3
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • $\begingroup$ 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

2 Answers 2

Reset to default
2
$\begingroup$

  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 $E_i^2=E_i$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ "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
1
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.