# What is a POVM?

I am having a hard time understanding what exactly a Measurement is by its definition? What I read is that a POVM $$M$$ is defined by its set of elements $$M_i$$. So is $$M$$ itself an operator? In circuit diagrams its a little "Measure" box, but I've never seen it written down in operator+braket formalism, at least I don't think I have.

If I have an unknown state $$|\psi\rangle$$ and its density operator $$\rho = |\psi\rangle\langle\psi|$$ then I know that the probability for it to be in the state defined by $$M_i$$ is given by $$\langle \psi|M_i|\psi\rangle = Tr(M_i \rho)$$ such that $$\sum_i Tr(M_i \rho) = 1$$. So I know how the individual elements act on a state, but what is the Measure itself? Is it effectively performing all possible measurement elements? Is writing something like $$\langle \psi|M|\psi\rangle$$ (no subscript here) meaningless?

My intuition for this comes from photon counting. Effectively you have a device that can count the number of photons coming in. The individual elements that make up the "measurement" would be $$|0\rangle\langle0|, |1\rangle\langle1|, |2\rangle\langle2| \dots$$ with some uniform normalization constant in front of each element.

• Measurements are Hermitian (they are equal to their own conjugate transpose) which guarantees that all of its eigenvectors are real. Operators are unitary (the inverse is their conjugate transpose.). X, Y, and Z happen to be both. Dec 11, 2022 at 6:16
• – glS
Dec 11, 2022 at 9:53

TL;DR: POVM $$M$$ is not an operator. It's more like a probability distribution over the set of all possible measurement outcomes, but parametrized by quantum states.

## Positive operator-valued measures

A Positive Operator-Valued Measure (POVM) $$M$$ is not an operator. It is typically defined as a collection of positive operators $$M_i$$ labeled by measurement outcomes $$i\in A$$ where $$A$$ is the set of all possible measurement results with the extra requirement that $$\sum_{i\in A}M_i=I$$.

We can also think of $$M$$ as a two-argument function that takes a measurement outcome $$i$$ and a quantum state $$\rho$$ and yields the probability of measuring $$i$$ on $$\rho$$ $$M: A\times D(\mathcal{H})\to[0,1]\tag1$$ where $$D(\mathcal{H})$$ denotes the set of density operators on some Hilbert space $$\mathcal{H}$$. We can write down an explicit formula for the function $$M$$ $$M(i,\rho)=\mathrm{tr}(M_i\rho)\tag2$$ which is just another statement of the Born rule. In certain ways, the function $$M$$ is similar to a good old probability distribution over $$A$$ with the caveat that the distribution is parametrized by quantum states $$\rho$$.

## Probability distributions

A probability distribution on $$A$$ is formalized as a probability measure defined as a function on a certain $$\sigma$$-algebra of subsets of $$A$$ called events. However, in the countable case, the fact that the definition requires a measure to be $$\sigma$$-additive implies that it is completely defined by its values on a partitioning of $$A$$. In particular, it is completely defined by specifying its value on singleton sets. For this reason, we can often forget about the $$\sigma$$-algebra and think of a probability distribution on $$A$$ as a function $$p:A\to[0,1]$$ that assigns a probability $$p(i)$$ to each $$i\in A$$.

Now, the POVM $$M$$ is not quite as direct. Instead of assigning to $$i\in A$$ a probability $$p(i)\in[0,1]$$, it assigns to $$i\in A$$ a positive operator $$M_i$$ (which explains the name Positive Operator-Valued Measure). To obtain an actual probability (of $$i$$ being the outcome of a measurement), a POVM needs one more piece of information: the quantum state $$\rho$$. This accounts for the fact that in quantum mechanics measurement outcome probability depends on both: the outcome $$i$$ and the quantum state $$\rho$$.

Further similarities between POVM $$M$$ and probability distribution $$p$$ can be seen in the constraints they are required to satisfy. The fact that $$p(i)\ge 0$$ for any $$i\in A$$ corresponds to the fact that $$M_i$$ is a positive operator. Similarly, the requirement that $$\sum_{i\in A}p(i)=1$$ corresponds to $$\sum_{i\in A}M_i=I$$.

## How to make sense of $$\langle\psi|M|\psi\rangle$$?

The discussion above suggests that we could make sense of an expression such as $$\langle\psi|M|\psi\rangle$$ by currying, i.e. by interpreting $$\langle\psi|M|\psi\rangle$$ as a function obtained from $$M$$ by fixing the quantum state. The result would be a single-argument function mapping a measurement outcome $$i\in A$$ to its probability when the measurement is performed on state $$|\psi\rangle$$ $$\langle\psi|M|\psi\rangle: i\mapsto\langle\psi|M_i|\psi\rangle\in[0,1].\tag3$$ Since $$M$$ is a positive operator-valued measure on $$A$$, the function $$\langle\psi|M|\psi\rangle$$ would be a probability measure on $$A$$ for every $$|\psi\rangle$$. That said, I have never seen anyone using an expression such as $$\langle\psi|M|\psi\rangle$$.

• If we take $M$ as a measure, then the integral $\int_A{dM\left( i \right)}=I$ is meaningless (I don't know the definition of the integral with respect to a measure which gives an operator instead of a number)? I think we can only talk about the integral of form $\int_A{\mathrm{tr}\left[ \rho dM\left( i \right) \right] f\left( i \right)}$ where $f$ is some function concerning $i$? Nov 26, 2023 at 6:17
• The usual constructions for the Lebesgue integral can be carried out in settings more general than scalar-valued measures. It's a nice exercise since it illuminates why we might want the operators the measure takes on as values to be positive semidefinite. However, even just the very first trivial step - defining the integral for indicator functions - is sufficient to evaluate $\int_AdM=\int \mathbb{1}_AdM=M(A)$ where we used the definition of integral over a set and then the definition of integral of an indicator function. Finally, $M$ is by definition normalized with $M(A)=I$. Nov 26, 2023 at 7:25