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A standard trick in probability manipulation is to take some joint distribution $P_{XY}$ and express it as $P_{Y|X}P_X$. This trick is useful because when one looks at things like the ratio of $\frac{P_{XY}}{P_XP_Y}$, one can rewrite it as $\frac{P_{Y|X}}{P_Y}$ and this is now independent of $P_X$.

What is the quantum analogue of this, if any? If I have a bipartite state $\rho_{AB}$, can I express it in terms of $\rho_A$ and some channel such that the output is $\rho_{AB}$? It is unclear what the correct quantum form of this classical trick is.

It is also not clear if this trick works for cq states i.e. where the $A$ register is actually classical.

glS
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user1936752
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1 Answers1

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The classical conditional distribution of $Y$ given $X$ is defined as the joint distribution divided by the marginal distribution. Note that we can reconstruct the joint distribution if we know all the conditional distributions (they're indexed by the value of $X=x$, essentially) and the marginal.

It's reasonable to consider $\rho_{AB}$ as a quantum analogy of a joint distribution, with $\rho_{A}=\text{Tr}_B(\rho_{AB})$ being a marginal distribution.

I think a reasonable way to define a quantum conditional state is by $$ \rho_{B~|~A = \sigma} = \frac{\text{Tr}_A(\sigma\otimes I_B \cdot \rho_{AB})}{\text{Tr}(\sigma\rho_A)} $$ You can check that this is a state on the system $B$ for every $\sigma$ (which is a state on $A$).

It's also not hard to prove that we can reconstruct $\rho_{AB}$ if we know $\rho_A$ and $\rho_{B~|~A = \sigma}$ for every $\sigma$. Actually, we just need a set of $\sigma$, the linear span of which covers all linear operators on $A$.

By the way, if $\sigma=|\phi\rangle\langle\phi|$ is pure, then $\rho_{B~|~A = \sigma}$ is the reduced state on $B$ after a participant at $A$ observed a collapse to $\sigma$.

Danylo Y
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