of them can manipulate her/his qubit. For instance, in measuring each qubit, there is a chance of seeing 0 and 1. The other qubit will be in the state of 0 and 1 accordingly. One can apply U I 2, I 2 U to a( two qubit ) states. For example, Alice can 1 1 apply the Hadamard gate H = 1 2 to her qubit so that the 1 1 entangled state ψ 0 is changed to 1 2 {( ) 0 + ( 0 1 ) 1 } = 1 { }. 2 One can also apply a unitary V M 4 to a two qubit states if the two qubits are brought together.

30 For example, the U CN gate defined by will have the effect change the basis of C 4 = C 2 C 2 as follows: 00 00, 01 01, 10 11, This can be described as x y x x y.

31 Application of entanglement: Superdense coding Suppose Alice and Bob share the entangled state ψ 0. Alice can manipulate her qubit to encode one of the two bits of classical information in {00, 01, 10, 11}, and send to Bob. Here is what she may do: (1) she does nothing to send 00; (2) she applies a phase flip σ z to send 01; (3) she applies a not gate σ x to send 10; (4) she applies iσ y to send 11. The resulting state of Bob will be: (1) ( )/ 2, (2) ( )/ 2, (3) ( )/ 2, (4) ( )/ 2, which form the Bell basis of C 4. By a suitable unitary gate V M 4, these will change to 00, 01, 10, 11 so that Bob can get the correct message upon measurement.

32 Teleportation Suppose Alice put another state ξ = α 0 + β 1 to the system, then Bob s qubit will be affected also. The resulting system with 3 qubits: ψ = ξ ψ 0 = 1 2 {α 0 ( ) + β 1 ( }. Alice can apply a CNOT gate to her qubits to get ψ 1 = 1 2 {α( )( ) + β 1 ( }. Then apply a Hadamard gate to the first qubit to get ψ 2 = 1 {α( )( ) + β( 0 1 )( }. 2

33 Regrouping yields ψ 2 = 1 { 00 (α 0 + β 1 ) + 01 (α 1 + β 0 ) (α 0 β 1 ) + 11 (α 1 β 0 )}. Now, measuring the two qubits of Alice gives one of the four possibilities: 00, 01, 10, 11. The corresponding qubit of Bob will become respectively. α 0 + β 1, α 1 + β 0, α 0 β 1, α 1 β 0, Based on the measurement of Alice, Bob applies σ 0, σ x, σ z, iσ y to convert his qubit to ψ.

34 Mixed States and Density Matrices A system is in a mixed state if there is a probability p i that the system is in state x i for i = 1,..., N. If there is only one possible state, i.e., p 1 = 1, then the system is in pure state. The mean value of the measurement of the system corresponding to the observable described by the Hermitian matrix A is A = N p i x j A x j = Tr(Aρ) where ρ = j=1 is a density operator (matrix). N p j x j x j (3) j=1

35 Description of a quantum system in mixed states A1 A physical state is specified by a density matrix ρ : H H, which is positive semidefinite with trace equal to one. A2 The mean value of an observable associate with the Hermitian matrix A is A = Tr(ρA). A3 The temporal evolution of the density matrix is given by the Liouville-von Neumann equation where H is the system Hamiltonian. ih d ρ = [H, ρ] = Hρ ρh, dt

36 Remarks (1) A density matrix ρ = pj xj xj corresponds to a mixed state, where j the vector states x 1,..., x N need not be orthonormal. (2) Each x j satisfies the Schrödinger equation ih d xj = H xj. dt One can derive the Liouville-von Neumann equation from these equations. (3) The set of density matrices is compact and convex.

37 Exercises 1) The following conditions are equivalent for a given state (density matrix) ρ. (a) ρ is pure. (b) ρ 2 = ρ. (c) Tr(ρ 2 ) = 1. 2) Show that every density matrix ρ M 2 has the form ρ = 1 2 (σ0 + xσx + yσy + zσz) with x2 + y 2 + z 2 1. The equality holds if and only if ρ is a pure state. Hence every density matrix in M 2 corresponds to a point in the unit sphere in R 3, known as the Bloch sphere; ρ is a pure state if and only if it corresponds to a point on the sphere.

38 Definition Suppose H = H 1 H 2. A state ρ is uncorrelated if ρ = ρ 1 ρ 2; it is separable if it is a convex combination of uncorrelated states, i.e., ρ = r q j ρ 1,j ρ 2,j with q 1,..., q r > 0, q q r = 1. j=1 Otherwise, it is inseparable. Remark Do not confuse this with the definitions of separability in the vector case. There will be discussion on this topic in depth. Definition Let ρ = r cjρ1,j ρ2,j act on H = H1 H2. The partial j=1 transpose of ρ with respect to H 2 is ρ pt = r ρ 1,j ρ t 2,j. j=1 Proposition If ρ is separable, then so is ρ pt. If ρ pt has negative eigenvalues, then it is not physical and ρ is not separable. The converse holds if H has dimension at most 6. Open problem Find effective way to determine separability.

39 Example Consider the Werner state and its partial transpose 1 p ρ = p 2p p 1 + p 0, p ρ pt = 1 4 The partial transpose has eigenvalues 1 p 0 0 2p p p 0 2p p. (1 + p)/4, (1 + p)/4, (1 + p)/2, (1 3p)/4. So, it is not separable if and only if p (1/2, 1].

40 The realignment matrices For any X = (x ij) M n, let vec(x) = (x 11, x 12,..., x nn). Suppose ρ = (ρ ij) 1 i,j m M mn is a density matrix such that ρ ij M n. The realignment matrix of ρ is the matrix ρ R = vec(ρ 11) vec(ρ 12). vec(ρ mm). Theorem Suppose m n and ρ M mn = M m M n is a density matrix. If ρ is separable, then the sum of the singular values of ρ R is at most one. In fact, the vector of singular values of ρ R majorizes the vector (α, β,..., β) R 1 m2, where α = 1/ mn and β = (1 α)/m 2. Open problem If ρ M 2 M 3 is separable, then ρ R cannot have its vector of singular values equal to (1, ( 6 1)/3, ( 6 1)/3, ( 6 1)/3)/ 6.

41 Partial Trace and Purification Let A be an operator acting on H = H 1 H 2. The partial trace of A over H 2 is an operator acting on H 1 defined by A 1 = Tr 2A = (I u )A(I u ). In particular, Tr 2(A 1 A 2) = A 1(TrA 2). We can also define Tr 1(A 1 A 2) = (TrA 1)A 2 and extend by linearity. u Remark The partial trace is the unique operation which gives rise to the correct description of observable quantities for subsystems of a composite system. Theorem (Purification) Suppose ρ 1 = n pj xj xj Mn. Let j=1 { y 1,..., y n } be an orthonormal basis of C n, and ψ = n pj x j=1 j y j. Then Tr 2( ψ ψ ) = ρ 1. Exercise Suppose ρ 1 = Tr 2( ψ 1 ψ 1 ) = Tr 2( ψ 2 ψ 2 ) M m with ψ 1, ψ 2 C m C n. Then there is a unitary U M n such that ψ 2 = (I m U) ψ 1.

42 Quantum channels and quantum operations A unitary time evolution of a closed system is determined by the quantum map E defined by E(ρ S) = U(t)ρ SU(t). Here, ρ S is the density matrix of a closed system at time t = 0 and U(t) is the time evolution operator. An open system is a system of interest (called the principal system) coupled with its environment. The total Hamiltonian is given by H T = H S + H E + H SE, where H S, H E and H SE are the system Hamiltonian, the environment Hamiltonian and their interaction Hamiltonian, respectively. The state of the total system, which is assumed to be closed, will be described by ρ acting on the Hilbert space H S H E such that one has the approximation ρ(0) = ρ S ρ E and ρ(t) = U(t)(ρ S ρ E)U(t) for t > 0. For simplicity, we may assume that H T is a constant matrix and U(t) = e ith T.

43 We study the system (H S) by taking the partial trace ρ S(t) = Tr E[U(t)(ρ S ρ E)U(t) ] = (I S ε a )[U(t)(ρ S ρ E)U(t) ](I S ε a ) a J for any complete orthonormal basis { ε a : a J} for H E. We may assume ρ E = ε 0 ε 0 by linearity or by purification. Let Then E a(t) = (I S ε a )U(t)(I S ε 0 ). ρ S(t) = E a(t)ρ SE a(t). a This is known as the operator-sum representation of the quantum operation. Note that E a(t) E a(t) = (I S ε 0 )U(t) (I S ε a )(I S ε a )U(t)(I S ε 0 ) a a = (I S ε 0 )U(t) (I S I E)U(t)(I S ε 0 ) = I S. This is the trace preserving condition for the quantum operation. For certain quantum operations or channels, one may relax this condition.

44 EPR and the Bell inequality Under the real locality theory of Einstein, Rosen, and Podolsky, Bell suggested the following inequality. Suppose a measurement of some quantity prepared by Charlie to that Alice measures Q or R, and Bob measures S and T, where Q, R, S, T each can assume the value 1 and 1. Then (Q + R)S = 0 or (R Q)T = 0 so that QS + RS + RT QT = ±2 and E(QS + RS + RT QT ) = E(QS) + E(RS) + E(RT ) E(QT ) 2.

45 However, in quantum world, Charlie prepares the state ψ = ( )/ 2. Alice performs the measurements Q = σ z and R = σ x, where as Bob performs the measurements S = (σ z + σ x)/ 2 and T = (σ z σ x)/ 2. Then so that QS = 1/ 2, RS = 1/ 2, RT = 1/ 2, QT = 1/ 2 QS + RS + RT QT = 2 2, which is confirmed by experiment. So, the real locality theory does not apply to quantum mechanics.

46 M. Nakahara and T. Ohmi, Quantum Computing: From Linear Algebra to Physical Realizations, CRS Press, New York, M. Nakahara, R. Rahimi, and A. SaiToh (editors), Mathematical Aspects of Quantum Computing 2007, World Scientific, Singapore, M.A. Nielsen and I.L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, Cambridge, 2000.

47 Exercises Exercises 1) If A is m n and B is n m, then Tr(AB) = Tr(BA). 2) If R is an n n matrix, and ψ C n, then ψ R ψ = Tr(R ψ ψ ).

48 Exercises Exercises 1) If A is m n and B is n m, then Tr(AB) = Tr(BA). 2) If R is an n n matrix, and ψ C n, then ψ R ψ = Tr(R ψ ψ ). Exercises 1) The Pauli matrices are trace zero Hermitian unitary matrices. 2) Let {i, j, k} = {x, y, z}, then σ iσ j = iγ ijσ k = σ jσ i, and [σ i, σ j] = 2iγ ij, where γ ij = 1 for (i, j) = (x, y), (y, z), (z, x).

49 Exercise Let A M m and B M n. Prove the following. If UAU and V BV are in upper triangular form, then (U V )(A B)(U V ) is in upper triangular form. det(a B) = det(a) n det(b) m. A B has eigenvalue λ iµ j corresponding to the eigenvector λ i µ j for 1 i m and 1 j n, if A M m has eigenvectors λ 1,..., λ m corresponding to the eigenvalues λ 1,..., λ m, and B M n has eigenvectors µ 1,..., µ n corresponding to the eigenvalues µ 1,..., µ n.

50 Exercises 1) The following conditions are equivalent for a given state (density matrix) ρ. (a) ρ is pure. (b) ρ 2 = ρ. (c) Tr(ρ 2 ) = 1. 2) Show that every density matrix ρ M 2 has the form ρ = 1 2 (σ0 + xσx + yσy + zσz) with x2 + y 2 + z 2 1. The equality holds if and only if ρ is a pure state. Hence every density matrix in M 2 corresponds to a point in the unit sphere in R 3, known as the Bloch sphere; ρ is a pure state if and only if it corresponds to a point on the sphere. Exercise Suppose ρ 1 = Tr 2( ψ 1 ψ 1 ) = Tr 2( ψ 2 ψ 2 ) M m with ψ 1, ψ 2 C m C n. Then there is a unitary U M n such that ψ 2 = (I m U) ψ 1.