2
$\begingroup$

We can represent a CP map $\mathcal{M}$ by a positive semidefinite matrix via Choi-Jamiolkowsky (CJ) isomorphism. The CJ matrix $M$ ∈ $L(H_1 \otimes H_2)$ corresponding to a linear map $\mathcal{M}$ : $L(H_1) \rightarrow L(H_2)$ is defined as $$M := I \otimes \mathcal{M} (|\psi\rangle\langle\psi|)$$ where $|\psi\rangle = \sum_{j=1}^{d}|jj\rangle ∈ H_1 ⊗ H_1$ is a maximally entangled state (not normalized), the set of states ${|j\rangle}_{j=1}^{d}$ is an orthonormal basis of $H_1$ and $I$ is the identity map.

In the appendix part of the paper 'Quantum correlations with no causal order', while calculating the probabilities for violation of the causal inequality the paper admits the following:

A CP map corresponding to the detection of a state $|\psi\rangle$ and repreparation of another state $|\phi\rangle$ has CJ matrix of the form: $|\psi\rangle\langle\psi| \ \otimes |\phi\rangle\langle\phi|$

I am sure this statement is general. But I am not able to see how do we get this form from the definition. Could someone provide some intuition?

$\endgroup$
  • $\begingroup$ Have you tried the derivation? Where have you failed? $\endgroup$ – Norbert Schuch Jul 27 '18 at 10:49
2
$\begingroup$

Since you asked for intuition, let me give you my intuition: The first part of the Choi state tells you what happens to the input, the second part what will be the output.

While this might be a bit vague in the scenario of an arbitrary channel, in the case of this specific channel it is fully precise: The first component is $|\psi\rangle\langle\psi|$ -- it describes that the input is projected onto the state $|\psi\rangle$ (or possibly $|\psi^*\rangle$, this would require more careful analysis). The second component is $|\phi\rangle\langle\phi|$ -- it describes the output which is the state $|\phi\rangle$.

This corresponds to the channel $\mathcal M(\rho) = \langle\psi|\rho|\psi\rangle\, |\phi\rangle\langle\phi|$.

$\endgroup$
  • $\begingroup$ So, the measurement is equivalent to projecting onto the state $|\psi\rangle$. 1) In this particular context, can we think of repreparation as projecting the measured state $|\psi\rangle$ onto the state $|\phi\rangle$? 2) In general, can we think of repreparation as projecting a general density matrix $\rho$ onto some state? $\endgroup$ – exp ikx Jul 29 '18 at 14:22
  • $\begingroup$ (1) No. (2) No. The reason is that this will have a success probability <1, while the given CP map has success rate 1. Preparation is just preparation. Whether you put "re-" in front of it doesn't really matter. Just think of a photon: You just prepare a different photon, and no-one will be able to tell the difference. $\endgroup$ – Norbert Schuch Jul 29 '18 at 15:14
  • $\begingroup$ To make this more precise: The "preparation" part has success probability 1. The projection part doesn't. Note that this means that this channel is not physical in the sense that it can only be implemented probabilistically. $\endgroup$ – Norbert Schuch Jul 29 '18 at 18:12

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.