The Mathematics of Entanglement - Summer 2013 29 May, 2013

Separable states, PPT and Bell inequalities

Lecturer: Fernando G.S.L. Brandao Lecture 7

Recall from yesterday this theorem.

Theorem 1. For any pure state |AB,

Ec(||AB) = ED(||AB) = S(A) = S(B),

where S() = tr log .

As a result, many copies of a pure entangled state can be (approximately) reversibly transformedinto EPR pairs and back again. Up to a small approximation error and inefficiency, we have|nAB LOCC|

+nS(A).

1 Mixed-state entanglement

For pure states, a state is entangled if its not a product state. This is easy to check, and we caneven quantify the amount of entanglement (using Theorem 1) by looking at the entropy of one ofthe reduced density matrices.

But what about for mixed states? Here the situation is more complicated.

Definition 1. Define the set of separable states Sep to be the set of all AB that can be written asi

pi|ii|A |ii|B. (1)

Definition 2. A state is entangled if it is not separable.

We should check that this notion of entanglement makes sense in terms of LOCC. And indeed,separable states can be created using LOCC: Alice samples i according to p, creates |i and sendsi to Bob, who uses it to create |i. On the other hand, entangled states cannot be created from aseparable state by using LOCC. In other words, the set Sep is closed under LOCC.

2 The PPT test

It is in general hard to test, given a state AB, whether Sep. Naively we would have to checkfor all possible decompositions of the form in (1). So it is desirable to find efficient tests that workat least some of the time.

One such test is the Positive Partial Transpose, or PPT, test. The partial transpose can bethought of as (T id), where T is the transpose map. More concretely, if

XAB =i,j,k,l

ci,j,k,l|ij|A |kl|B

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then the partial transpose is

XTAAB =i,j,k,l

ci,j,k,l(|ij|A)T |kl|B

=i,j,k,l

ci,j,k,l|ji|A |kl|B

The PPT test asks whether TA is positive semidefinite. If so, we say that is PPT.Observe that all separable states are PPT. This is because if =

i pi|ii|A |ii|B,

thenTA =

i

pi|ii|A |ii|.

This is still a valid density matrix and in particular is positive semidefinite (indeed, it is also inSep).

Thus, Sep implies PPT. The contrapositive is that 6 PPT implies 6 Sep. Thisgives us an efficient test that will detect entanglement in some cases.

Are there in fact any states that are not in PPT? Otherwise this would not be a very interestingtest.

Examples

1. |+AB = |0,0+|1,12 . Then

|++|TAAB =1

2(|0, 00, 0|+ |1, 00, 1|+ |0, 11, 0|+ |1, 11, 1|)

=

1/2 0 0 00 0 1/2 00 1/2 0 00 0 0 1/2

= 12SWAP.This has eigenvalues (1/2, 1/2, 1/2,1/2), meaning that |++|AB 6 PPT. Of course, wealready knew that |+ was entangled.

2. Lets try an example where we dont already know the answer, like a noisy version of |+.Let

= p|++|+ (1 p)I4.

Then one can calculate min(TA) = p2 +

1p4 which is < 0 if and only if p > 1/3.

Maybe PPT = Sep? Unfortunately not. In D(C2 C3) (i.e. density matrices in which onesystem has 2 dimensions and the other has 3) then all PPT states are separable. But for largersystems, e.g. 3x3 or 2x4, then there exist PPT states that are not separable.

2.1 Bound entanglement

Theorem 2. If AB PPT then ED() = 0.

To prove this we will establish two properties of the set PPT:

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1. PPT is closed under LOCC.Consider a general LOCC protocol. This can be thought of as Alice and Bob alternating gen-eral measurements and sending each other the outcomes. When Alice makes a measurement,this transformation is

AB 7(M I)AB(M I)tr((M M I)AB)

.

After Bob makes a measurement as well, depending on the outcome, the state is proportionalto

(M N)AB(M N ),and so on. The class SLOCC (stochastic LOCC) consists of outcomes that can be obtainedwith some positive probability, and we will see later that this can be characterized in termsof (M N)AB(M N ).We claim that if AB PPT then (M N)AB(M N ) PPT. Indeed

((M N)AB(M N ))TA = (MT N)TAAB(MT N).

Now TAAB 0 and XYX 0 whenever Y 0, implying that ((MN)AB(M N ))TA 0.

2. PPT is closed under tensor product. If AB, AB PPT, then (AB AB) PPT. Why?Because

(AB AB)TAA = TAAB TAAB 0.

Proof of Theorem 2. Assume towards a contradiction that PPT and ED() > 0. Then for any > 0 there exists n such that nAB can be transformed to |+ using LOCC up to error . Since PPT, n is also PPT and so is the output of the protocol, which we call . Then TA 0 and |++|1 . If we had = 0, then this would be a contradiction, because is in PPTand |++| is not. We can use an argument based on continuity (details omitted) to show thata contradiction must appear even for some sufficiently small > 0.

If is entangled but ED() = 0, then we say that has bound entanglement meaning thatit is entangled, but no pure entanglement can be extracted from it. By Theorem 2, we know thatany state in PPT but not Sep must be bound entangled.

Open question: A major open question (the NPT bound entanglement question) is whetherthere exist bound entangled states that have a non-positive partial transpose.

3 Entanglement witnesses

Sep is a convex set, meaning that if , Sep and 0 1 then + (1 ) Sep. Thus theseparating hyperplane theorem implies that for any 6 Sep, there exists a Hermitian matrix Wsuch that

1. For all Sep, tr(W) 0

2. tr(W) < 0.

Example: consider the state = |++|. Let W = I 2|++|. As an exercise, show thattr(W) 0 for all Sep. We can also check that tr(W) = 1.

Observe that an entanglement witness W needs to be chosen with a specific in mind. An anexercise, show that no W can be a witness for all entangled states of a particular dimension.

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4 CHSH game

One very famous type of entanglement witness is called a Bell inequality. In fact, these bounds ruleout not only separable states but even classically correlated distributions over states that couldbe from a theory more general than quantum mechanics. Historically, Bell inequalities have beenimportant in showing that entanglement is an inescapable, and experimentally testable, part ofquantum mechanics.

The game is played by two players, Alice and Bob, together with a Referee. The Referee choosebits r, s at random and sends r to Alice and s to Bob. Alice then sends a bit a back to the Refereeand Bob sends the bit b to the Referee.

A B

R

a

r

b

s

Alice and Bob win if a b = r s, i.e. they want a b to be chosen according to this table:

r s desired a b0 0 00 1 01 0 01 1 1

One can show that if Alice and Bob use a deterministic strategy, their success probability willbe 3/4. However, using entanglement they can achieve a success probability of cos2(/8) 0.854 . . . > 3/4. This strategy, together with the payoff function (+1 if they win, -1 if they lose),yields an entanglement witness, and one that can be implemented only with local measurements.

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Mixed-state entanglementThe PPT testBound entanglement

Entanglement witnessesCHSH game