# Equivalence of POVM and projective measurement

Suppose I have a POVM whose elements are given by $$\{M_i^\dagger M_i\}$$ such that $$\sum_i M_i^\dagger M_i = I_A$$. Let it act on some state $$\rho_A$$. Everything here happens in the Hilbert space $$A$$.

By Neumark's theorem, it is known that one can write this POVM as a PVM by using

1. An extension of the state to include an ancilla i.e. $$\rho_A\otimes \vert 0_B\rangle\langle 0_B\vert$$
2. A unitary operator $$U_{AB}$$
3. Projective measurement on the ancilla.

The unitary operator and the POVM elements are related in the following way

$$M_i = \langle 0_B\vert U_{AB}\vert i_B \rangle$$

How does one show that the unitarity of $$U$$ guarantees that $$\sum_i M^\dagger_i M_i = I_A$$? My attempt below is stuck at the first step and I'm not sure how to proceed.

$$\sum_i M^\dagger_i M_i = \sum_i \langle i_B\vert U^\dagger_{AB}\vert 0_B\rangle\langle 0_B\vert U_{AB}\vert i_B \rangle$$

There is a mistake (or rather: inconsistency) in how you define the $$M_i$$ (which makes the condition you want to prove incorrect, so there cannot be a proof!).
To be consistent, given the Kraus representation $$\rho\mapsto \sum M_i\rho M_i^\dagger$$, you need to define $$M_i = \langle i_B| U |0_B\rangle$$ (with $$U$$ the unitary in the Stinespring dilation and $$|i_B\rangle$$ the outcome of the projective measurement). In that case, the trace-preserving condition corresponds to $$\sum M_i^\dagger M_i = I\ .$$
This can then indeed be immediately proven from \begin{align} \sum_i M_i^\dagger M_i &= \sum_i \langle 0_B| U^\dagger |i_B\rangle \langle i_B| U |0_B\rangle \\ &= \langle 0_B| U^\dagger U |0_B\rangle \\ &= \langle 0_B| I_{AB} |0_B\rangle = I_A\ . \end{align}
Note that with the convention you chose above, the condition $$\sum M_i^\dagger M_i=I$$ (with my convention, $$\sum M_i M_i^\dagger = I$$) corresponds to a unital channel, and is thus not satisfied in general!
• Thank you for the answer. Just a typo in the last line where you say $I_B$ -- I assume you meant to type $I_A$? – user1936752 Aug 10 '19 at 15:37