# Projective vs general measurements - a missing piece

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.

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 $$a$$.

• 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. – Danylo Y Jun 2 at 9:30
• Sorry, you're right. That's me not being careful enough. Will try to sort it later... – DaftWullie Jun 2 at 9:34
• Still not entirely happy with the answer, but it's a step in the right direction.... – DaftWullie Jun 2 at 13:04
• 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. – Danylo Y Jun 2 at 16:26
• 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. – Danylo Y Jun 2 at 16:26

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.