# Moving between $\sum_{I}E_{i}=I$ and $\sum_{i}M^{\dagger}M=I$ for non-hermitian $M$

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.

• 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 $U$. May 20 at 8:23

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 $$M_i^\dagger M_i = \sqrt{E_i}U^\dagger U \sqrt{E_i} = \sqrt{E_i}\sqrt{E_i} = E_i$$.
• 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 $M_{i}=\sqrt{E_{i}}$? May 20 at 10:40
• Yep, because then the relationship reduces to $E_i = M_i^2$. Just keep in mind that the square root is not unique. May 20 at 11:16