Quantum Entanglement and Error Correction Fall 2016 Bei Zeng University of Guelph
Course Information Instructor: Bei Zeng, email: beizeng@icloud.com TA: Dr. Cheng Guo, email: cheng323232@163.com Wechat Group: Course materials and discussions. Download Scichat: http://scichat.com/ Scichat broadcast (beta): group number 2060 Please do NOT share the scichat video link (view in group please). Registration and evaluation.
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 https://arxiv.org/abs/1508.02595
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 = 1 0. 0 n c i Ψ i i=1 Ψ 2 0 1. 0 A basis. Ψ n 0 0. 1
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
Example Ψ = 1 i 1 Φ 2i 1 i 0 Ψ =?, Ψ Φ =?
Observables ÂΨ i = a i Ψ i, ˆBΨ i = b ij Ψ j, Â ˆB ˆB = ˆB T a 1 0 0 0 a 2 0.... 0 0 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
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
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
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
Operators ÂΨ A Ψ = AΨ AΨ = Ψ A The average value Â Ψ = Ψ ÂΨ Ψ A Ψ A basis Ψ 1 1 0. 0 Ψ 2 0 1. 0 Ψ n 0 0. 1 Ψ Ψ i
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=1 1 0 0 0 1 0.... 0 0 1
Example For and ψ 1 = 1 3 ( 0 + 1 + 2 ) ψ 2 = 1 3 ( 0 i 1 + 2 ) and an operator A = i 0 1 i 1 0 + 2 2 compute ψ 2 ψ 1 and ψ 1 A ψ 1.
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
Two-level System Electron Spin: The Stern-Gerlach Experiment Figure: From Nielsen & Chuang
The Stern-Gerlach Experiment Figure: From Nielsen & Chuang The interpretation: the electron spin, a two-level system + Z 0 + X 1 2 ( 0 + 1 ) Z 1 X 1 2 ( 0 1 )
Qubit Two-level system 0, 1 Qubit: ψ = α 0 + β 1, α 2 + β 2 = 1. ( ψ = e iγ cos θ 2 0 + eiφ sin θ ) 2 1. Figure: Bloch Sphere
The Pauli Matrices ( ) ( ) 1 0 0 1 I =, σ 0 1 1 = 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
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 = = 0 1 + 1 0 1 0 ( ) 1 0 Phase flip: Z = = 0 0 1 1 0 1 Hadamard ( Transform: ) H = 1 1 1 2 = 1 1 1 2 ( 0 0 + 0 1 + 1 0 1 1 ).
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
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.
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
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
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
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
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 0 0 0 i 0 0 i 0 0 i 0 0 i 0 0 0 X Y X 1 Y 2 X 1 Y 2 XY
Multiple Qubits A single qubit: ψ = a 0 0 + a 1 1. { 0, 1 }, a basis. Two qubits: { 0 0, 0 1, 1 0, 1 1 }, a basis. Other notations: 0 0 0 0 00 A two-qubit state: ψ = a 00 00 + a 01 01 + a 10 10 + a 11 11 For ψ 1 = α 0 0 + α 1 1 and ψ 2 = β 0 0 + β 1 1, ψ 1 ψ 2 = α 0 β 0 00 + α 0 β 1 01 + α 1 β 0 10 + α 1 β 1 11
Two Qubit Gates Example X Y controlled-not = controlled-z = 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
Quantum Entanglement ψ = 1 2 ( 00 + 11 ) ψ ψ 1 ψ 2 ψ 1 = α 0 0 + α 1 1 ψ 2 = β 0 0 + β 1 1
Quantum Entanglement ψ = 1 2 ( 00 + 11 ) = 1 2 ( + + + ) ± = 1 2 ( 0 ± 1 ) eigenvectors of X
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 0 0 0 0 U 1 0 1 1 U(α 0 + β 1 ) 0 α 0 0 + β 1 1 α 0 0 + β 1 1 (α 0 + β 1 )(α 0 + β 1 )
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
The Reduced Density Operator Example For ψ = 1 2 ( 00 + 11 ), find ρ 1. ρ 1 = tr 2 ρ = 0 0 + 1 1 2 = I 2
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