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If $\sum_{i}E_{i}=I$ is a set of POVM's and $\sum_{i}M^{\dagger}M=I$ is a set of general measurement operators, I have always been confused on how to move from one to the other, in regards to the expression of $M=\sqrt{E_{i}}$

If $E_{i}$ is a projector, then $\langle\Psi|E_{i}\Psi\rangle=\langle\Psi|\sqrt{E_{i}}\sqrt{E_{i}}\Psi\rangle=\langle\Psi|M^{\dagger}M\Psi\rangle=\langle\Psi|P^{\dagger}P\Psi\rangle=\langle\Psi|P\Psi\rangle$, due to the properties of the projector.

If $E_{i}$ is not a projector, then $\langle\Psi|E_{i}\Psi\rangle=\langle\Psi|\sqrt{E_{i}}\sqrt{E_{i}}\Psi\rangle=\langle\Psi|M^{\dagger}M\Psi\rangle$

But how can the second equality aboce hold if $M$ isn't hermitian, such as $M_{0}=|0\rangle\langle0|, M_{1}=|0\rangle\langle1|$? In this case, $\sum_{i}M^{\dagger}M=I$ still holds, and this is used as an example in N&C as a set of measurement operators. However, both N&C and other books frequently use the notation of $E_{i}=\sqrt{E_{i}}\sqrt{E_{i}}$. But it's clear from the above example that $M_{1} \neq M_{1}^{\dagger}$, as it's not hermitian.

If $M_{1}=\sqrt{E_{i}}$, but it isn't hermitian, then $M_{1}^{\dagger} \neq \sqrt{E_{1}}$ and so the product of them $\neq E_{1}$

I feel like I am missing something obvious here, but apart from it being related to the non-uniqueness of sqaure roots for matrices, I am not sure what.

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    $\begingroup$ Indeed, the square root is ambiguous here ... All possible square roots have the same singular values and the right eigenvectors are unique (namely EV of $E$). Mateus' comment gives the correct solution: The unitary $U$ corresponds to a change of left eigenvectors. To see this, consider the diagonalisation of the psd operator $E = V D V^\dagger$. $D$ is psd, so $\Sigma=\sqrt{D}$ is diagonal and psd. Now we can write $E = V \Sigma \Sigma^\dagger V^\dagger = (V \Sigma U^\dagger) (U \Sigma^\dagger V^\dagger)$ which gives us a SVD of a square root for any unitary $U$. $\endgroup$ May 20 at 8:23
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Indeed, $M_i=\sqrt{E_{i}}$ is wrong. The correct relationship is $E_i = M_i^\dagger M_i$. The possible $M_i$ for a given $E_i$ are $M_i = U\sqrt{E_i}$ for any unitary $U$, as $M_i^\dagger M_i = \sqrt{E_i}U^\dagger U \sqrt{E_i} = \sqrt{E_i}\sqrt{E_i} = E_i$.

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  • $\begingroup$ Fantastic. Am I to assume that, in the case of $M_{i}$ being hermitian, then this expression would simply translate into $U = I$, and then each $M_{i}=\sqrt{E_{i}}$? $\endgroup$ May 20 at 10:40
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    $\begingroup$ Yep, because then the relationship reduces to $E_i = M_i^2$. Just keep in mind that the square root is not unique. $\endgroup$ May 20 at 11:16

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