# How do we show that a measurement is a projective measurement

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?

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

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.