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I know the definition of projective measurement, generalized measurement, POVM.

I understand the usage of generalized measurement for the reason that it can model experiments "easier" (for example measurement of a photon that will be destructive so that measuring again the state just after the first measurement will give me another answer).

However, I am still kinda confused by why we have introduced the notion of P.O.V.M. For me we have everything we want from generalized & projective measurement.

Would you agree with me if I say that POVM is just an axiomatic way to define statistics of measurement. There is nothing much to understand/overthink.

In the sense, we ask the minimal mathematical properties that our measurement operator must fullfil with respect to statistical behavior which is:

  • they are semidefinite positive (to have positive probabilities)
  • they sum up to identity (to have probability summing up to $1$)

and we relate our measurement operator with the physics:

$$p(m)=\mathrm{Tr}(E_m \rho)$$ where $m$ is the outcome, $E_m$ the associated POVM.

glS
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Marco Fellous-Asiani
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3 Answers3

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If you want to prove a no-go theorem in quantum mechanics, such as the impossibility of distinguishing two non-orthogonal states given a single unknown state, you want to prove that even if you add ancillae, apply unitaries, do projective measurements in weird bases, apply unitaries conditional on previous measurements, etc, you still can't distinguish two non-orthogonal states. One nice way to do that is instead of considering all possible combinations of operations you might perform on your system, you realize that every possible sequence of operations is really a POVM, and every POVM can be realized by some convoluted combination of elementary operations. Then you just prove that no POVM can distinguish two non-orthogonal states, and you're done!

Jahan Claes
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  • As a comment, this is the measurement version of considering all CPTP maps when trying to prove something like the no-cloning theorem. You could say "CPTP maps are just a generalization of unitaries and projective measurements," but it's a lot easier to prove something about all CPTP maps than to prove something for every combination of unitaries and measurements. – Jahan Claes Oct 05 '21 at 15:57
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For me, generalised measurements cover everything (obviously, that's why they're generalised), with projective measurements being a simple case that covers what we usually want to be doing.

So, yes, why introduce POVMs which are basically the generalised measurements but without the output state? Because they describe what actually happens in some experiments. If you're using optics, for example, where measurement of a photon is destructive, i.e. you do not have the photon afterwards, the POVMs perfectly describe what's going on.

What I think is a reasonable follow-up question is why so many courses/texts etc put as much weight as they do on POVMs. I believe that the reason is simply that, when performing many quantum information protocols, we don't care about the state after measurement, we only care about the probabilities of the different outcomes. For example, if you're trying to identify what state you have, you only care about which measurement result you get, not the final state. Then it's a mathematical convenience that you're dealing with $E_i$ instead of having to calculate $M_i^\dagger M_i$ first.

DaftWullie
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  • So in the end, would you agree with my view of saying it is just an axiomatic way to define measurement statistics ? Indeed maybe the operator of measurement will be $M_i^{\dagger} M_i$ but as we only care about the probability we call $E_i=M_i^{\dagger}M_i$ and as it is a measurement it follows POVM definition and properties. Nothing much more to understand. – Marco Fellous-Asiani Jan 07 '20 at 12:23
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    I suppose I'd agree that you could choose to put it like that, although it's not a way that I would be likely to express it! – DaftWullie Jan 07 '20 at 12:56
  • Ok thanks. Well if for you it makes sense at least I will stick to this understanding then because it is the one that I'm confortable with ^^ Thank you. – Marco Fellous-Asiani Jan 07 '20 at 12:57
  • @StarBucK Well, physicists and mathematicians don't make axioms for no reason. While the POVM formalism is equivalent to generalized measurement, it is the better way to think about measurements. The way you frame your sentence: "is just an axiomatic way" and "nothing much more to understand", makes it appear like you haven't yet developed an intuition for why the POVM formalism is more preferable and you view it as mere mathematical hairsplitting. – Sanchayan Dutta Jan 07 '20 at 15:57
  • @SanchayanDutta well for me the equivalence almost bring nothing significant here. If it is just a mapping $E_i=M_iM_i^{\dagger}$ where on the rhs it is generalized measurement operators then it is only a slightly more condensed way to express the same idea. On the other hand the equivalence between generalized and projective measurement brings something usefull because first the equivalence is not trivial (you need to use entanglement unitary to see it), and it gives you new angles of interpretation (destructive measurement can be seen as resulting of entanglement process for example) – Marco Fellous-Asiani Jan 07 '20 at 23:53
  • For the povm the equivalence is "trivial" and doesnt give much of a new insight from my current understanding. So i can understand the point only under the angle of "expressing the same thing in a more condensed way". Thus axiomatic approach of measurement which gives definition that may make proof writing more readable. But I dont see much more from it at the moment. Maybe I missed something indeed – Marco Fellous-Asiani Jan 07 '20 at 23:55
  • For me, the point of axioms is to have a minimal set of axioms that you use to describe the entirety of quantum theory, the hope being that those axioms then give you some insight about what it takes to make up the theory (and you can ask "what if I changed..."). For the complete theory, you need the axiom about generalised measurements (or, as you say, projective+entangling unitaries) because you need to be able to talk about the outcome after measurement. Once you have that axiom, an axiom about POVMs is unnecessary. It's merely a definition within your theory. – DaftWullie Jan 08 '20 at 08:21
  • @DaftWullie I agree but what I meant is that if you only focus on statistics of measurements, POVM can be seen as a set of axioms defining it. Indeed they give you the minimal set of conditions you need to have for a well defined measurement. Now, in themselves they don't give much insight about what a measurement is because the equivalence with generalized measurements (I exclude the post measurement state property) is trivial and it doesn't really give you a new angle of interpretation for measurement. – Marco Fellous-Asiani Jan 08 '20 at 10:45
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POVMs model the most general way to measure something about a quantum state. I'd argue POVMs differ from "generalized measurements" in that they are tools built to answer different questions.

A POVM tells you about how quantum states are converted in probability distributions. The focus is on that measurement probabilities, nothing more and nothing less.

Generalized measurements also deal with a possible post-measurement state. Aside from the fact that in many situations you do not have a post-measurement state, the fact of the matter is that you might not care about post-measurement states for whatever you are studying. If you do, then it makes sense to use generalized measurements, or equivalently, the corresponding quantum-to-classical channels (which codify the same identical information as generalized measurements anyway). You often don't, and in these cases it's therefore easier/preferable to work directly with POVMs.

glS
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