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.

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.

Danylo Y

Posted 2020-06-01T17:18:24.840

Reputation: 3 940


A good reference for this is Watrous' notes where he discusses Naimark's theorem -

– user1936752 – 2020-06-01T21:01:00.600

Yes, these notes also explain equivalence between general measurements and indirect projective measurements, but not what I'm asking. – Danylo Y – 2020-06-01T21:18:25.627


BTW, the clearest explanation of the Naimark's theorem I found in

– Danylo Y – 2020-06-02T08:27:15.320



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 $a$.


Posted 2020-06-01T17:18:24.840

Reputation: 35 722

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 – 2020-06-02T09:30:07.693

Sorry, you're right. That's me not being careful enough. Will try to sort it later... – DaftWullie – 2020-06-02T09:34:09.137

Still not entirely happy with the answer, but it's a step in the right direction.... – DaftWullie – 2020-06-02T13:04:25.677

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 – 2020-06-02T16:26:33.837

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 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 – 2020-06-02T16:26:45.893

OK, so this was my starting point in order to try and get to the bottom of what you're actually asking. I'm not sure I understand the "what if $P_i$ are general?" part. Do you mean you want to replace the $P_i$ with a generalised measurement (which, again, would just mean going to a larger Hilbert space and projecting)? Or do you want to project also the original system as well as $a$? – DaftWullie – 2020-06-03T07:33:40.637

By general ${ P_i }$ I mean PVM on the whole $H \otimes H_a$ space. So we can't assume that $P_i = I \otimes |i\rangle\langle i|$. In some sence we also project the original system along with ancilla in this case. The effect of it wasn't clear to me. – Danylo Y – 2020-06-03T07:58:56.857

But you still need to restriction that the outcome is always separable. That means that the projectors are separable. If they also project on the original system, then you might as well only project on the original system. Then you're back where you started with projective measurements (although I suppose you might get less information depending on what $U$ does). – DaftWullie – 2020-06-03T08:08:54.730

So, the only case when ${ M_i }$ exist is when $P_i = Q_i \otimes R_i$? Though it's not quite a composition of projective measurements on subsystems. – Danylo Y – 2020-06-03T08:48:51.243


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.

Danylo Y

Posted 2020-06-01T17:18:24.840

Reputation: 3 940