In order to show that a measurement is a projective measurement, is it sufficient to prove that the measurement operators $\{M_{m}\}$ satisfy the properties:

  1. Hermitian: $M_{m}^{T*} = M_{m}$
  2. Indempotent: $M_{m} ^{2} = M_{m}$
  3. The operators satsify the completeness relation
  4. The operators have orthogonal eigenstates

Is this sufficient? Are all of these conditions necessary? Is there a better way to ascertain whether a measurement is a projective measurement?

  • 2
    $\begingroup$ Number 4 is a consequence of numbers 2 and 3. Number 2 is what specifically identifies projective measurements. You always need 1 and 3. $\endgroup$
    – DaftWullie
    Jul 10, 2023 at 10:55

1 Answer 1


A generalized measurement $M = \{E_k\}_k$ will be projective if all its POVM elements are projectors, implying that for all $k$ you will have $E_k^2 = E_k$. Guaranteed that you have a list of positive operators summing to identity, i.e., $\sum_k E_k = \mathbb{I}$, it suffices to test the idempotent property.


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