I'm trying to understand how different authors assign measurement operators to POVM operators so we can define the post-measurement state. In particular, it's not clear for me if both assignments are in fact the same or one is a generalisation of the other.

The POVM operators are typically defined as follows so that we can assign probabilities to particular measurement outcomes without defining the post-measurement state:

$$ \sum_{n} \hat{E}_n = \hat{\text{I}},\quad p(n) = \text{Tr}(\hat{E}_{n}\rho) $$ with $p(n)$ denoting the probability of outcome $n$ for some density matrix $\rho$.

If we then want to talk about post-measurement states we need to express our POVM operators in terms of measurement operators, this is where my confusion arises.

In some texts the POVM operators are assigned to a single measurement operator: $$ \hat{E}_n = M^{\dagger}_{n}M_n $$ This is the case in the following examples:

  • Nielsen & Chuang - Quantum Computation and Quantum Information
  • Jacobs & Steck - A straightforward introduction to continuous quantum measurement

However, in some texts the POVM operators are assigned as a sum over measurement operators: $$ \hat{E}_n = \sum_{k}\hat{\text{A}}^{\dagger}_{nk}\hat{\text{A}}_{nk} $$ This is seen in:

I'm not clear if these two formulations are equivalent. My impression is that $\hat{E}_n = M^{\dagger}_{n}M_n$ is a special case of the latter definition of $\hat{E}_n$ given only a single measurement operator in the sum. Defining $\hat{E}_n = \sum_{k}\hat{\text{A}}^{\dagger}_{nk}\hat{\text{A}}_{nk}$ suggests to me that all of the measurement operators $\hat{A}_{nk}$ give the result $n$ and so should be included in the sum.

So far the only discussion I have seen about this is in [Schleich & Walther - Elements of Quantum Information - Pg.403] where $\Pi_b$ is the notation used here for POVM associated with outcome $b$:

We assume that for each result $b$ there is only a single operator $A_{b1}$ with $A^{\dagger}_{b1}A_{b1} = \Pi_b$. Following the interpretation of Sect. 20.2.1 this means that the measurement device does not conceal internally available results

Regarding the above quote, what is meant by "does not conceal internally available results"?

I would really appreciate any clarity or help with understanding! As far as I understand there is overlap with 2 but the reasoning still isn't clear to me, for example why does "forgetting" imply the operator-sum representation given?

  • 1
    $\begingroup$ related on qc.SE: quantumcomputing.stackexchange.com/q/27673/55, and links therein. The gist is that a POVM doesn't specify the post-measurement states, and multiple choices of post-measurement states are compatible with a given POVM. If you really want to talk about post-measurement states, using "measurement channels" is generally more suitable. Eg "quantum instruments" formalise the most general form of a channel compatible with a given POVM. $\endgroup$
    – glS
    Commented Dec 7, 2023 at 11:48
  • $\begingroup$ It really depends on the physical mechanism behind your measurement. There is no "right" set of measurement operators if you only know the POVM operator $E_n$. $\endgroup$ Commented Dec 7, 2023 at 16:24

1 Answer 1


One can consider the POVM operator with the sum as a coarse graining operation that takes each measurement result labeled by both $n$ and $k$ and labels them all by $n$ alone.

For example, if $n$ tells you the number of sides of a polygon and $k$ tells you the colour, your measurement device might actually measure both properties but only tell you the number of sides. Or if you are looking at DNA and wondering what species it belongs to, your measurement will give you lots of information that could let you identify the individual organism but you might end up just classifying the result as "dog."

Specifically, the formalisms are not equivalent; rather, $M_n^\dagger M_n$ is a subset of $\sum_k A_{nk}^\dagger A_{nk}$.

  • $\begingroup$ I see, so defining your POVM operator with a single measurement operator is labelling one very specific outcome. In the case of the sum, you consider many measurement operators that contribute to the outcome "n", so the sum is coarse graining in the sense that you lose the information about "k"? In general there's nothing stopping you from defining your POVM as a sum of measurement operators as long as it still fulfils the POVM definition I assume? $\endgroup$
    – ConorP
    Commented Dec 7, 2023 at 11:42
  • $\begingroup$ And just to add, when I read "does not conceal internally available results" in [Schleich & Walther] I now understand that to mean we don't sum over "k" and we have access to every measurement outcome. $\endgroup$
    – ConorP
    Commented Dec 7, 2023 at 11:45
  • 1
    $\begingroup$ It should be noted that these two measurement do not produce the same post-measurement state -- in one case superpositions of colors are preserved, in the other case they aren't. $\endgroup$ Commented Dec 7, 2023 at 16:25

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