Quantum Hadamard channels (I)

Transcription

1 .... Quantum Hadamard channels (I) Vlad Gheorghiu Department of Physics Carnegie Mellon University Pittsburgh, PA 15213, U.S.A. July 8, 2010 Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

2 Outline...1 Introductory notions...2 Quantum Hadamard channels Reference: Bradler et al, PRA 81, (2010) A summary of this talk is available online at Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

3 . Introduction Introductory notions Noisy quantum channel, Kraus representation N (σ) = N i σn i, N i N i = I. i i (1) Isometric extension N A B (σ) = Tr E {U N σu N }, U N U N = I. (2) Figure: Isometric extension Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

4 Introductory notions The Kraus operators provide a straightforward method for constructing an isometric extension U A BE N = i N A B i i E, (3) where { i E } is an orthonormal set of states. Useful relation U N σu N = i,j (N i σn j )B i j E. (4) Complementary channel, unique up to isometries on the system E N c (σ) := Tr B {U N σu N }. (5) Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

5 Introductory notions Useful fact: a valid complementary channel for the channel introduced before N c (σ) = i,j Tr{N i σn j } i j E. (6) Degradable channel: if there exists a degrading channel D B E that simulates the action of the complementary channel (N c ) A E, i.e. D B E s.t. σ, D B E N A B (σ) = (N c ) A E (σ). (7) Degradable channels are nice! Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

6 Introductory notions More tips and tricks: suppose Alice possesses an ensemble {p X (x), ρ A x }, where p X (x) is the probability density function for a random variable X. She can augment this ensemble by correlating a classical variable (a label) with each ρ A x, and produce {p X (x), x x X ρ A x }, with { x X } an orthonormal set. Classical-quantum state: ρ XA = x p X (x) x x X ρ A x. (8) Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

7 Introductory notions Let ϕ x AA be the purification of ρ A x. Then the following is also a classical-quantum state ρ XAA = x p X (x) x x X ϕ x ϕ x AA. (9) Suppose now Alice transmits the A subsystem through a noisy quantum channel N A B. The output state will be ρ XAB := N A B (ρ XAA ) = x p X (x) x x X N A B ϕ x ϕ x AA = x p X (x) x x X ρ AB x. (10) Define ϕ x ABE as ϕ x ABE := U N ϕ x AA purify ρ AB x. and observe that ϕ x ABE Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

8 . Entanglement-breaking channels Entanglement-breaking channels: N is entanglement-breaking if it outputs a separable state whenever half of any entangled state is the input to the channel. Reference: Horodecki, Shor and Ruskai, Rev. Math. Phys. 15, 629 (2003). Enough to check only for MES inputs ϕ AA := 1 D 1 i A i A, (11) D i=0 i.e. N A B is entanglement-breaking if and only if N A B ( ϕ AA ) = x p X (x)ρ A x ρ B x. (12) Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

9 It turns out that the action of an entanglement-breaking channel can always be written as N EB (ρ) = x Tr{Λ x ρ}σ x, (13) where {Λ x } is a POVM and the states σ x depend on the channel. Additionally, any entanglement-breaking channel admits a Kraus representation whose Kraus operators are unit rank: N i = ξ B ζ A. The sets { ξ B } and { ζ A } are not necessarily orthonormal. The classical capacity of an entanglement-breaking channel admits a single-letter formula. It s quantum capacity is zero. Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

10 . Hadamard channels A Hadamard channel is a quantum channel whose complementary channel is entanglement breaking. Why Hadamard? Its output can always be written as the Hadamard product (elementwise multiplication) of a representation of the input density operator with another operator. How does this work? Let U N c be an isometric extension of the complementary channel (N c ) A E. The Kraus operators of the complementary channel (which is entanglement-breaking) (N c ) A E are unit rank, ξ i E ζ i A. Then U N c σu N c = i,j ξ i E ζ i A σ ζ j A ξ j E i B j B = i,j ζ i A σ ζ j A ξ j ξ i E i B j B. (14) Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

11 Now trace over E to get the original channel from A to B N A B H (σ) = i,j ζ i A σ ζ j A ξ j ξ i E i B j B. (15) Now let [Σ] i,j = ζ i A σ ζ j A, a representation of the input state σ, and let [Γ] i,j = ξ j ξ i E. Then, w.r.t. the basis { i B } N A B H (σ) = Σ Γ. (16) Hadamard channels are degradable! Why so? Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

12 Bob performs a von Neumann measurement on his state in the basis { i B } and prepares the state ξ i E conditional on the outcome i of the measurement. This procedure simulates the complementary channel (N c ) A E and also implies that the degrading map D B E is entanglement breaking. The Kraus operators of the degrading map D B E are { ξ E i B } so that [ ] D B E N A B H (σ) = ξ i E i B N A B (σ) i B ξ i E i = i i A σ i A ξ ξ E, (17) hence this degrading map effectively simulates the complementary channel (N ) A E. Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

13 The degrading map can be viewed as a composition of two maps:..1. A first map D1 B Y performs the von-neumann measurement, leading to a classical variable Y...2. A second map D2 Y E performs the state preparation, conditional on the value of the classical variable Y. Hence D B E = D Y E 2 D B Y 1. (18) Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

14 . Examples of Hadamard channels Generalized dephasing channels: model physical processes where there is no loss of energy but there is loss of quantum coherence (noise that dominates, for example, in superconducting qubits). The input A and the output B have the same dimension. Let { i A } and { i B } be orthonormal bases (the first is called the dephasing basis). The channel does not affect any state that is diagonal in the dephasing basis, but it mixes coherent superpositions of these basis states. An isometric extension U A BE N GD has the form N A B GD of a generalized dephasing channel U A BE N GD := i i B i A υ i E, (19) where the set { υ i E } is not necessarily orthonormal. Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

15 The output of a generalized dephasing channel is N GD (σ) = i,j υ j υ i E i σ j A i j B. (20) If the states { υ i E } are orthonormal, the channel is the completely dephasing channel A B A B (σ) := i i B i A σ i A i B. (21) The complementary channel of a generalized dephasing channel is Tr B {U NGD σu N GD } = i = i i A σ i A υ i υ i E Tr{ i i A σ} υ i υ i E (22) and is indeed entanglement breaking. Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16

16 Simple example: qubit dephasing channel: N (σ) = (1 p)σ + p (σ), (σ) = 1 (σ + ZσZ). (23) 2 An isometric extension of it has the form U A BE N = 1 p 2 I 0 E + p 2 Z 1 E. (24) Its complementary channel is N c (σ) = p E + + ( 1 p 2 ) 1 1 E ( 1 p ) p 2 2 Tr{σZ}( 0 1 E E ), (25) and one observes that a bit flip on the input state does not change the eigenvalues of the resulting environment output state. Vlad Gheorghiu (CMU) Quantum Hadamard channels (I) July 8, / 16