I would like to know how one finds the Kraus operators of a channel corresponding to a POVM.

Consider a POVM of the form $M_i$ such that $\sum_i M_i = \mathbb{I}$. I can represent this by a quantum channel with Kraus operators $\{ A_i\}$ such that $A_i = \vert i\rangle \otimes B_i$ (up to unitaries on $B$). In order for this to be consistent, it must be that $B^\dagger_i B_i = M_i$

I guess (correct me if this is wrong) that every positive operator $M_i$ can be written in the form $B^\dagger_i B_i$ but how can I explicitly show this and compute $B_i$ given the $M_i$?


For one: the $B_i$ will never be unique, because changing them to $B_i'=U_i B_i$, for $U_i$ an arbitrary unitary, does not affect the square: $$ B_i'^\dagger B_i' = B_i^\dagger U_i^\dagger U_i B_i = B_i^\dagger B_i. $$ Moreover, the set of possible choices is huge, given how widely that $U_i$ can roam. (In fact, there's very little that requires that the $B_i$ be operators within the same Hilbert space - they can go elsewhere too! This is reflected in the fact that the $U_i$ do not need to be rectangular, and they only need to satisfy the left-inverse relationship $U_i^\dagger U_i = \mathbb I$. The other product, $U_i U_i^\dagger$, must be an orthogonal projector, but it does not need to have full rank.)

This means that giving a single recipe is a hopeless task, but you can nevertheless still provide one way to get a suitable $B_i$, as a proof of principle. (This won't cover all the solutions, nor does it necessarily work well in practice or get the same operator that you'll get with easier-in-practice methods. But none of that matters.)

To get a solution, simply start by realizing that since the $M = M^\dagger \geq 0$ are hermitian and positive-semidefinite (dropping the subindex $i$ for simplicity), they always have an orthonormal eigenbasis, $$ M = \sum_m m |m⟩⟨m|, $$ with $m\geq 0$ for all $m$. Then $$ B = \sum_{m\neq 0} \sqrt{m} |m⟩⟨m|, $$ often denoted $B = \sqrt{M}$, is well-defined, and it is a suitable solution to the problem.


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