Suppose one has a classical channel $W(y|x)$ that is a conditional probability distribution. Can one define a Choi state for this channel?

My guess is that one should think of it as a special case of a quantum channel where we measure the state in the computational basis and then given that the outcome of the measurement is $x\in\mathcal{X}$, we obtain $y\in\mathcal{Y}$ with probability $W(y|x)$. The Choi state itself is therefore some kind of correlated classical distribution? Any pointers with understanding this would be very helpful!


1 Answer 1


I'd say the closest thing to a "classical channel" corresponding to a conditional probability distribution $p(y|x)$ is a quantum channel of the form $$\Phi(\rho) = \sum_{a,b} \langle a|\rho|a\rangle p(b|a) \, |b\rangle\!\langle b| \equiv \sum_a \langle \Pi_a,\rho\rangle \sigma_a, \quad \sigma_a\equiv\sum_b p(b|a) \Pi_b,$$ where $\Pi_a\equiv |a\rangle\!\langle a|$. This is the channel that, conditionally to the input state being in the $a$-th state (when measured in an appropriate basis), gives the outcome $b$ (again, wrt some choice of measurement basis) with probability $p(b|a)$. You can notice this is a special case of an entanglement-breaking channel. As mentioned e.g. in this answer, these have Choi equal to $$J(\Phi) = \sum_a\sigma_a\otimes \Pi_a^T.$$

Just as an addendum, one effectively considers similar channels e.g. when discussing Holevo's bound. In there, one is interested in figuring out how much of the (classical) information encoded in a bunch of quantum states $\sigma_a$ can be later recovered through some measurement $\mu_b$ on said states. One thus wonders about the mutual information of the state $$\sum_{a,b} \langle \mu_b,\sigma_a\rangle\, ( \Pi_a\otimes \Pi_b).$$ Notice the similarity with the $\Phi$ in the first equation.

  • $\begingroup$ Just to clarify, $\Pi_a^T$ is the same as $\Pi_a$ isn't it? Or is the $T$ there explicitly for some other reason? Secondly, your answer suggests that $J(\Phi)$ is basically a quantum state with classical correlations. Can one then propose that $J(W)$ is simply some bipartite probability distribution with those correlations? Thank you for your answer :) $\endgroup$
    – EVMaverick
    Commented Apr 10, 2022 at 14:27
  • 1
    $\begingroup$ @EVMaverick it depends on the representation. For computational basis states, yes, you can assume $\Pi_a^T=\Pi_a$, but for more general bases no. The point is that in general $|v\rangle\!\langle v|$ is hermitian, but not necessarily symmetric. Regarding the second point, it depends on what you mean with "bipartite distribution with those correlations". But I'd say yes, measuring the state corresponding to $J(\Phi)$ in the computational bases for each system gives correlated results compatible with $p(b|a)$ (with an additional balanced distribution for the marginal $p(a)$) $\endgroup$
    – glS
    Commented Apr 10, 2022 at 14:38
  • $\begingroup$ Thank you for the clarifications! $\endgroup$
    – EVMaverick
    Commented Apr 10, 2022 at 14:46

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