Quantum Entanglement and Error Correction

Transcription

1 Quantum Entanglement and Error Correction Fall 2016 Bei Zeng University of Guelph

2 Course Information Instructor: Bei Zeng, TA: Dr. Cheng Guo, Wechat Group: Course materials and discussions. Download Scichat: Scichat broadcast (beta): group number 2060 Please do NOT share the scichat video link (view in group please). Registration and evaluation.

3 The Book Part of the course will be based on the book Quantum Information Meets Quantum Matter From Quantum Entanglement to Topological Phase of Matter Bei Zeng, Xie Chen, Duan-Lu Zhou, Xiao-Gang Wen In Springer Book Series - Quantum Information Science and Technology

4 Quantum Mechanics in Finite Dimensional Systems An arbitrary state can be expanded in the complete set of eigenvectors of  (ÂΨ i = a i Ψ i ). When n is finite: c 1 c 2 Ψ. c n A column vector. Ψ 1 Ψ = n c i Ψ i i=1 Ψ A basis. Ψ n

5 Inner Products c 1 d 1 c 2 Ψ. Φ d 2 n. (Ψ, Φ) = c i d i c n d n i=1 The matrix form: d 1 (Ψ, Φ) = ( c 1 c 2 c n ) d 2. d n The conjugate transpose: Ψ = Ψ T, (Ψ ) = Ψ, (Ψ, Φ) = Ψ Φ

6 Example Ψ = 1 i 1 Φ 2i 1 i 0 Ψ =?, Ψ Φ =?

7 Observables ÂΨ i = a i Ψ i, ˆBΨ i = b ij Ψ j, Â ˆB ˆB = ˆB T a a a n b 11 b 12 b 1n b 21 b 22 b 2n.... b n1 b n2 b nn b 11 b 21 b n1 b 12 b 22 b n2.... b 1n b 2n b nn

8 Observables Average value: Ψ = n c i Ψ i i=1 (Ψ, ˆBΨ) Ψ ˆBΨ = ˆB Ψ b 11 b 12 b 1n ( c 1 c 2 c ) b 21 b 22 b 2n n.... b n1 b n2 b nn c 1 c 2. c n

9 Dirac Notation Ψ c 1 c 2. Ψ ket i=1 c n n n Ψ = c i Ψ i Ψ = c i Ψ i i=1 Ψ ( c 1 c 2 c ) n Ψ bra

10 Inner Product Ψ = c 1 c 2., Φ = d 1 d 2. c n (Ψ, Φ) = d n n c i d i Ψ Φ i=1

11 Operators ÂΨ A Ψ = AΨ AΨ = Ψ A The average value Â Ψ = Ψ ÂΨ Ψ A Ψ A basis Ψ Ψ Ψ n Ψ Ψ i

12 Operators A basis for matrix i j a 11 a 12 a 1n a 21 a 22 a 2n A =.... a n1 a n2 a nn n n a ij i j i=1 j=1 The identity operator I = n i i = i=

13 Example For and ψ 1 = 1 3 ( ) ψ 2 = 1 3 ( 0 i ) and an operator A = i 0 1 i compute ψ 2 ψ 1 and ψ 1 A ψ 1.

14 Evolution Ψ(r, t) i = ĤΨ(r, t) t Ψ(r, t) i = Ĥ Ψ(r, t). t If H is time independent [ ] ih(t2 t 1 ) Ψ(t 2 ) = exp Ψ(t 1 ) = U(t 1, t 2 ) Ψ(t 1 ) Evolution is unitary Ψ U Ψ U U = UU = I

15 Two-level System Electron Spin: The Stern-Gerlach Experiment Figure: From Nielsen & Chuang

16 The Stern-Gerlach Experiment Figure: From Nielsen & Chuang The interpretation: the electron spin, a two-level system + Z 0 + X 1 2 ( ) Z 1 X 1 2 ( 0 1 )

17 Qubit Two-level system 0, 1 Qubit: ψ = α 0 + β 1, α 2 + β 2 = 1. ( ψ = e iγ cos θ eiφ sin θ ) 2 1. Figure: Bloch Sphere

18 The Pauli Matrices ( ) ( ) I =, σ = σ x = X =, 1 0 ( ) ( ) 0 i 1 0 σ 2 = σ y = Y =, σ i 0 3 = σ z = Z =. 0 1 form a basis for 2 2 matrices. σ = (σ x σ x σ z ) Any 2 2 matrix A can be written as A = A 0 I + A σ = A 0 I + A x X + A y Y + A z Z ( ) A0 + A z A x ia y A x + ia y A 0 A z

19 Evolution of Qubit ( ) u00 u A 2 2 unitary operator U = 01 with u 10 u 11 U U = UU = I is a (single qubit) quantum gate Example ( ) 0 1 Bit flip: X = = ( ) 1 0 Phase flip: Z = = Hadamard ( Transform: ) H = = ( ).

20 The Density Operator If with probability p i we have the pure state ψ i, where i p i = 1, we have an ensemble of pure states {p i, ψ i }. We use the density operator ρ to describe this ensemble: ρ = i p i ψ i ψ i For a pure state ψ, the corresponding density operator is ρ = ψ ψ, and ρ 2 = ( ψ ψ )( ψ ψ ) = ψ ψ ψ ψ = ψ ψ

21 The Density Operator For the ensemble {p i, ψ i }, we have the density operator ρ = i p i ψ i ψ i. For any observable A, its average value is A ρ = i p i ψ i A ψ i = i p i tr( ψ i A ψ i ) = i p i tr(a ψ i ψ i ) = i tr(ap i ψ i ψ i ) = tr(aρ). Properties of density operators Trace Condition tr(ρ) = i p i tr( ψ i ψ i ) = i p i = 1 Positivity Condition: for any state φ, φ ρ φ = i p i φ ψ i ψ i φ = i p i φ ψ i 2 0.

22 The Density Operator Example For a single qubit, any density operator ρ can be written as ρ = 1 2 (I + r σ) = 1 2 (I + r xx + r y Y + r z Z) Figure: Bloch Sphere

23 Composite Systems The state space of a composite physical system is the tensor product of the state spaces of the component physical systems For two systems (two vector spaces) V 1 and V 2, we have two states ψ 1 V 1, ψ 2 V 2 The joint state of the total system is ψ 1 ψ 2 For n systems with states ψ i, i = 1, 2,..., n ψ 1 ψ 2 ψ n

24 Composite Systems Properties of tensor products bilinear ψ 1 (α ψ 2 + β φ 2 ) = α ψ 1 ψ 2 + β ψ 1 φ 2 (α ψ 1 + β φ 1 ) ψ 2 = α ψ 1 ψ 2 + β φ 1 ψ 2 inner product ( ψ 1 ψ 2 )( φ 1 φ 2 ) = ψ 1 φ 1 ψ 2 φ 2 operators A 1 A 2 (A 1 A 2 )( ψ 1 ψ 2 ) = (A 1 ψ 1 ) (A 2 ψ 2 ) = A 1 ψ 1 A 2 ψ 2

25 Composite Systems Matrix representation: Kronecker Product For two matrices A (m n) and B (p q): A = a 11 a 12 a 1n a 21 a 22 a 2n.... a m1 a m2 a mn B = b 11 b 12 b 1q b 21 b 22 b 2q.... b p1 b p2 b pq their tensor product a 11 B a 12 B a 1n B a 21 B a 22 B a 2n B A B =.... a m1 B a m2 B a mn B mp nq

26 Kronecker Product Example( ) 0 1 For X = and Y = 1 0 X Y = ( ) 0 i, i 0 ( ) 0 Y 1 Y = 1 Y 0 Y i 0 0 i 0 0 i 0 0 i X Y X 1 Y 2 X 1 Y 2 XY

27 Multiple Qubits A single qubit: ψ = a a 1 1. { 0, 1 }, a basis. Two qubits: { 0 0, 0 1, 1 0, 1 1 }, a basis. Other notations: A two-qubit state: ψ = a a a a For ψ 1 = α α 1 1 and ψ 2 = β β 1 1, ψ 1 ψ 2 = α 0 β α 0 β α 1 β α 1 β 1 11

28 Two Qubit Gates Example X Y controlled-not = controlled-z =

29 Quantum Entanglement ψ = 1 2 ( ) ψ ψ 1 ψ 2 ψ 1 = α α 1 1 ψ 2 = β β 1 1

30 Quantum Entanglement ψ = 1 2 ( ) = 1 2 ( ) ± = 1 2 ( 0 ± 1 ) eigenvectors of X

31 No-Cloning Theorem W.K. Wootters and W.H. Zurek, A Single Quantum Cannot be Cloned, Nature 299 (1982), pp. 802?03. Suppose we can do ψ ψ ψ, then we have a unitary U acting on ψ 0, such that Therefore, but we know that U U U(α 0 + β 1 ) 0 α β 1 1 α β 1 1 (α 0 + β 1 )(α 0 + β 1 )

32 The Reduced Density Operator For a two-particle density operator ρ, the quantum state of the first particle is given by ρ 1 = tr 2 ρ V 1 of dimension d 1, with basis a 1, a = 0, 1,..., d 1 1 V 2 of dimension d 2, with basis b 2, b = 0, 1,..., d 2 1 tr 2 ( a 1 a 1 ) ( b 2 b 2 ) = a 1 a 1 tr( b 2 b 2 ) If ρ = σ 1 σ 2, then = a 1 a 1 b 2 b 2 = a 1 a 1 δ bb ρ 1 = tr 2 ρ = tr 2 (σ 1 σ 2 ) = σ 1 tr(σ 2 ) = σ 1

33 The Reduced Density Operator Example For ψ = 1 2 ( ), find ρ 1. ρ 1 = tr 2 ρ = = I 2

34 Why Partial Trace Consider a two particle state ψ = a ij ij, and an operator A 1 which only acts on the first qubit. Then the average value ψ A 1 ψ = ( ) a ij ij A 1 a kl kl ij kl = ijkl a ija kl i A 1 k j l = ijk a ija kj i A 1 k = tr( ijk a ija kj i A 1 k ) = tr( ijk a ija kj A 1 k i ) = tr(a 1 a ija kj k i ) = tr(a 1 ρ 1 ) ijk