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It is often claimed that POVMs represent the most general measurement statistics possible. But what is the justification for this claim? Textbooks and university courses generally try to build up to the claim starting from the measurement postulate, which can either be from the perspective of measurement of observables/in a basis (most courses and books) or general measurements (to use Nielsen and Chuang's term). However, I am not sure how to conclusively demonstrate the generality of POVMs. Here are some attempts, and why I think they fail:

  1. From the perspective of measurement in a basis being fundamental, a general measurement $\{ M_m \}$ (and the POVM $\{ M_m^{\dagger} M_m \}$ it implements) can be implemented on $|\psi\rangle_A$ by finding a unitary $U$ such that $U |\psi\rangle_A |0\rangle_B = \sum_m (M_m |\psi\rangle_A) |m\rangle_B$ and measuring the $B$ subsystem of this state in the basis $\{|m\rangle_B\}$. However, this does not rule out the possibility that non-POVM statistics could be obtained from similar protocols.
  2. From the perspective of general measurements being fundamental, this would seem to "define away" the problem, but it still does not rule out the possibility of elaborate protocols involving larger Hilbert spaces producing non-POVM statistics.
  3. More generally, the framework of completely positive linear maps may not capture the evolution of a system that is entangled with its environment: $\rho_A \mapsto \mathrm{Tr}_B[ U \rho_{AB} U^{\dagger} ]$ may not be possible to describe as a completely positive linear map (see e.g. here or here for discussions). So we can't just rely on arguments from complete positivity of evolution (e.g. expressing CP maps as Kraus operators, Stinespring dilation) to constrain the statistics.

How can we then demonstrate that POVMs are indeed the most general possible type of measurement on a system?

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  • $\begingroup$ You might get a better response to this question at the Quantum Computing Stack Exchange: quantumcomputing.stackexchange.com $\endgroup$
    – hft
    Sep 13, 2023 at 18:47
  • $\begingroup$ Note that 1 applies to arbitrary quantum states, not just pure states, as can be seen by convexity. This should cover 3 as well, then. It also says nothign about the dimension of the Hilbert spaces ($U$ can be anything) so I'm not sure what you mean in 2 $\endgroup$ Sep 13, 2023 at 19:31
  • $\begingroup$ @QuantumMechanic edited for clarity; these are broad approaches to proving the generality of POVMs, and the issues I see with each approach $\endgroup$
    – aquohn
    Sep 14, 2023 at 11:03

2 Answers 2

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A measurement with outcomes $i$ associates probabilities $p_i(\rho)$ to every input density matrix $\rho$.

If you assume that $p_i$ is linear (and a probability distribution, $p_i\ge0$ and $\sum p_i=1$), this immediately implies that POVMs are the most general measurement possible.

Thus, the only non-trivial requirement is linearity, which is indeed a basic tenet of quantum mechanics, and without which things like the ensemble interpretation of density matrices and really the entire theory don't make sense any more.


The reason, in short, is that any such map can then be written as $$ p_i(\rho) = \mathrm{tr}(F_i\rho)\ . $$ $p_i(\rho)\ge0$ implies $F_i\ge0$ (or can be chosen such), and $\sum p_i=1$ implies $\sum F_i=I$. Thus, $\{F_i\}$ describes a POVM. (Which you can decompose as $F_i = M_i^\dagger M_i$ if you like.)

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  • $\begingroup$ This is good, thanks! I think we also need to state that the functionals $\mathrm{tr}(E_{ij} \cdot)$ (where $E_{ij} = |i\rangle \langle j|$) form a basis for the space of all linear functionals (which also works in infinite dimensions for a separable space, I think). $\endgroup$
    – aquohn
    Sep 14, 2023 at 11:09
  • $\begingroup$ However, the reduced dynamics of the subsystem of an entangled state may not be linear (see e.g. arxiv.org/abs/2003.00460 for recent work on this). I'm wondering if that loophole can lead to non-POVM statistics. $\endgroup$
    – aquohn
    Sep 14, 2023 at 11:15
  • $\begingroup$ My answer is only for finite dimensions. $\endgroup$ Sep 14, 2023 at 14:46
  • $\begingroup$ I don't think this "the effective dynamics is nonlinear" makes much sense (it basically only happens if you distort the perspective, and linearity is there on the fundamental level). It is certainly not a loophole. If quantum mechanics were properly non-linear, you would get all kind of obscure things (postselection, solving NP-hard problems, the whole theory of mixed states breaks down, ... ) $\endgroup$ Sep 14, 2023 at 14:46
  • $\begingroup$ @NorbertSchuch I agree that the overall dynamics must be linear to avoid all the badness you've mentioned - what I'm asking about is precisely whether the "distorted perspective" you've mentioned can make it seem like a subsystem has been measured with a non-POVM measurement. $\endgroup$
    – aquohn
    Sep 14, 2023 at 19:12
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A measurement is an interaction that produces a record that can be copied as many times as you like. When you measure a system you often do this by letting it interact with a measuring instrument or the environment that stores a record of the result of that interaction but you don't interact with the system itself. In this situation the information you have about the system you wanted to measure is described by a POVM, see the note on p.9 of:

https://arxiv.org/abs/quant-ph/0408125

If you interact with the system as well as the measuring apparatus or environment then the measurement won't necessarily be described by a POVM.

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  • $\begingroup$ Thanks for the link! I'm hoping to find an explicit example of measurements that cannot be described by a POVM, but the link claims Eq. 100 of arxiv.org/abs/quant-ph/0205039 demonstrates a non-POVM measurement obtained by measuring a state after it has undergone some evolution, but the cited equation does not seem to support that notion - any idea what that's about? $\endgroup$
    – aquohn
    Sep 14, 2023 at 19:10
  • $\begingroup$ I haven't managed to find such a non-povm measurement example in that paper. It's possible that I have missed the example, or that Zurek saw an earlier draft of the paper or something like that. $\endgroup$
    – alanf
    Sep 15, 2023 at 6:58

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