# Tag Info

1 vote

### What does distinguishability mean in this case?

In this context the idea is that Alice wants to send classical information, by encoding it in quantum states. In other words, Alice encodes her choice of bit to send $i$ into the state $\rho_i$, and ...
• 18.1k
1 vote

### What does distinguishability mean in this case?

Imagine they are almost the same, i.e. $\rho_0 \approx \rho_1$. It's almost impossible to distinguish them, yet they are different. So that, your success probability is $\approx 1/2$. Also, in the ...
• 5,423
1 vote

### What is the relation between POVMs and observables (as Hermitian operators)?

In various respects, the POVM formalism is much more fundamental than Hermitian observables. The notion of generalized observables was developed precisely to weaken and generalize the notion of ...
Accepted

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

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 ...
• 1,593

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

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 ...
1 vote
Accepted

### Characterise, via Naimark's theorem, the POVM corresponding to a PVM in a dilated space

You could define $$V_F^S u \equiv \sum_a |a\rangle\otimes (S^a\sqrt{F^a}\,u), \qquad u \in\mathcal X,$$ where $S^a$ is any unitary. Then $V_F^S$ is also an isometry and  F^a=(V_F^S)^\dagger (\...
• 5,423
You can assume without loss of generality that the POVM's are rank one because $\sum_i A_i=I$, so it's not necessary just more convenient. The enlargement of the space in the Naimark dilation theorem ...
Any POVM can be interpreted as a projective measurement in a higher-dimensional space. Derivation of the dilated representation More precisely, let $\{\mu(a)\}_a\subset\operatorname{Pos}(\mathcal X)$ ...