Pure-state Entanglement d A d B | ψ � AB = � � c i , j | i � ⊗ | j � . i =1 j =1 Faithful squashed entanglement The SVD gives us orthonormal bases {| a i �} i , {| b i �} i and a prob. dist. { p i } such that √ p i | a i � ⊗ | b i � . | ψ � AB = � Aram Harrow i group meeting The reduced states are ψ A = ψ B = Apr 28, 2011 � � p i | a i �� a i | and p i | b i �� b i | . i i The entropy of entanglement of | ψ � is p i log 1 E ( ψ ) := S ( ψ A ) = − tr ψ A log ψ A = � . p i i An analogy Entropy of entanglement Let E ( ψ ) = S ( ψ A ). Then LOCC (local operations and classical communication) can on average only decrease E ( ψ ). Mixed-state entanglement E ( ψ 1 ⊗ ψ 2 ) = E ( ψ 1 ) + E ( ψ 2 ). E ( ψ ) > 0 if and only if ψ is entangled. Pandora’s Box The fact that pure-state Using LOCC we can asymptotically and approximately convert entanglement can be de- Hope ψ ⊗ n ↔ Φ ⊗ nE ( ψ ) scribed by the entropy of 2 entanglement. where | Φ 2 � = | 0 , 0 � + | 1 , 1 � . √ 2
Mixed-state measures of entanglement Squashed entanglement Entanglement cost: Introduced by Matthias Christandl and Andreas Winter in quant-ph/0308088. →≈ ρ ⊗ n for sufficiently large n } . E { Φ nE E c ( ρ ) := inf 2 Definition (squashed entanglement) � 1 � Distillable entanglement: 2 I ( A ; B | E ) σ : tr E σ ABE = ρ E sq ( ρ A : B ) = inf { ρ ⊗ n →≈ Φ nE D ( ρ ) := sup for sufficiently large n } . 2 E Definition (conditional mutual information) I ( A ; B | E ) = S ( A | E ) + S ( B | E ) − S ( AB | E ) Neither is known to be computable in finite time. = S ( AE ) + S ( BE ) − S ( ABE ) − S ( E ) Bound-entangled states exist with D ( ρ ) = 0 and E c ( ρ ) > 0. ≥ 0 (by strong subadditivity) Since 2005, we at least know that E c ( ρ ) = 0 iff ρ ∈ Sep: Why 1/2? Because if ρ = | ψ �� ψ | , then I ( A ; B ) ρ = S ( ψ A ) + S ( ψ B ) − S ( ψ AB ) = 2 S ( ψ A ). Sep = conv { α ⊗ β : α, β are density matrices } Squashed examples Properties of E sq Desirable Example: pure entangled state Additive: E sq ( ρ 1 ⊗ ρ 2 ) = E sq ( ρ 1 ) + E sq ( ρ 2 ) ρ 1 = | 00 � + | 11 � � 00 | + � 11 | √ √ Monogamous: E sq ( ρ A : B 1 B 2 ) ≥ E sq ( ρ A : B 1 ) + E sq ( ρ A : B 2 ) 2 2 Bounded: E sq ( ρ A : B ) ≤ log dim A . Any extension σ is of the form σ = ρ AB ⊗ ω E . 1 I ( A ; B | E ) = 0 implies “quantum Markov property”, which ∴ E sq ( ρ 1 ) = 1. implies membership in Sep. LOCC monotone, asymptotically continuous. Example: correlated state Less desirable ρ 2 = | 00 �� 00 | + | 11 �� 11 | Not known to be computable (optimization over extensions 2 may involve unbounded dimension). Choose σ = ( | 000 �� 000 | + | 111 �� 111 | ) / 2 to obtain E sq ( ρ 2 ) = 0. Not known to be faithful: may have E sq ( ρ ) = 0 for some (Note: σ = ( | 000 � + | 111 � )( � 000 | + | 111 � ) / 2 doesn’t work.) ρ �∈ Sep.
How faithful, exactly? Faithfulness and Monogamy go well together If d is a distance measure, then define Goal: optimize tr M ρ over ρ A : B ∈ Sep for M a LOCC measurement d ( ρ, Sep) := min { d ( ρ, σ ) : σ ∈ Sep } . Relaxation: Instead optimize over states ρ AB 1 that can be extended to ρ AB 1 ··· B k that is symmetric under permutation of B 1 , . . . , B k . Trace distance? � log dim A Everyone’s favorite distance measure is d tr ( ρ, σ ) = 1 2 � ρ − σ � 1 = Claim: This gives error O ( ). k maximum ρ -vs- Sep bias of any measurement . Proof: 1 | i , j � − | j , i � � i , j | − � j , i | � log dim A ≥ E sq ( ρ A : B 1 ··· B k ) √ √ boundedness ρ = � n � 2 2 2 1 ≤ i < j ≤ n k � E sq ( ρ A : B i ) ≥ monogamy has d ( ρ, Sep) ≥ 1 / 2, but E sq ( ρ ) ≤ const / n . [0910.4151] i =1 = k · E sq ( ρ A : B 1 ) symmetry LOCC distance! 1 4 ln 2 d LOCC ( ρ A : B 1 , Sep) 2 Define d LOCC ( ρ, σ ) to be the maximum ρ -vs- Sep bias of any LOCC ≥ k faithfulness measurement . Now [1010.1750] proves 1 const log 2 dim A � � 4 ln 2 d LOCC ( ρ, Sep) 2 . E sq ( ρ ) ≥ Note: optimization can be performed in time exp . ǫ 2 Additional definitions needed for proof Hypothesis testing Setting: We have n samples from classical distribution p or from q , 1 Relative entropy of entanglement: and want to accept p and reject q . E R ( ρ ) = min σ ∈ Sep S ( ρ � σ ) = min σ ∈ Sep tr ρ (log ρ − log σ ). The test: We choose a test that depends only on p , and that is guaranteed to accept p with probability ≥ 0 . 99. 2 Regularized relative entropy of entanglement: More concretely: If our samples are i 1 , . . . , i n and they have type E ∞ R ( ρ ) = lim n →∞ 1 n E R ( ρ ⊗ n ). t 1 , . . . , t d then we demand that t i ≈ np i . The rate function: The probability of accepting q ⊗ n is 3 Hypothesis testing: We are given ρ ⊗ n or an arbitrary separable � n � � d i =1 q np i ≈ ≈ exp( − nD ( p � q )). Thus the rate function np 1 ,..., np d i state. In the former case, we want to accept with probability is D ( p � q ). ≥ 1 / 2; in the latter with probability ≤ 2 − nD . Example: Chernoff bound: Pinsker’s inequality states that 4 Rate function for hypothesis testing: If M is a class of 1 2 ln 2 � p − q � 2 D ( p � q ) ≥ 1 . measurements (e.g. LOCC, LOCC → , ALL), then D M ( ρ ) is the largest D achievable above. Therefore, if we are sampling from q , the probability of observing a distribution p with � p − q � 1 = δ is exp( − const n δ 2 ).
Quantum Hypothesis testing Proof outline Quantum relative entropy: For states ρ, , σ define S ( ρ � σ ) := tr ρ (log ρ − log σ ) . 1 I ( A ; B | E ) ρ ≥ E ∞ R ( ρ A : BE ) − E ∞ R ( ρ A : E ) 2 E ∞ R ( ρ A : BE ) − E ∞ R ( ρ A : E ) ≥ D LOCC ← ( ρ A : B ) Still equals optimal rate function: For any ρ , we can design a test that accepts ρ ⊗ n with probability ≥ 0 . 99 and for any σ , accepts σ ⊗ n with probability � exp( − nS ( ρ � σ )). And this is optimal. 3 If M is a class of measurements that preserves Sep, then Distinguishing against convex sets: If K is a convex set, then our 1 D M ( ρ A : B ) ≥ M ∈M d M ( ρ, σ ) 2 2 ln 2 min σ ∈ Sep max rate function for distinguishing ρ against a collection of arbitrary σ ∈ S is σ ∈ K S ( ρ � σ ) . min Application to entanglement testing: E ∞ R ( ρ ) = D ALL ( ρ ). 1. Proof that 2. Proof that I ( A ; B | E ) ρ ≥ E ∞ R ( ρ A : BE ) − E ∞ R ( ρ A : E ) E ∞ R ( ρ A : BE ) ≥ E ∞ R ( ρ A : E ) + D LOCC ← ( ρ A : B ) Lemma [ H 3 O ]: E ∞ R is not lockable; i.e. tracing out Q qubits Lemma: E ∞ R ( ρ ) = D ALL ( ρ ). decreases E R by at most 2 Q . Lemma [State redistribution; Yard-Devetak]: B can be sent from A Proof: Construct an optimal unrestricted measurement M 1 to E using 1 2 I ( A ; B | E ) qubits (cf. Anup’s work). distinguishing ρ A : E from Sep and an optimal LOCC ← measurement M 2 distinguishing ρ A : B from Sep. Apply M 2 and then M 1 . By gentle measurement, we are still likely to accept ρ . And since M 2 Proof: Purify ρ ABE to get ψ ABEE ′ . Tracing out B means sending it doesn’t create entanglement between A : E , passing M 2 doesn’t to E ′ , which requires 1 2 I ( A ; B | E ′ ) qubits, which reduces E ∞ R by at increase the probability of passing M 1 . most I ( A ; B | E ′ ) = I ( A ; B | E ).
3. Proof that Implications D M ( ρ A : B ) ≥ 2 ln 2 min σ ∈ Sep max M ∈M d M ( ρ, σ ) 2 1 Let M be an LOCC measurement on k systems, each with n qubits. Estimating the optimal acceptance probability on product-state inputs is complete for QMA LOCC ( k ). This class n Von Neumann’s minimax theorem means that can be reduced to QMA n 2 k /ǫ 2 (1) by introducing error ǫ . The class QMA SEP δ := min σ ∈ Sep max M ∈M d M ( ρ, σ ) = max M ∈M min σ ∈ Sep d M ( ρ, σ ) √ n (2) (with constant error) contains 3-SAT. [1001.0017] If this result could be improved to apply to QMA LOCC Thus, there exists a separable M with √ n (2) or the simulation were improved to apply to SEP measurements, tr M ρ = p then we would have a matching bound. tr M σ ≤ p − δ ∀ σ ∈ Sep This problem is useful not only for simulating multiple unentangled Merlins, but for calculating tensor norms, such as Since M is separable, it preserves Sep. Thus, applying M n times will distinguish ρ ⊗ n from Sep with � � n � n � n � k =1 A i , j , k x i y j z k � i =1 j =1 � error ≤ exp( − const · n δ 2 ). � A � tensor := max � � � . � x � � y � � z � x , y , z ∈ C d � � � Details at MSR.
Recommend
More recommend
