1
$\begingroup$

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.

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

$\endgroup$
0
2
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ Thank you for the answer. Just a typo in the last line where you say $I_B$ -- I assume you meant to type $I_A$? $\endgroup$ – user1936752 Aug 10 '19 at 15:37
  • $\begingroup$ @user1936752 Indeed, thanks. $\endgroup$ – Norbert Schuch Aug 10 '19 at 15:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.