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.

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


1 Answer 1


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!

  • $\begingroup$ Thank you for the answer. Just a typo in the last line where you say $I_B$ -- I assume you meant to type $I_A$? $\endgroup$ Commented Aug 10, 2019 at 15:37
  • $\begingroup$ @user1936752 Indeed, thanks. $\endgroup$ Commented Aug 10, 2019 at 15:38

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