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A positive-operator valued measure (POVM) is a set $\{F_i\}$ whose elements are each Hermitian and positive semidefinite ($F_i^\dagger=F_i$; $F_i\succeq 0$) and which together sum to unity ($\sum_i F_i=\mathbb{I}$). A projection-valued measure (PVM) is a POVM whose elements $\{P_i\}$ are each orthogonal projectors ($P_i P_j=P_i\delta_{ij}$).

From these two definitions, repeated entries should not be allowed in PVMs, since those repeated entries would not be orthogonal.

However, there are situations where a POVM with repeated entries seems to be physically equivalent to a PVM. Consider the POVM with the three elements $F_1=P_1$, $F_2=P_2/2$, and $F_3=P_2/2=F_2$, where $\{P_1,P_2\}$ form a PVM. The POVM has three different measurement outcomes, but two of them are identical. According to the state-update rule of quantum measurement, the PVM has state $\rho$ transforming to each of states $$\frac{P_1\rho P_1 }{\mathrm{Tr}(P_1\rho P_1 )}\quad\mathrm{or}\quad \frac{P_2\rho P_2 }{\mathrm{Tr}(P_2\rho P_2 )}\quad\mathrm{with~probabilities} \quad\mathrm{Tr}(P_1\rho P_1 ) \quad\mathrm{or}\quad \mathrm{Tr}(P_2\rho P_2 ).$$ Choosing identical Kraus operators for the POVM $K_2=K_3=\sqrt{P_2}/\sqrt{2}=P_2/\sqrt{2}$ together with $K_1=P_1$, the POVM's state-update rule transforms $\rho$ into each of states $$\frac{P_1\rho P_1 }{\mathrm{Tr}(P_1\rho P_1 )}\quad\mathrm{or}\quad \frac{P_2\rho P_2 }{\mathrm{Tr}(P_2\rho P_2 )}\quad\mathrm{or}\quad \frac{P_2\rho P_2 }{\mathrm{Tr}(P_2\rho P_2 )}\quad\mathrm{with~probabilities} \quad\mathrm{Tr}(P_1\rho P_1 ) \quad\mathrm{or}\quad \mathrm{Tr}(P_2\rho P_2 )/2\quad\mathrm{or}\quad \mathrm{Tr}(P_2\rho P_2 )/2.$$ Overall, the possible outcomes for the state update are the same in the PVM and POVM descriptions and occur with the same probabilities; can the measurements be considered the same?

Further physical considerations are that the device performing the PVM has two outcome labels and the device performing the POVM has three labels of which two are identical. So maybe the distinction must be made at the level of the measurement device's possible outputs, as opposed to the possible transformations of the state?

As a side note, the crucial property here is really that the Kraus operators are repeated. Then we know that we can unitarily rearrange the Kraus operators to find new ones satisfying $K_2^\prime=(K_2+K_3)/\sqrt{2}=\sqrt{2}K_2$ and $K_3^\prime=(K_2-K_3)/\sqrt{2}=0$. So maybe the point is that the Kraus operators underlying the POVM can be rearranged such that one (or more) of them vanishes. I would think that if a Kraus operator vanishes then we can ignore it and consider there to be fewer Kraus operators, but maybe this is again a distinction between asking what happens to the state (fewer Kraus operators suffice) versus asking what happens to the measurement device (rearranging Kraus operators changes the device).

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Having repeated POVM elements, or more generally some POVM element that is a scalar multiple of another, introduces some trivial redundancies in the resulting probability distributions. You could imagine that the different elements correspond to different physical outcomes though. Consider for example a measurement strategy involving evolving an input qubit through the isometry $$V=\frac12\begin{pmatrix}1&1\\1&1\\1&-1\\1&-1\end{pmatrix},$$ and then measuring in the (enlarged) computational basis. This corresponds to performing the POVM on the input state whose elements are the projections onto the rows of $V$, that is, $$F_1= F_2 = \frac12|+\rangle\!\langle +|, \qquad F_3 = F_4 = \frac12|-\rangle\!\langle -|.$$ This POVM is, to all intent and purposes, equivalent to the PVM in the eigenbasis of $X$, but the post-measurement outcomes corresponding to different identical POVM elements are all different.

You might find this at odds with your reasoning based on the standard formula for post-measurement states, but I think the problem there is that POVMs do not uniquely specify post-measurement states. Even sticking with the "best" choice of post-measurement outcomes, following e.g. the argument I put forward in this answer, the rule is not quite that if the PVM has operators $(P_k)_k$ then post-measurement states are $P_k \rho P_k/\operatorname{tr}(P_k\rho)$. I'd say it's more appropriate to state that given a POVM $(F_k)_k$, you take a set of "square roots", $A_k$ such that $F_k=A_k^\dagger A_k$, and then take as post-measurement outcomes $A_k \rho A_k^\dagger/\operatorname{tr}(A_k^\dagger A_k \rho)$.

The difference is that even if the POVM is a PVM, these "square roots" are only defined up to isometries: from $(P_k)_k$ you can take $A_k=V_k P_k$ for any set of isometries $V_k$. These won't change the POVM, but will change the post-measurement states which now become $V_k P_k\rho P_kV_k^\dagger/\operatorname{tr}(P_k\rho)$, and this explains how you can get to the situation in my example above, where identical POVM element can correspond to different measurement outcomes.

As to the titular question Can a POVM with repeated elements be considered a PVM? I think it's really more a matter of semantics than anything else. If you strictly define the terms as you did, then sure, such POVMs are not really PVMs. But they still produce effectively identical results from the point of view of the generated measurement statistics. Whether the potentially different post-measurement states are important depends entirely on the context that is being considered.

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  • $\begingroup$ I absolutely agree about the square roots, that's why I had to switch to using Kraus operators and ask what happens when Kraus operators are scalar multiples of each other. The question is thus whether having two identical Kraus operator maps, one corresponding to a POVM and another corresponding to a PVM, should actually be considered identical. And I guess the answer is that Kraus operators don't care about the measurement device, but POVMs/PVMs explicitly care about the measurement device $\endgroup$ Jul 28, 2023 at 12:42
  • $\begingroup$ @QuantumMechanic I think it's really a matter of semantic at this point. With "identical Kraus operators" I think identical sets of operators, in which case they both necessarily produce the same identical types of measurements. I assume you mean different sets of Kraus operators containing different numbers of copies of the same operators, or something of this sort. In which case, I agree there can be differences in the ways they are implemented etc, as per my example. $\endgroup$
    – glS
    Jul 31, 2023 at 18:30
  • $\begingroup$ It's true, there really isn't much left here. There are POVMs for which one may bin the outcomes to essentially have a PVM, but never where one must treat it as a PVM $\endgroup$ Jul 31, 2023 at 18:34

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