2
$\begingroup$

The POVM element $E_{i}$ is associated with the measurement outcome $i$, such that the probability of obtaining it when making a measurement on the quantum state $\rho$ is given by: $p(i)=tr(\rho E_i)$;

So what does the outcome $i$ refer to? It looks like it's just an index symbol and $\sum_{i=1}^n E_i=I$.

$\endgroup$

1 Answer 1

1
$\begingroup$

The outcome $i$ is just an index/label. After a POVM measurement corresponding to a set say $$\mathcal{M} = \{ E_i \}\,,$$ where $|\mathcal{M}|=i$, the reading on your measurement device could be any of the $i$ different outcomes corresponding to this set $\mathcal{M}$.

Say you perform POVM measurement on some initial state $|\psi\rangle$. Then, after the POVM measurement, you get one of the $i$ possible readings on your measurement device, with the probability of getting this $i^{\text{th}}$ outcome being $p_i$. Where,

$$p_i = \langle \psi | E_i |\psi\rangle\,.$$

$\endgroup$
5
  • 2
    $\begingroup$ Careful! Your statement would be correct for any measurement operators $\{M_i\}$ such that $\sum_iM_i^\dagger M_i=I$. However, POVMs are not that. They do not have final states associated with the outcomes. One way that you can see this is via the non-uniqueness of $\sqrt{E_i}$: for any $M_i$ such that $E_i=M_i^\dagger M_i$, you can also have $\tilde M_i=UM_i$ for any unitary $U$. $\endgroup$
    – DaftWullie
    Oct 30, 2023 at 7:33
  • 1
    $\begingroup$ @DaftWullie Yes I agree. We can write $M_i$ as $$M_i = U_i \sqrt{E_i}$$ upto an arbitrary choice of $U_i$. So, if we only have $\{E_i\}$ specified, then we are free to choose $U_i$ and for every such choice, we would get a different set of $\{M_i\}$, right? and in that case, the post-measurement state would what I have specified above. Is this the correct line of reasoning? $\endgroup$
    – FDGod
    Oct 30, 2023 at 9:18
  • 1
    $\begingroup$ @DaftWullie But yes, I agree that practically, since this $U_i$ is just an arbitrary choice, the post-measurement state is not uniquely defined; it's upto an arbitrary unitary transformation, so for us in a measurement setup, the post-measurement state after a POVM is unspecified. Is that correct? And also, thanks for pointing out my mistake. Appreciate it! $\endgroup$
    – FDGod
    Oct 30, 2023 at 9:22
  • 2
    $\begingroup$ Well, my more fundamental point is that if you are talking about POVMs, you should not be talking about output states. POVMs are used precisely in the context of destructive measurements when there is no output state. $\endgroup$
    – DaftWullie
    Oct 30, 2023 at 9:26
  • 1
    $\begingroup$ @DaftWullie I see; I did not know that! Thank you! I have updated my answer accordingly. $\endgroup$
    – FDGod
    Oct 30, 2023 at 9:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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