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I want to know a definition of a von Neumann measurement. Because I can't find this concept referenced correctly in internet, and what differentiates it from a POVM, that by definition is the measurement with operators $\{E_i\}$ that

  • Are positive definite
  • $\sum_i E_i = \mathbb{I}$
  • Probability preserving.

But what condition do we have to add to this POVM measurement to have a von Neumann measurement?

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  • $\begingroup$ Just looking at this I gather that the operators $E_i$ in the POVM have to be self-adjoint (Hermitian). Von Neumann requires only a single operator that is self-adjoint and does not have to be positive definite. $\endgroup$
    – Kurt G.
    Commented Apr 25, 2022 at 12:23

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  1. You only need the first two requirements for a POVM. You need Hermitian operators $\{E_i\}$ such that $E_i\ge0$ and $\sum_i E_i=I$. I'm not sure what you mean with them being "probability preserving".

  2. Presumably, by a "von Neumann measurement" you mean a projective-valued measurement, in this context often abbreviated with PVM. These are POVMs whose elements are projections. In other words, collections $\{E_i\}$ such that $E_i\ge0$, $\sum_i E_i=I$, and $E_i^2=E_i$.

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  • $\begingroup$ "Probability preserving" = "sum of outcome probabilities = 1". $\endgroup$
    – Norbert Schuch
    Commented Apr 29, 2022 at 12:57
  • $\begingroup$ A key property which is worth pointing out separately is that the $E_i$ must project onto orthogonal subspaces. While these follows from the conditions you list, it is quite fine-tuned, as all three of those are required. E.g., a SIC-POVM has $E_i$ which are only proportional to projectors, which alllows them to no longer be orthogonal. $\endgroup$
    – Norbert Schuch
    Commented Jan 28 at 12:24
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A von Neumann measurement is defined by an observable (Hermitian operator), the spectral decomposition of which gives a set of mutually orthogonal projectors (Nielsen Sec. 2.2.5) .

So the additional condition is that the POVM effects are orthogonal projectors, i.e. $E_i E_j = \delta_{ij} E_i$.

Update thanks to the comment: The condition that $E_i E_i = E_i$ is equivalent with $E_i E_j = \delta_{ij} E_i$, which is explained also in projectors sum to identity.

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  • $\begingroup$ Since the $E_i$ are hermitian and sum to the identity, the condition $E_i^2=E_i$ indeed implies orthogonality. (Otherwise, their sum would have an eigenvalue >1.) -- P.S.: If you have a mistake in an answer, generally edit it rather than deleting and re-posting it. $\endgroup$
    – Norbert Schuch
    Commented Jan 28 at 12:22

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