# Can any rank-$n$ POVM be realized as a rank-one POVM?

Let, $$\mathcal{M}$$ is a POVM measurement whose elements are $$M_i=\sum_{k=1}^np_{ki}|\phi_{ki}\rangle\langle\phi_{ki}|$$

with $$p_{ki}\geq 0$$ and $$\sum_{i=1}^sM_i=I$$ where $$|\phi_{ki}\rangle$$ is a normalized basis.

So basically, $$\forall M_i \in \mathcal{M}$$ are rank-n positive operators.

Is it possible to design the POVM in such a way that the individual post measurement states entanglement of the rank one POVM are equal and correspond exactly with the outcomes of the original POVM?

• E. Davies, "Information and quantum measurement," in IEEE Transactions on Information Theory, vol. 24, no. 5, pp. 596-599, September 1978, doi: 10.1109/TIT.1978.1055941. Apr 23, 2022 at 11:11
• What does equivalent mean? Apr 23, 2022 at 11:47
• If you talk about post-measurement states, the POVM operators you describe are not the correct formalism (in particular, they lack information). Apr 24, 2022 at 21:01

Is this an exercise? The answer is yes, you can do it, but I'm not going to tell you what the equivalent rank-1 POVM is.

In general, any higher-rank POVM is equivalent to a rank-1 POVM if you relabel and remix the outcomes of the latter. Remix means assigning outcome $$a$$ or $$a'$$ with some probability $$p$$ to the former POVM when you get outcome $$b$$ in the latter.

This is explained in detail in this paper.

EDIT: To answer your edited question, no, it's not possible to simulate a rank-$$n$$ POVM via a rank-1 POVM such that the post-measurement states are the same. It's a bit difficult to answer this for POVMs directly, because a POVM doesn't determine what the post-measurement states are, but we can answer it by focussing on the particular case of PVMs. Consider the PVM $$\{|0\rangle\langle 0| + |1\rangle\langle 1|, |2\rangle \langle 2|\}.$$ The possible post-measurement states when measuring state $$\alpha|0\rangle + \beta|1\rangle + \gamma|2\rangle$$ are $$\frac1{\sqrt{|\alpha|^2+|\beta|^2}}(\alpha|0\rangle + \beta|1\rangle)$$ and $$|2\rangle$$. The crucial thing to notice is that coherence in the $$\{|0\rangle,|1\rangle\}$$ subspace is preserved. It's impossible to preserve this coherence with a rank-1 PVM. For example, one could implement the former PVM via the rank-1 PVM $$\{|0\rangle\langle 0|,|1\rangle\langle 1|, |2\rangle \langle 2|\},$$ but now the post-measurement states are $$|0\rangle, |1\rangle$$, and $$|2\rangle$$.

• Thank you. No, it's not a homework problem. I want to know post-measurement states will be the same? Both post-measurement states will have the same amount of entanglement? Apr 23, 2022 at 12:16
• Thanks. got it. Apr 25, 2022 at 5:58
• So essentially my answer is correct as well? Apr 26, 2022 at 7:11

I originally intended this to be a comment, but it got too long for a comment.

I think it depends on what you want to do with the POVM, if you are more interested in maximising the amount of information gained from measuring the system, then the optimal measurement may be regarded as a POVM comprising rank-one elements only.

However, if you want to treat the POVM as a quantum channel or you are interested in the post-measurement state of the system for each outcome of the POVM, I don't think there's a way to replace the POVM with another POVM of rank-one elements so that the two are equivalent. Take my answer with a pinch of salt though, let's wait to see what the experts have to say.

E. Davies, "Information and quantum measurement," in IEEE Transactions on Information Theory, vol. 24, no. 5, pp. 596-599, September 1978, doi: https://doi.org/10.1109/TIT.1978.1055941