Pure-state Entanglement d A d B | AB = c i , j | i | j . i - - PowerPoint PPT Presentation

Faithful squashed entanglement

Aram Harrow

Apr 28, 2011

Pure-state Entanglement

|ψAB =

• i=1

• j=1

ci,j |i ⊗ |j . The SVD gives us orthonormal bases {|ai}i, {|bi}i and a prob.

• dist. {pi} such that

|ψAB =

• i

√pi |ai ⊗ |bi . The reduced states are ψA =

• i

pi|aiai| and ψB =

• i

pi|bibi|. The entropy of entanglement of |ψ is E(ψ) := S(ψA) = − tr ψA log ψA =

• i

pi log 1 pi .

An analogy

Mixed-state entanglement Pandora’s Box The fact that pure-state entanglement can be de- scribed by the entropy of entanglement. Hope

Entropy of entanglement

Let E(ψ) = S(ψA). Then LOCC (local operations and classical communication) can on average only decrease E(ψ). E(ψ1 ⊗ ψ2) = E(ψ1) + E(ψ2). E(ψ) > 0 if and only if ψ is entangled. Using LOCC we can asymptotically and approximately convert ψ⊗n ↔ Φ⊗nE(ψ)

where |Φ2 = |0,0+|1,1

.

Mixed-state measures of entanglement

Entanglement cost: Ec(ρ) := inf

→≈ ρ⊗n for sufficiently large n}. Distillable entanglement: D(ρ) := sup

{ρ⊗n →≈ ΦnE

for sufficiently large n}. Neither is known to be computable in finite time. Bound-entangled states exist with D(ρ) = 0 and Ec(ρ) > 0. Since 2005, we at least know that Ec(ρ) = 0 iff ρ ∈ Sep: Sep = conv {α ⊗ β : α, β are density matrices}

Squashed entanglement

Definition (squashed entanglement) Esq(ρA:B) = inf 1 2I(A; B|E)σ : trE σABE = ρ

• Definition (conditional mutual information)

I(A; B|E) = S(A|E) + S(B|E) − S(AB|E) = S(AE) + S(BE) − S(ABE) − S(E) ≥ 0 (by strong subadditivity) Why 1/2? Because if ρ = |ψψ|, then I(A; B)ρ = S(ψA) + S(ψB) − S(ψAB) = 2S(ψA).

Squashed examples

Example: pure entangled state ρ1 = |00 + |11 √ 2 00| + 11| √ 2 Any extension σ is of the form σ = ρAB

⊗ ωE. ∴ Esq(ρ1) = 1. Example: correlated state ρ2 = |0000| + |1111| 2 Choose σ = (|000000| + |111111|)/2 to obtain Esq(ρ2) = 0. (Note: σ = (|000 + |111)(000| + |111)/2 doesn’t work.)

Properties of Esq

Desirable Additive: Esq(ρ1 ⊗ ρ2) = Esq(ρ1) + Esq(ρ2) Monogamous: Esq(ρA:B1B2) ≥ Esq(ρA:B1) + Esq(ρA:B2) Bounded: Esq(ρA:B) ≤ log dim A. I(A; B|E) = 0 implies “quantum Markov property”, which implies membership in Sep. LOCC monotone, asymptotically continuous. Less desirable Not known to be computable (optimization over extensions may involve unbounded dimension). Not known to be faithful: may have Esq(ρ) = 0 for some ρ ∈ Sep.

How faithful, exactly?

If d is a distance measure, then define d(ρ, Sep) := min{d(ρ, σ) : σ ∈ Sep}. Trace distance? Everyone’s favorite distance measure is dtr(ρ, σ) = 1

maximum ρ-vs-Sep bias of any measurement . ρ = 1 n

• 1≤i<j≤n

|i, j − |j, i √ 2 i, j| − j, i| √ 2 has d(ρ, Sep) ≥ 1/2, but Esq(ρ) ≤ const/n. [0910.4151] LOCC distance! Define dLOCC(ρ, σ) to be the maximum ρ-vs-Sep bias of any LOCC measurement . Now [1010.1750] proves Esq(ρ) ≥ 1 4 ln 2dLOCC(ρ, Sep)2.

Faithfulness and Monogamy go well together

Goal: optimize tr Mρ over ρA:B ∈ Sep for M a LOCC measurement Relaxation: Instead optimize over states ρAB1 that can be extended to ρAB1···Bk that is symmetric under permutation of B1, . . . , Bk. Claim: This gives error O(

• log dim A

). Proof: log dim A ≥ Esq(ρA:B1···Bk) boundedness ≥

• i=1

Esq(ρA:Bi) monogamy = k · Esq(ρA:B1) symmetry ≥ k 1 4 ln 2dLOCC(ρA:B1, Sep)2 faithfulness Note: optimization can be performed in time exp

• const log2 dim A

• .

Additional definitions needed for proof

ER(ρ) = minσ∈Sep S(ρσ) = minσ∈Sep tr ρ(log ρ − log σ).

E ∞

• state. In the former case, we want to accept with probability

≥ 1/2; in the latter with probability ≤ 2−nD.

measurements (e.g. LOCC, LOCC→, ALL), then DM(ρ) is the largest D achievable above.

Hypothesis testing

Setting: We have n samples from classical distribution p or from q, and want to accept p and reject q. The test: We choose a test that depends only on p, and that is guaranteed to accept p with probability ≥ 0.99. More concretely: If our samples are i1, . . . , in and they have type t1, . . . , td then we demand that ti ≈ npi. The rate function: The probability of accepting q⊗n is ≈

• n

d

≈ exp(−nD(pq)). Thus the rate function is D(pq). Example: Chernoff bound: Pinsker’s inequality states that D(pq) ≥ 1 2 ln 2p − q2

Therefore, if we are sampling from q, the probability of observing a distribution p with p − q1 = δ is exp(−const nδ2).

Quantum Hypothesis testing

Quantum relative entropy: For states ρ,, σ define S(ρσ) := tr ρ(log ρ − log σ). 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(ρσ)). And this is optimal. Distinguishing against convex sets: If K is a convex set, then our rate function for distinguishing ρ against a collection of arbitrary σ ∈ S is min

Application to entanglement testing: E ∞

Proof outline

DM(ρA:B) ≥ 1 2 ln 2 min

• 1. Proof that

I(A; B|E)ρ ≥ E ∞

decreases ER by at most 2Q. Lemma [State redistribution; Yard-Devetak]: B can be sent from A to E using 1

Proof: Purify ρABE to get ψABEE ′. Tracing out B means sending it to E ′, which requires 1

most I(A; B|E ′) = I(A; B|E).

• 2. Proof that

E ∞

Proof: Construct an optimal unrestricted measurement M1 distinguishing ρA:E from Sep and an optimal LOCC← measurement M2 distinguishing ρA:B from Sep. Apply M2 and then M1. By gentle measurement, we are still likely to accept ρ. And since M2 doesn’t create entanglement between A : E, passing M2 doesn’t increase the probability of passing M1.

• 3. Proof that

DM(ρA:B) ≥

Thus, there exists a separable M with tr Mρ = p tr Mσ ≤ p − δ ∀σ ∈ Sep Since M is separable, it preserves Sep. Thus, applying M n times will distinguish ρ⊗n from Sep with error ≤ exp(−const · nδ2).

Implications

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 QMALOCC

(k). This class can be reduced to QMAn2k/ǫ2(1) by introducing error ǫ. The class QMASEP

[1001.0017] If this result could be improved to apply to QMALOCC

(2) or the simulation were improved to apply to SEP measurements, then we would have a matching bound. This problem is useful not only for simulating multiple unentangled Merlins, but for calculating tensor norms, such as Atensor := max

• n

n

n

x y z

• .

Details at MSR.