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I know that a general quantum measurement is described by a set of measurement operators $\{M_j\}$ satisfying the completeness relation $\sum_j M_j^\dagger M_j = 1$ and that the outcome j occurs with probability $$p(j)=Tr(M_j^\dagger M_j \rho)$$ Furthermore, the post-measurement state after obtaining this outcome is $$\rho_j = \frac{M_j \rho M_j^\dagger}{p(j)}$$ However, in the POVM formalism we describe the measurment using only the POVM elements $\{A_j\}$ that correspond to the operator products $M_J^\dagger M_j$. Obviously, there are different sets of measurement operators that realize the same POVM.

Here's my question: Given (1) the POVM elements of a quantum measurement and (2) the post-measurement states $\{\sigma_j\}$ for some fixed pre-measurement state $\sigma$, are there measurement operators $\{M_j\}$ such that $$A_j = M_j^\dagger M_j$$ and $$\sigma_j = \frac{M_j \sigma M_j^\dagger}{Tr(A_j \sigma )}$$

Moreover: If there is such a measurement, is it unique?

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  • $\begingroup$ I'm not sure I understand this question. The post-measurement states are typically functions of the pre-measurement state. Are you trying to define an operation similar to quantum measurement on a fixed pre-measurement state? Or do you have the measurement results for some known state(s) and want to constrain what the measurement itself is? $\endgroup$ – qwyxivi Jun 28 at 18:00
  • $\begingroup$ Its the latter one. Basically, the idea is to construct a measurement using two pieces of information, namely (1) the probabilities of obtaining the different results and (2) the post-measurement states when the operation is performed on a fixed pre-measurement state. (I was wondering if these would uniquely define a measurement but your comment seems to suggest this is not the case) $\endgroup$ – Larss96 Jun 28 at 19:35
  • $\begingroup$ I find the question unclear. What do you want to know: Given X and Y, does there exist a Z such that ... ? (Then you should phrase it like that!) $\endgroup$ – Norbert Schuch Jun 28 at 20:44
  • $\begingroup$ (Reading your comment, this is more clear. But you should edit the question to add clarifications!) -- Most likely, it not only can be non-unique, but there might not exist a POVM for certain choices. $\endgroup$ – Norbert Schuch Jun 28 at 20:45
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I constructed a counterexample to the uniqueness property of your first question. Take a 3-state system with $$\rho = \begin{bmatrix}1/4 & 0 & 0 \\ 0 & 1/4 & 0 \\ 0 & 0 & 1/2\end{bmatrix}$$ and our post-measurement states each with probabilities $1/2$ each. $$\rho_0 = \begin{bmatrix}1/2 & 0 & 0\\ 0 & 1/2 & 0 \\ 0 & 0 & 0\end{bmatrix},~ \rho_1 = \begin{bmatrix}0 & 0 & 0\\ 0 &0&0 \\ 0&0& 1\end{bmatrix}$$ and corresponding POVMs $$M_0^\dagger M_0 = \begin{bmatrix}1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix},~ M_1^\dagger M_1 = \begin{bmatrix}0 & 0 & 0\\ 0 &0&0 \\ 0&0& 1\end{bmatrix}$$

There are at least two possible sets of values for $M_0,M_1$ with the same POVM and corresponding final states, namely: $$M_0 = \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}, M_1 = \begin{bmatrix}0 & 0 & 0 \\ 0& 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$$ and $$M_0 = \begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}, M_1 = \begin{bmatrix}0 & 0 & 0 \\ 0& 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$$

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