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Can local projections increase entanglement?

Quantum Computing Asked on December 4, 2020

Consider a generic bipartite pure state $newcommand{ket}[1]{lvert #1rangle}ketPsiequiv sum_k sqrt{p_k}ket{u_k}otimesket{v_k}inmathcal Xotimesmathcal Y$,
where $p_kge0$ are the Schmidt coefficients, and ${ket{u_k}}_ksubsetmathcal X,{ket{v_k}}_ksubsetmathcal Y$ are orthonormal sets of states.

We known that, for any pair of unitary operations $U,V$, the state $(Uotimes V)ket{Psi}$ has the same amount of entanglement as $ketPsi$, as reflected by the invariance of the Schmidt coefficients under such operation.

Consider now a local projection operation. More precisely, suppose $mathcal X$ also has a bipartite structure, $mathcal X=mathcal X_1otimesmathcal X_2$, take some state $ketgammainmathcal X_1$, and consider the postselected state $ket{Psi’}equiv langle gammarvertPsirangle/|langle gammarvertPsirangle|inmathcal X_2otimesmathcal Y$.
If I were to describe this as an operation, I guess this would amount to applying some non-unitary linear operator $A$ to $ketPsi$.

Can the amount of entanglement of $ket{Psi’}$ in the "residual bipartition" $mathcal X_1otimesmathcal Y$ be larger than the initial entanglement in $ketPsi$?
If so, is there some kind of known characterisation of when this is possible?

Intuitively, this would mean that an initially low amount of entanglement can be "enhanced", or somehow "activated", conditionally to some observation (i.e. finding $ketgamma$) of one party. Such a situation seems strange to me but I’m not sure how to rule out the possibility.

One Answer

Of course this is possible. By LOCC you can probabilistically convert any (pure) state to a maximally entangled state with the same Schmidt rank, using a "filtering" POVM.

When this can be done, and at which optimal rate, is covered by the theory of majorization. This is covered very well in e.g. Nielsen and Chuang, or there is a review by Nielsen and Vidal.

(I realize you don't talk about POVMs but projections, but Stinespring/Naimark tells us that this is equivalent.)

Correct answer by Norbert Schuch on December 4, 2020

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