Entanglement vs Discord: Who Wins?

Transcription

1 Entanglement vs Dscord: Who Wns? Vlad Gheorghu Department of Physcs Carnege Mellon Unversty Pttsburgh, PA 15213, U.S.A. Januray 20, 2011 Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11

2 Outlne 1 Introducton to Quantum Dscord 2 Entanglement vs Dscord Reference: A. Brodutch and D. Terno, PRA 83, (Rapd) (2011) A summary of ths talk s avalable onlne at Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11

3 Introducton to Quantum Dscord Quantfyng (quantum) correlatons Mutual Informaton (Classcal): I (A : B) = H(A) + H(B) H(A, B) Yet another equvalent defnton (Classcal) I (A : B) = H(B) H(B A) How does one thnk operatonally about a condtonal entropy? H(B A) = a Pr(a)H(B A = a) = a Pr(a) b Pr(b a) log(pr(b a)) = a,b Pr(a, b) [log(pr(a, b)) log(pr(a))] = H(A, B) H(A) Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11

4 Introducton to Quantum Dscord Quantum doman: replace H by S and classcal random varables by quantum states! How to defne the quantum condtonal entropy? There are 2 ways: 1 A smple way (but whch lacks a nce operatonal nterpretaton, and can also be negatve!!!) S(B A) = S(A, B) S(A) 2 An operatonal way, whch nvolves measurements {Π a A } on Alce s sde: S(B Π A ) = a Pr(a)S(ρ a B), Pr(a)ρ a B = Tr A (Π a A I B ρ AB Π a A I B ) and s always postve. Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11

5 Introducton to Quantum Dscord They gve rse to 2 (nequvalent) ways of wrtng the quantum mutual nformaton: 1 S(A : B) = S(B) S(B A) = S(ρ A ) + S(ρ B ) S(ρ AB ), and 2 χ(a : B, Π A ) = S(B) S(B Π A ) = S(ρ B ) a Pr(a)S(ρa B ), wth a choce of {Π a A } n mnd. The second defnton s not symmetrc and depends on who s dong the measurement and also on the measurement operators! Quantum dscord (ntroduced by H. Ollver and W. H. Zurek n PRL 88, (2001)): D A (ρ AB, Π A ) = S(A : B) χ(a : B, Π A ). Sometmes an optmzaton s made over all possble measurement strateges D A (ρ AB ) = S(A : B) max {Π A } χ(a : B, Π A). Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11

6 Introducton to Quantum Dscord Propertes of Quantum Dscord In a sense, dscord removes the classcal correlatons from a bpartte quantum state. D A (ρ AB ) 0 D A (ρ AB ) = 0 f and only f there exsts an orthonormal bass { ψ k } on Alce s sde such that ρ AB = k p k ψ k ψ k A ρ k B, see A. Datta, arxv: [quant-ph]. Recently Vedral et al proved n PRL 105, (2010) that D A (ρ AB ) = 0 f and only f [L n, L m ] = 0, n, m, where ρ AB = n c nl n R n s a Schmdt operator form. Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11

7 Introducton to Quantum Dscord There are separable states that have non-zero quantum dscord! Example: ρ AB = 1 [ ]. Hence dscord s not an LOCC monotone! Zero dscord envronment states s a necessary and suffcent condton for CPTP evoluton! Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11

8 Entanglement vs Dscord Entanglement vs Dscord Defnton A blocal mplementaton G of a gate U on some fnte set of separable states L = {ρ n } N =1 (and ther convex combnatons) s a CPTP map that s mplemented by local operatons on the subsystems A and B, asssted by unlmted classcal communcaton such that for any state ρ n L G(ρ n ) = k K k ρ n K k Uρn U = ρ out The dual map G (ρ) := k K k ρk k satsfes for all pure nput states ρ n ρ out = G(ρ n ) = Uρ n 1 = ρ out G (ρ out ) = ρ n L. Why? U s pure, hence ρ out = G(ρ n ) ρ out = ρ n G (ρ out ) Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11

9 Entanglement vs Dscord If the set L s locally dstngushable, then the gate can be mplemented by LOCC (obvous). If the acton creates entanglement, then agan the mplementaton must fal! However, the absence of entanglement s not suffcent! If one restrcts the local operatons to projectve measurements and untares, then zero dscord becomes a necessary crteron for such mplementaton success. That s because local measurements on a state of non-zero dscord ncreases t s entropy, see PRA 81, (2010), whereas the gate U does not! Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11

10 Entanglement vs Dscord Result 1: If a set L contans one product state ( 00 ) and the maxmally mxed state (I I )/4, and the acton of U s realzed by LOCC, then all other allowed nputs (and ther arbtrary convex combnatons) must have zero dscord! Example: the tles Fgure: Bennett et al states Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11

11 Entanglement vs Dscord Result 2: If the set L contans two pure non-orthogonal states, and the untary operaton s such that D(ρ) D(UρU ), where ρ = p ψ 1 ψ 1 + (1 p) ψ 2 ψ 2, for some 0 < p < 1, then t cannot be mplemented on L by LOCC alone. Conclusons: The absence of entanglement n both nput and output does not automatcally enable a remote mplementaton by LOCC. A dscrepancy between local and global nformaton content of separable states (whch s captured by the dscord) requres entanglement for ther processng. Entanglement s requred for any gate whch changes the dscord of the states! Recent results (arxv: , arxv: ) suggest that a change n dscord rather than entanglement s the requred resource n computatonal speedup. Vlad Gheorghu (CMU) Entanglement vs Dscord: Who Wns? Januray 20, / 11