The Mathematics of Entanglement - Summer 2013 29 May, 2013

Quantum de Finetti theorem

Lecturer: Aram Harrow Lecture 4

1 de Finetti

Let us remind ourselves that the quantum de Finetti theorem states that for all |ψ〉 ∈ Symn+k(Cd),then trn |ψ〉〈ψ| ≈

∫dµ(σ)σ⊗k.

The intuition here is that measuring the last n systems and finding that they are each in stateσ implies that the remaining k systems are also in state σ.

Let us now do the math. Recall∫dφ|phi〉〈phi|⊗m = ΠD,m/

(D +m− 1

D − 1

)then

trn|ψ〉〈ψ| = trnid⊗k ⊗ΠD,n|ψ〉〈ψ|id⊗k ⊗ΠD,n (1)

= trn

∫dφdφ′

(D + n− 1

D − 1

)(id⊗k ⊗ |φ〉〈φ|⊗n|ψ〉〈ψ|(id⊗k ⊗ |φ〉〈φ|⊗n)

(D + n− 1

D − 1

)(2)

=

∫ ∫dφdφ′|vφ〉〈vφ||〈φ||φ′〉|n (3)

≈∫dφ|vφ〉〈vφ| (4)

(5)

where we defined(D+n−1D−1

)(id⊗k ⊗ 〈φ|⊗n)|ψ〉 =: |vφ〉 and where in the approximation we assume

that n is large.If we had not inserted the projector on the left, in fact we obtain the equality

trn|ψ〉〈ψ| =∫dφ|vφ〉〈vφ| =

∫dφpφ|vφ〉〈vφ| (6)

We now claim that |vφ〉 ≈ |φ〉⊗k on average:

∫dφpφ|〈vφ||φ〉⊗k|2 =

∫dφ|〈vφ||φ〉⊗k|2 (7)

=

(D + n− 1

D − 1

)∫dφ|〈ψ|(id⊗k ⊗ |φ〉⊗n)|φ〉⊗k|2 (8)

=

(D + n− 1

D − 1

)tr(|ψ〉〈ψ|

∫dφ|φ〉〈φ|⊗n+k) (9)

=

(D + n− 1

D − 1

)/

(D + n+ k − 1

D − 1

)tr(|ψ〉〈ψ|Π) =

(D + n− 1

D − 1

)/

(D + n+ k − 1

D − 1

)(10)

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We can now get a good lower bound from this by expanding the binomial coefficients intofactorials:(D + n− 1

D − 1

)/

(D + n+ k − 1

D − 1

)=

(n+ 1) · · · (n+D − 1)

(n+ k + 1) · · · (n+ k +D − 1)≥ (1−k/(n+1))D−1 ≥ 1−kD/n.

Note that this bound is polynomial in n. This is tight. There exists, however, an improvementto an exponential dependence in n at the cost of replacing product states by almost product states.

In order to conclude the proof of the quantum de Finetti theorem, we need to relate the tracedistance to the average we computed.

For this, we consider the fidelity |〈α||β〉| between states |α〉 and |β〉. If now |〈α||β〉| = 1− ε andwe expand |β〉 =

√1− ε|α〉+

√ε|α〉

|||α〉〈α| − |β〉〈β|||1 = ||id− ....||1 = 2√ε

2 Quantum Key Distribution

A surprising application of entanglement is quantum key distribution. Suppose Alice and Bob sharean EPR pair |φ〉 = 1√

2|00 + 11〉, then the joint state of Alice Bob and a potential eavesdropper Eve

is |ψ〉ABE s.th. trE |ψ〉〈ψ|ABE = |φ〉〈φ|AB it follows that |ψ〉ABE = |φ〉AB ⊗ |γ〉EBy measuring in their standard basis, Alice and Bob thus obtain a secret random bit r. They

can use this bit to send a bit securely with help of the Vernam one-time pad cipher: Let’s callAlice’s message m. Alice sends the cipher c = m ⊕ r to Bob. Bob then recovers the message byadding r: c⊕ r = m⊕ r ⊕ r = m.

How can we establish shared entanglement between Alice and Bob? Alice could for instancecreate the state locally and send it to Bob using a quantum channel (i.e. a glas fibre).

But how can we now verify that the joint state that Alice and Bob have after the transmissionis an EPR state?

Protocol 1) Alice sends halves of n EPR pairs to Bob2) They choose randomly half of them and perform the CHSH tests. 3) They get key from theremaining halves

There are many technical details that I am glossing over here. One is, how can you be confidentthat the other halves are in this state? The de Finetti theorem! (the choice was permutationinvariant)

In order to make this applicable in actual implementations, one may use the exponential deFinetti theorem (Renner) or the post-selection technique (Christandl, Knig, Renner).

Other issues: there may be noise on the line. It is indeed possible to quantum key distributioneven in this case, but here one needs some other tools mainly relating to classical information theory(information reconciliation or privacy amplification).

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