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.


1 Answer 1


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 $\langle\psi|M|\psi\rangle$.

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    $\begingroup$ Thank you for your answer, this helped a fair bit. $\endgroup$
    – TTa
    Commented Dec 11, 2022 at 19:21
  • $\begingroup$ 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 $i$? $\endgroup$
    – narip
    Commented Nov 26, 2023 at 6:17
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    $\begingroup$ 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 $M(A)=I$. $\endgroup$ Commented Nov 26, 2023 at 7:25

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