Decay of the Singlet Conversion Probability in One Dimensional Quantum Networks

Transcription

1 Decay of the Singlet Conversion Probability in One Dimensional Quantum Networks Scott Hottovy Advised by: Dr. Janek Wehr University of Arizona Applied Mathematics December 18, 2009

2 Networks A network is a way of setting up nodes to transmit information Figure: A one dimensional lattice

3 Networks A network is a way of setting up nodes to transmit information Figure: A one dimensional lattice Figure: A two dimensional lattice

4 Quantum Information A quantum network, is a network that uses quantum phenomena to transmit information.

5 Quantum Information A quantum network, is a network that uses quantum phenomena to transmit information. In this talk, we will

6 Quantum Information A quantum network, is a network that uses quantum phenomena to transmit information. In this talk, we will Give a brief introduction to quantum mechanics in the quantum computation setting.

7 Quantum Information A quantum network, is a network that uses quantum phenomena to transmit information. In this talk, we will Give a brief introduction to quantum mechanics in the quantum computation setting. Explain a method that improves communication in 2-D quantum networks.

8 Quantum Information A quantum network, is a network that uses quantum phenomena to transmit information. In this talk, we will Give a brief introduction to quantum mechanics in the quantum computation setting. Explain a method that improves communication in 2-D quantum networks. Show that this method is not reliable in large 1-D quantum networks.

9 A Note about Notation Notation in quantum computation consists of

10 A Note about Notation Notation in quantum computation consists of A vector v C n is called a ket and is written v. The vector w C n is a bra, v.

11 A Note about Notation Notation in quantum computation consists of A vector v C n is called a ket and is written v. The vector w C n is a bra, v. So the inner product of v and w is w v, a bracket.

12 A Note about Notation Notation in quantum computation consists of A vector v C n is called a ket and is written v. The vector w C n is a bra, v. So the inner product of v and w is w v, a bracket. An operator (matrix), A C n m acting on a ket is written A v. and the inner product of the result with w is w A v.

13 Quantum Mechanics Figure:

14 Postulates of QM Postulate I: Associated to any isolated physical system is a complex vector space with an inner product called the state space. The system is described by the state vector.

15 Postulates of QM Postulate I: Associated to any isolated physical system is a complex vector space with an inner product called the state space. The system is described by the state vector. Example: A quantum bit, called a qubit, is a physical system with state space C 2 and is described by the state vector ψ = α 0 + β 1.

16 Postulates of QM Postulate II: The evolution of a closed quantum system is described by a unitary operation U. ψ(t + s) = U ψ(t).

17 Postulates of QM Postulate II: The evolution of a closed quantum system is described by a unitary operation U. ψ(t + s) = U ψ(t). Postulate III: Let {M m } be measurement operations. The {M m } must satisfy m M mm m = I. If the quantum system is in state ψ immediately before the measurement is taken, then the probability that the result of the measurement is m is given by p(m) = ψ M mm m ψ, and the state of the system immediately after the measurement is given by M m ψ p(m).

18 Measurement Example Example: A qubit is in state ψ = α 0 + β 1, Measurement described by the operators 0 0, 1 1. Then,

19 Measurement Example Example: A qubit is in state ψ = α 0 + β 1, Measurement described by the operators 0 0, 1 1. Then, p(0) = ψ ( 0 0 ) ( 0 0 ) ψ = ψ 0 2 = α 2, p(1) = β 2. Note: α 2 + β 2 = 1.

20 Example: A qubit is in state Measurement Example ψ = α 0 + β 1, Measurement described by the operators 0 0, 1 1. Then, p(0) = ψ ( 0 0 ) ( 0 0 ) ψ = ψ 0 2 = α 2, p(1) = β 2. Note: α 2 + β 2 = 1. ψ 0 = 0 0 ψ p(0) = α α 0. ψ 1 = β β 1.

21 Postulates of Quantum Mechanics Postulate IV: The state space of a composite physical system is the tensor product of the state spaces of the component physical systems. Example: The joint state of two qubits ψ 1 = α β 1 1 and ψ 2 = α β 2 is ψ 1 ψ 2 = α 1 α α 1 β β 1 α β 1 β ψ 1 ψ 2 = α 1 α α 1 β β 1 α β 1 β ψ 1 ψ 2 C 2 C 2 or C 4.

22 Schmidt Decomposition Map the two qubit system to a matrix, ψ 1 ψ 2 C 2 2,

23 Schmidt Decomposition Map the two qubit system to a matrix, ψ 1 ψ 2 C 2 2, [ ] α1 α 2 α 1 β 2 = UΣV. β 1 α 2 β 1 β 2

24 Schmidt Decomposition Map the two qubit system to a matrix, ψ 1 ψ 2 C 2 2, [ ] α1 α 2 α 1 β 2 = UΣV. β 1 α 2 β 1 β 2 A change of basis gives the joint system as, ψ 1 ψ 2 = λ 0 + λ 1. Where λ 0 λ 1 and λ 0 + λ 1 = 1.

25 Schmidt Decomposition Map the two qubit system to a matrix, ψ 1 ψ 2 C 2 2, [ ] α1 α 2 α 1 β 2 = UΣV. β 1 α 2 β 1 β 2 A change of basis gives the joint system as, ψ 1 ψ 2 = λ 0 + λ 1. Where λ 0 λ 1 and λ 0 + λ 1 = 1. Without loss of generality, any two qubit system is of the form ψ = λ λ 1 11.

26 Entangled Pairs A joint system of two qubits are referred to as an entangled pair.

27 Entangled Pairs A joint system of two qubits are referred to as an entangled pair. An important basis in two qubit systems (C 4 ) is the Bell basis, Φ ± = 00 ± 11, Ψ ± 01 ± 10 =. 2 2

28 1-D Quantum Networks All entangled pairs are in the state φ = λ λ 1 11.

29 1-D Quantum Networks All entangled pairs are in the state φ = λ λ Want to send information from A to B. Need singlets.

30 1-D Quantum Networks All entangled pairs are in the state φ = λ λ Want to send information from A to B. Need singlets. Probability of converting an entangled pair to a singlet, or Singlet Conversion Probability (SCP), is 2λ 1 (Vidal). Is there a better way?

31 Entanglement Swapping Perform the Bell measurements on qubits BC. The initial state is, φ AB φ BC = λ λ 0 λ λ 0 λ λ (Acín, Cirac, Lewenstein)

32 Entanglement Swapping Four different outcomes for states BC including, ψ ABCD = I 1 ( Φ + Φ + ) 23 I 4 φ p(φ + ) 1/2 = p(φ + ) 1/2 ( λ1 2 ( 0 Φ + 0 ) + λ 2 2 ( 1 Φ + 1 ). Where p(φ + ) = (λ λ2 1 )/2 is the probability of getting the result Φ + on qubits BC.

33 Entanglement Swapping Four different outcomes for states BC including, ψ ABCD = I 1 ( Φ + Φ + ) 23 I 4 φ p(φ + ) 1/2 = p(φ + ) 1/2 ( λ1 2 ( 0 Φ + 0 ) + λ 2 2 ( 1 Φ + 1 ). Where p(φ + ) = (λ λ2 1 )/2 is the probability of getting the result Φ + on qubits BC. The advantage: Using the density language, and the partial trace, ρ AD = tr BC (ρ ABCD ) = ψ AD ψ AD, ψ AD = λ λ λ2 1 λ 1 λ λ

34 Entanglement Swapping What is the probability of obtaining a singlet between AD?

35 Entanglement Swapping What is the probability of obtaining a singlet between AD? 4 different states with Schmidt coefficients λ k 1 with probability p k. The weighted average Singlet Probability, or mean SCP,

36 Entanglement Swapping What is the probability of obtaining a singlet between AD? 4 different states with Schmidt coefficients λ k 1 with probability p k. The weighted average Singlet Probability, or mean SCP, 4 p k λ k 1 = 2λ 1. No loss in mean SCP! k=1

37 Entanglement Swapping What is the probability of obtaining a singlet between AD? 4 different states with Schmidt coefficients λ k 1 with probability p k. The weighted average Singlet Probability, or mean SCP, No loss in mean SCP! 4 p k λ k 1 = 2λ 1. k=1 Bad News: Same trick doesn t work with 2 repeaters. Mean SCP exponentially decays for Bell measurements (Perseguers et al.).

38 One Repeater Ensemble of states, ψab k = λ k 0 λ 00 + k 1 11, with probability p k, ψcd l = µ l 0 µ 00 + k 1 11, with probability q l.

39 One Repeater Ensemble of states, ψab k = λ k 0 λ 00 + k 1 11, with probability p k, ψcd l = µ l 0 µ 00 + k 1 11, with probability q l. Perform a measurement, { u m u m } 4 m=1, on BC where 1 1 u m = a i,j,m ij, a i,j,m 2 = 1, and u n u m = δ nm. i,j=0 i,j=0

40 The resulting state is = One Repeater ψ ABCD = I 1 ( u m u m ) 23 I 4 ψab k ψl CD p(um ) ( ) 1 p(um ) (a 0,0,m λ k 0 µl 0 0 u m 0 + a 0,1,m λ k 0 µl 1 0 u m 1 + a 1,0,m λ k 1 µl 0 1 u m 0 + a 1,1,m λ k 1 µl 1 1 u m 1 ).

41 The resulting state is = One Repeater ψ ABCD = I 1 ( u m u m ) 23 I 4 ψab k ψl CD p(um ) ( ) 1 p(um ) (a 0,0,m λ k 0 µl 0 0 u m 0 + a 0,1,m λ k 0 µl 1 0 u m 1 + a 1,0,m λ k 1 µl 0 1 u m 0 + a 1,1,m λ k 1 µl 1 1 u m 1 ). ( 1 ( ψ AD = a 0,0,m λ0 µ a 0,1,m λ0 µ 1 01 p(um )) ) +a 1,0,m λ0 µ a 1,1,m λ1 µ 1 11.

42 One Repeater Using the Schmidt Decomposition Theorem, ψ AD = s 0(A(λ k, µ l, m)) p(um ) + s 1(A(λ k, µ l, m)). p(um )

43 One Repeater Using the Schmidt Decomposition Theorem, ψ AD = s 0(A(λ k, µ l, m)) p(um ) + s 1(A(λ k, µ l, m)). p(um ) Now we compute the mean SCP between AD, 2 k,l p k q l ( ) 2 4 s 1 (A(λ k, µ l, m)) p(u m ). p(um ) m=1

44 One Repeater Using the Schmidt Decomposition Theorem, ψ AD = s 0(A(λ k, µ l, m)) p(um ) + s 1(A(λ k, µ l, m)). p(um ) Now we compute the mean SCP between AD, 2 k,l p k q l ( ) 2 4 s 1 (A(λ k, µ l, m)) p(u m ). p(um ) m=1 The smallest singular value of a matrix A C 2 2 has the property, s 1 (A) 2 = min x 2 =1,x 0 Ax 2 2.

45 Mean SCP s 1 (A(λ k, µ l, m)) 2 A(λ k, µ l, m)(0, 1) T 2 2 = a 0,1,m λ k 2 0 µl 1 a 1,1,m λ k = a 0,1,m 2 λ k 0µ l 1 + a 1,1,m 2 λ k 1µ l 1. 1 µl 1 Note: u m are an orthonormal basis in C 4. 2 a 0,0,m a m = a 0,1,m a 1,0,m. a 1,1,m Then the matrix A = [a 1 a 2 a 3 a 4 ] is orthogonal AA = A A = I.

46 Mean SCP So the rows are orthonormal as well, 4 a i,j,m 2 = 1, for i,j fixed. m=1

47 Mean SCP So the rows are orthonormal as well, 4 a i,j,m 2 = 1, for i,j fixed. m=1 so the mean SCP, SCP AD 2 k,l = 2 k,l 4 p k q l a 0,1,m 2 λ k 0µ l 1 + a 1,1,m 2 λ k 1µ l 1 m=1 p k q l µ l 1(λ k 0 + λ k 1) = 2 p k q l µ l 1. k,l

48 Mean SCP Also, the singular values of A(λ k 1, µl 1, m) are the singular values of A(λ k 1, µl 1, m). So s 1 (A (λ k, µ l, m)) 2 A(λ k, µ l, m) (0, 1) T 2 2 = a 1,0,m λ k 1 µl 0 a 1,1,m λ k 1 µl 1 = a 1,0,m 2 λ k 1µ l 0 + a 1,1,m 2 λ k 1µ l

49 Mean SCP Also, the singular values of A(λ k 1, µl 1, m) are the singular values of A(λ k 1, µl 1, m). So s 1 (A (λ k, µ l, m)) 2 A(λ k, µ l, m) (0, 1) T 2 2 = a 1,0,m λ k 1 µl 0 a 1,1,m λ k 1 µl 1 = a 1,0,m 2 λ k 1µ l 0 + a 1,1,m 2 λ k 1µ l So the mean SCP is also bounded by, SCP AD 2 k,l = 2 k,l 4 p k q l λ k 1 µ l 0 a 1,0,m 2 + µ l 1 a 1,1,m 2 m=1 p k q l λ k 1(µ l 0 + µ l 1) = 2 p k q l λ k 1. k,l

50 Mean SCP So the mean SCP is, SCP AD 2 p k q l min{λ k 1, µ l 1}. k,l

51 Decay After each arbitrary measurement, The SCP is bounded by the smaller Schmidt coefficient in the states.

52 Decay After each arbitrary measurement, The SCP is bounded by the smaller Schmidt coefficient in the states. The splitting of states that occurs results in a decay of Schmidt coefficients.

53 Decay After each arbitrary measurement, The SCP is bounded by the smaller Schmidt coefficient in the states. The splitting of states that occurs results in a decay of Schmidt coefficients. Therefore, the mean SCP decays to zero as the number of repeaters grows.

54 Decay After each arbitrary measurement, The SCP is bounded by the smaller Schmidt coefficient in the states. The splitting of states that occurs results in a decay of Schmidt coefficients. Therefore, the mean SCP decays to zero as the number of repeaters grows. But what is the rate of the decay?

55 Future Work

56 Future Work Prove exponential decay rigorously for large 1-D quantum networks.

57 Future Work Prove exponential decay rigorously for large 1-D quantum networks. Analyze different protocols in 2-dimensional quantum networks.

58 Future Work Prove exponential decay rigorously for large 1-D quantum networks. Analyze different protocols in 2-dimensional quantum networks. Look at geometrical structure in 2-dimensional quantum networks. Figure: Figure 3, Nature Physics, VOL 3, 25 Feb. 2007

59 Questions?