Why are POVMs useful? Are they just an axiomatic way to define measurement?



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.


Posted 2020-01-06T16:43:37.830

Reputation: 1 004


For a start, see Martin's answer here.

– Sanchayan Dutta – 2020-01-06T16:57:26.837



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.


Posted 2020-01-06T16:43:37.830

Reputation: 35 722

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. – StarBucK – 2020-01-07T12:23:19.967

1I 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 – 2020-01-07T12:56:39.203

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. – StarBucK – 2020-01-07T12:57:46.353

@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 – 2020-01-07T15:57:16.350

@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) – StarBucK – 2020-01-07T23:53:15.050

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 – StarBucK – 2020-01-07T23:55:47.673

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 – 2020-01-08T08:21:30.237

@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. – StarBucK – 2020-01-08T10:45:59.603