Are the eigenvectors of the Choi-Jamiolkowski state maximally entangled?

Question

Let $\phi: M_n\rightarrow M_n$ be a quantum channel (completely positive trace preserving). Via the Choi-Jamiolkowski isomorphism we can transform this into a state

$$J(\phi) = (I_n\otimes\phi)(M) = \sum_{ij}E_{ij}\otimes\phi(E_{ij})$$ where $M$ denotes the maximally entangled state and $E_{ij}$ the matrix with a 1 at the $ij$ position and zero's everywhere else.

This state is positive definite if and only if $\phi$ is completely positive. This means that it has an eigenvalue decomposition: $$J(\phi) = \sum_i \lambda_i P_i$$ for some 1-dimensional projections $P_i\in M_n\otimes M_n$. These projections can be called maximally entangled when Tr$_1(P_i) = I_n$ and Tr$_2(P_i) = I_n$.

Can the $P_i$ be chosen such that they all are maximally entangled?

I know this is true when $\phi$ is a unitary conjugation and when $n=2$. Is it true in general?

I have checked it for the depolarizing maps, for which it is true. I don't know for any others. Are you implying that there is an easy counter-example? If so, could you point me towards it? — John, Nov 09 '16 at 15:31
Just posted an answer. In fact, I am right now thinking about posting a canonical "which (counter)examples for channels should I check" question/answer. A few come to mind: Depolarizing, dephasing, "discard + make a new state" (the example below), the Holevo-Werner channel, entanglement-breaking channels. If it's true for all of those, it's probably a theorem. — Norbert Schuch, Nov 09 '16 at 15:33
Such an answer would definitely be helpful for me at least, especially since the definitions of these channels are scattered all over the place. — John, Nov 09 '16 at 16:01
Related: What's the intuition behind the Choi-Jamiolkowski isomorphism? — Emilio Pisanty, Nov 09 '16 at 16:04
@NorbertSchuch That sounds like a particularly useful Q&A, please do write it. I'm looking for good questions to sink some rep into, if it'll help ;-). — Emilio Pisanty, Nov 09 '16 at 16:09
@EmilioPisanty Here you go: http://physics.stackexchange.com/questions/291810/canonical-examples-of-quantum-channels/ BTW, my first question here ;-) — Norbert Schuch, Nov 09 '16 at 18:58

score 6 · Accepted Answer · edited Apr 13 '17 at 12:40

6

No.

A simple counterexample is the qubit channel $$ \phi:\rho\mapsto \mathrm{tr}(\rho)|0\rangle\langle0|\ . $$ Its Choi state is $J(\phi)=\tfrac12\mathbb{I}\otimes |0\rangle\langle0|$, whose eigenvalue decomposition satisfies $\mathrm{tr}_1(P_i)=|0\rangle\langle0|$.

EDIT: I have now compiled a list of canonical examples to check.

edited Apr 13 '17 at 12:40

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answered Nov 09 '16 at 15:31

Norbert Schuch

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Is there also an easy example of a unital such map? I guess some entanglement-breaking channels would do right? – John Nov 09 '16 at 15:50
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Any unital channel that is not a convex combination of unitary conjugations will provide a unital counter-example. The Werner-Holevo channels for odd $d\geq 3$ are specific examples. – John Watrous Nov 11 '16 at 03:10

Are the eigenvectors of the Choi-Jamiolkowski state maximally entangled?

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