A POVM is typically defined as a collection of operators $\{\mu(a)\}_{a\in\Sigma}$ with $\mu(a)\in\mathrm{Pos}(\mathcal X)$ positive operators such that $\sum_{a\in\Sigma}\mu(a)=I$, where I take here $\Sigma$ to be some finite set and $\mathcal X$ some vector space (using the notation from Watrous' TQI).

On the other hand, when discussing observables in a measure-theoretic framework, given a nonempty set $\Omega$, a $\sigma$-algebra $\mathcal F$ on $\Omega$ (in other words, given a measurable space $(\Omega,\mathcal F)$), and denoting with $\mathcal E(\mathcal X)$ the set of effects on the space $\mathcal X$, that is, the set of Hermitian operators $E$ such that $0\le E\le I$, we say that a mapping $\mathrm A:\mathcal F\to\mathcal E(\mathcal X)$ is an observable if, for any state $\psi\in\mathcal X$, the function $$\mathcal F\ni X\mapsto \langle \psi|\mathrm A(X)\psi\rangle\in\mathbb R$$ is a probability measure. Here I'm taking definition and notation from Heinosaari et al. (2008). In other words, $\mathrm A$ is an observable iff, for all $\psi\in\mathcal X$, the triple $(\Omega,\mathcal F,\mathrm A_\psi)$ is a probability space, with $\mathrm A_\psi$ defined as $\mathrm A_\psi(X)=\langle \psi|\mathrm A(X)\psi\rangle$.

I'm trying to get a better understanding of how these two different formalisms match. In particular, an observable as thus defined is closer to a POVM than an "observable" as usually defined in physics (which is just a Hermitian operator), right?

Are these observables equivalent to POVMs? That is, does any such observable correspond to a POVM, and vice versa?

I can see that a POVM can be thought of as/is the map $\mu:\Sigma\to\mathrm{Pos}(\mathcal X)$, which then extends to a map $\tilde\mu:2^{\Sigma}\to\mathrm{Pos}(\mathcal X)$ such that $\tilde\mu(2^\Sigma)=1$, which is then an observable. However, I'm not sure whether any observable also corresponds to such a POVM.

  • $\begingroup$ As far as i understand, wikipedia's POVM definition and this definition of an observable are the same. $\endgroup$
    – tsgeorgios
    Jan 21, 2021 at 21:46
  • $\begingroup$ well, in one case you only impose $\mu(a)\ge0$ and $\sum_a \mu(a)=I$, while in the other you ask for $F$ s.t. $F(\Omega)=I$ and $\mathcal F\ni E\mapsto \langle F(E)\psi|\psi\rangle$ is a non-negative countably additive measure on $\mathcal F$ for all $\psi$. I guess the question is pretty much why these are equivalent $\endgroup$
    – glS
    Jan 21, 2021 at 22:02

2 Answers 2


These two definitions define the same concept: the POVM measurement. The observable definition is how POVM is defined for use in the case of infinite index set and dimension (see e.g. POVM) and POVM definition in the question is how it is simplified for use in the finite case. If you are working in finite dimensions, the two constructions are equivalent.

POVM from observable

We will construct a POVM $\{\mu(a)\}_{a\in\Sigma}$ from the observable $A: \mathcal{F} \to \mathcal{E}(\mathcal{X})$. Begin by choosing a finite $\Sigma \subset \mathcal{F}$ such that $\cup_{a\in\Sigma}a = \Omega$ and $a \cap b = \emptyset$ for every $a, b\in\Sigma$. This serves as the index set for the POVM and establishes the connection between the two definitions. Define

$$ \mu(a) := A(a). $$

Note that this defines valid POVM elements because $0 \le A(X) \le I$ for every $X\in\mathcal{F}$ and

$$ \sum_{a\in\Sigma} \mu(a) = \sum_{a\in\Sigma} A(a) = A\left(\bigcup_{a\in\Sigma} a\right) = A(\Omega) = I $$

where in the second equality we used $\sigma$-additivity of $A_\psi$ for every $|\psi\rangle$ and in the last step we used the fact that $A_\psi$ is a probability measure for every $|\psi\rangle$.

Observable from POVM

(This merely fills in some details in the construction you sketched in the question.)

We will construct an observable from the POVM $\{\mu(a)\}_{a\in\Sigma}$. Let $\Omega = \Sigma$ and $\mathcal{F} = \mathcal{P}(\Sigma)$. For $X\in\mathcal{F}$ define

$$ A(X) := \sum_{a\in X}\mu(a) $$

where we are using the fact that $X$ is finite. Note that the right-hand side is a valid effect. We need to show that for every $|\psi\rangle$, $A_\psi$ is a probability measure. It is clear that for every $X\in\mathcal{F}$, $A_\psi(X) \in [0, 1]$ and $A_\psi(\emptyset) = 0$. Now, $\mathcal{F}$ is finite, because $\Omega$ is finite. Thus, to show $\sigma$-additivity we just need to show additivity. Let $X_{i=1,\dots,k}$ be a collection of disjoint subsets of $\Omega$, then

$$ A_\psi\left(\bigcup_{i=1}^k X_i\right) = \sum_{a\in\cup_{i=1}^k X_i} \mu_\psi(a) = \sum_{i=1}^k \sum_{a\in X_i} \mu_\psi(a) = \sum_{i=1}^k A_\psi(X_i) $$

where $\mu_\psi(a) = \langle\psi|\mu(a)\psi\rangle$ and in the second equality we used the fact that $X_i$ are disjoint.

Remark: Note that the definition of observable appears to be more flexible: it doesn't just give rise to a single POVM. Instead, every partitioning of $\Omega$ using the elements of the $\sigma$-algebra gives rise to a potentially different POVM. However, this fact corresponds to the trivial observation that you can obtain a new POVM by grouping and adding elements of a given POVM.

  • $\begingroup$ ah, I ended up self-answering at the same time you answered. I like this one better though, thanks. Regarding the last remark: doesn't the "added flexibility" just amount to being able to write sums $\sum_{a\in Y}\mu(a)$ as some (additive) function acting directly on $Y$? I'm not sure that I see any material advantage in this (apart from noticing that the resulting object is a measure, and therefore being able to use known results from measure theory, I guess) $\endgroup$
    – glS
    Jan 22, 2021 at 0:01
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    $\begingroup$ Thanks! Heh, perhaps I should have polished less and just send the key fact that the observable definition hides the index set whereas the POVM definition makes it explicit. It turns out the index set is just some partitioning of $\Omega$ using the sets in the $\sigma$-algebra. $\endgroup$ Jan 22, 2021 at 0:03
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    $\begingroup$ I wrote the remark to explain why in the "POVM from observable" construction we do not end up with one-to-one mapping from observable to POVM. That flexibility really corresponds to the flexibility we already have in the finite POVM definition, so I agree it isn't any real new advantage to the observable definition. $\endgroup$ Jan 22, 2021 at 0:07
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    $\begingroup$ Regarding the ability to deploy measure theory - that is key! AFAIK, the motivation behind the observable definition is to enable its use in the infinite case. Measure theory provides tools to do that. The concepts such as $\sigma$-algebra and $\sigma$-additivity capture the sort of "tamed" infinity which we can work with and still end up with a reasonable concept of observable in the infinite case. $\endgroup$ Jan 22, 2021 at 0:10

Upon some more reflection, the answer is probably as follows.

Let $\mathrm A$ be an observable according to the definition in the question, and assume $\Omega$ is finite. Then any $X\in\mathcal F$ is also some finite subset of $\Omega$. By definition of observable, we require the mapping $\mathrm A_\psi$ to be additive and non-negative, and therefore $$\langle \psi|\mathrm A(X)\psi\rangle = \sum_{\omega\in X}\langle \psi|\mathrm A(\{\omega\})\psi\rangle \ge0.$$ It follows that, for any $\omega\in\Omega$, the operator $\mathrm A(\{\omega\})$ is positive semi-definite: $\mathrm A(\{\omega\})\ge0$.

Furthermore, asking $\mathrm A_\psi$ to be a probability measure also means that $\mathrm A_\psi(\Omega)=1$ for all $\psi$, that is, $\sum_{\omega\in\Omega}A(\{\omega\})$ to be an operator whose expectation value is $1$ on any state $\psi$. This implies $\sum_{\omega\in\Omega}A(\{\omega\})=I$.

It follows that, defining $\mu(\omega)\equiv A(\{\omega\})$, we get the definition of POVM as collection of positive semi-definite operators summing to the identity.

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    $\begingroup$ This construction is correct. The other answer is a generalization, based on the fact that all finite partitionings of $\Omega$ - not just the one that breaks it up into the single-element sets $\{\omega\}$ - yield valid POVMs. This allows us to drop the assumption that $\Omega$ is finite. $\endgroup$ Jan 22, 2021 at 0:47
  • $\begingroup$ @AdamZalcman good point. Thanks for the remark $\endgroup$
    – glS
    Jan 22, 2021 at 9:12

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