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This may be a very basic and common question (also discussed a lot), but strikingly enough I couldn't find the answer in the books or elsewhere.

The projective measurement is given by the PVM on the space $H$: $$\sum P_i = I,$$ where $P_i$ are mutually orthogonal projections. The post-measurement state of a density matrix $\rho$ is $$P_i \rho P_i ~/~ \text{Tr}(P_i \rho P_i),$$ with the probability $\text{Tr}(P_i \rho P_i)=\text{Tr}(\rho P_i)$.

The general measurement is given by the set of operators $M_i$ that corresponds to the POVM on $H$: $$\sum M_i^\dagger M_i = I.$$

The post-measurement state of a density matrix $\rho$ is $$M_i \rho M_i^\dagger ~/~ \text{Tr}(M_i \rho M_i^\dagger),$$ with the probability $\text{Tr}(M_i \rho M_i^\dagger) = \text{Tr}(\rho M_i^\dagger M_i)$.

Note that POVM itself doesn't describe the post-measurement state, because $M_i^\prime = UM_i$ for some unitary $U$ gives the same POVM but different post-measurement results (I mean states, though the probability will be the same).

It's known that, roughly speaking, general measurements correspond to projective measurements on a larger space. But the best exact statement I could find is that general measurement corresponds to an indirect projective measurement! The indirect measurement is when we add some ancilla state to a target system, perform a unitary evolution of a joint state followed by a projective measurement on that ancilla space and finally trace out the ancilla system.

So, the question is $-$ what if we perform PVM on the whole joint system, not just on the ancilla? Will the post-measurement results correspond to some general measurement?

Formally, let $H$ is the target system, $H_a$ - ancilla space with some fixed density matrix $\rho_0$ on it, $U$ is a unitary on $H \otimes H_a$ and $\sum P_i = I$ is a PVM on the whole $H \otimes H_a$. The post-measurements states of this scheme are $$ \text{Tr}_a ( P_i U \cdot \rho \otimes \rho_0 \cdot U^\dagger P_i) ~/~ n_i,$$ with the probability $n_i$ where $n_i$ is just the trace of the numerator. The question is $-$ are there operators $M_i$ such that those post-measurement states equal to $$M_i \rho M_i^\dagger ~/~ \text{Tr}(\rho M_i^\dagger M_i) ?$$

I know how to prove that there exists a unique corresponding POVM $\sum F_i=I$ on $H$ that can be used to compute probabilities, i.e. $n_i = \text{Tr}(\rho F_i)$, but it's not clear how to derive the exact $M_i$ or even prove that they exist.

Update
Also, we can consider a related quantum channel $$ \Phi(\rho) = \sum_i \text{Tr}_a ( P_i U \cdot \rho \otimes \rho_0 \cdot U^\dagger P_i) $$ and derive Kraus decomposition $$ \Phi(\rho) = \sum_j K_j \rho K_j^\dagger, $$ but it still doesn't answer the question. It's not even clear if Kraus decomposition has the same number of summands.

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Let me start with a clarification:

Note that POVM itself doesn't describe the post-measurement state, because $M′_i=UM_i$ for some unitary U gives the same POVM but different post-measurement results.

The formalism you're talking about here is not POVMs. POVMs is when you only use the operators $E_i=M_i^\dagger M_i$, the point being that with these you can calculate the probability of the measurement outcome but cannot calculate the final state because, given $E_i$, I cannot find $M_i$ because any $M'_i$ would do just as well. If you are given the $\{M_i\}$, then the post-measurement state, as you state, is well defined: $$ \frac{M_i\rho M_i^\dagger}{\text{Tr}(\rho M_i^\dagger M_i)}. $$ The fact that other $M_i'$ give different outcomes is irrelevant. They're due to different measurements!

As I understand your actual question, you're wanting to understand the correspondence between $$ \text{Tr}_a ( P_i U \cdot \rho \otimes \rho_0 \cdot U^\dagger P_i) ~/~ n_i $$ and $$M_i \rho M_i^\dagger ~/~ \text{Tr}(\rho M_i^\dagger M_i) ?$$ In particular, you want to go from $\{P_i\}$ and $U$ to finding $\{M_i\}$.

Let me start the other way around. If you're given a set $\{M_i\}$, then you can introduce an ancilla in the $|0\rangle$ state, and define a $U$ such that $$ U|\psi\rangle|0\rangle=\sum_i(M_i|\psi\rangle)\otimes|i\rangle, $$ in which case $P_i=I\otimes |i\rangle\langle i$. Note that, if we were given $U$ and $\{P_i\}$ of this form, we could easily calculate the $M_i$: $$ M_i=I\otimes\langle i|\cdot U\cdot I\otimes|0\rangle. $$

Now, generically, if we're given $U$ and $\{P_i\}$, can we write down $\{M_i\}$? No, because they don't exist. Note that when $M_i$ acts on a pure state (every pure state), it must give a pure state output. That is extremely constraining on the possible forms of $U$ and $P_i$: $P_iU|\psi\rangle|0\rangle$ must be separable for all $|\psi\rangle$ and all $i$ that have non-zero outcome probabilities. To all intents and purposes, this reduces you to the previous case, up to a local unitary on system $a$.

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  • $\begingroup$ But your last expression depends on both $i$ and $j$. That is, the $i$-th post-measurement state can be written as $\sum_{j} M_{ij} \rho M_{ij}^\dagger ~/~ n_i$. I tend to think that it's maximum that we can get in this general situation. $\endgroup$ – Danylo Y Jun 2 at 9:30
  • $\begingroup$ Sorry, you're right. That's me not being careful enough. Will try to sort it later... $\endgroup$ – DaftWullie Jun 2 at 9:34
  • $\begingroup$ Still not entirely happy with the answer, but it's a step in the right direction.... $\endgroup$ – DaftWullie Jun 2 at 13:04
  • $\begingroup$ If we have $P_i = I \otimes |i\rangle \langle i|$ then for any unitary $U$ that formula for $M_i$ is indeed the correct one. Also, $P_i |v\rangle$ is always separable in such case because it's just $|w\rangle \otimes |i\rangle$. So, there are no constrains on $U$. In the other way, if we are given $M_i$ then we can set $P_i = I \otimes |i\rangle \langle i|$ and set $U$ as you've wrote. $\endgroup$ – Danylo Y Jun 2 at 16:26
  • $\begingroup$ But such $P_i$ act on the ancilla. This is what I call indirect projective measurement. And that equivalence called the Naimark's theorem, these notes explain it well arxiv.org/abs/1110.6815. The question was what if $P_i$ are general. Now I see that if $P_iU|\psi\rangle |0\rangle$ is not separable for some $|\psi\rangle$ and $i$ then, indeed, it proves that there are no corresponding $\{ M_i \}$. But there are $\{ M_{ij} \}$ as you wrote before. $\endgroup$ – Danylo Y Jun 2 at 16:26
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I'll try to explain DaftWullie's answer as I see it. We assume $\rho_0 = |0\rangle\langle0|$.

If we have $P_i = I \otimes |i\rangle \langle i|$ then for any unitary $U$ on $H \otimes H_a$ operators $M_i$ can be computed by the formula $$ M_i=I\otimes\langle i|\cdot U\cdot I\otimes|0\rangle.$$ It shows that indirect projective measurement (in which PVM acts on the ancilla only) can be seen as a general measurement on the target system.
This also works in the other direction $-$ general measurement $\{M_i\}$ on the target system can be seen as a unitary evolution $U$ of $\rho \otimes |0\rangle\langle0|$ followed by a PVM on the ancilla. The unitary can be derived from the equation $$U|\psi\rangle|0\rangle=\sum_i(M_i|\psi\rangle)\otimes|i\rangle.$$

Such equivalence between measurements also known as Naimark's theorem.

Now, if $P_i$ is a PVM on the whole $H \otimes H_a$ then there are no $\{M_i\}$ in general.
To see this consider $\rho = |\psi\rangle \langle \psi|$. In general, the state $P_iU|\psi\rangle|0\rangle$ will not be separable. In such case the state $$\text{Tr}_a ( P_i U \cdot |\psi\rangle \langle \psi| \otimes |0\rangle \langle 0| \cdot U^\dagger P_i) ~/~ n_i $$ will be mixed. But $$M_i |\psi\rangle \langle \psi| M_i^\dagger ~/~ \text{Tr}(|\psi\rangle \langle \psi| M_i^\dagger M_i)$$ is a pure state $-$ a contradiction, so there are no such $\{ M_i \}$.

But we can write that $$\text{Tr}_a ( P_i U \cdot \rho \otimes |0\rangle \langle 0| \cdot U^\dagger P_i) ~/~ n_i = $$ $$ = \sum_j I \otimes \langle j| \cdot P_i U \cdot \rho \otimes |0\rangle \langle 0| \cdot U^\dagger P_i \cdot I \otimes |j\rangle ~/~ n_i = $$ $$ = \sum_j \big(I \otimes \langle j| \cdot P_i U \cdot I \otimes |0\rangle \big) \rho \otimes 1 \big(I \otimes \langle 0| \cdot U^\dagger P_i \cdot I \otimes |j\rangle \big) ~/~ n_i = $$ $$ = \sum_j M_{ij} \rho M_{ij}^\dagger ~/~ n_i,$$ where $$ M_{ij} = I \otimes \langle j| \cdot P_i U \cdot I \otimes |0\rangle.$$

So, the $i$-th post measurement state can be seen as an output of some quantum channel (that depends on $i$). Though, this was natural to expect, according to the general theory.

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