# How to find the Kraus operators of a channel corresponding to a POVM?

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.