# Are POVM elements invertible?

A POVM is a set $$\mathcal{M} = \{A_i : A_i \geq 0, \sum{A_i }= \mathbf{I}\}_{i=1}^m$$ on a Hilbert space $$\mathcal{H}^d$$ of dimension $$d$$, I want to know whether $$A_i$$ can be invertible linear map?

• Yes, it can. A very simple example is when $\mathcal M$ only has one element $I$. Commented Oct 30, 2023 at 12:07

Generally no, but they can be invertible.

Example:

$$\mathcal{M} = \{ A_1, A_2, A_3 \}\,,$$

where \begin{align} A_1 &= a|1\rangle \langle1|\,, \\ A_2 &= a|-\rangle\langle-| \,,\\ A_3 &= I- A_1 - A_2\,. \end{align}

You can clearly see that $$E_1^{-1}, E_2^{-1}$$ does not exist but $$E_3^{-1}$$ does exist.

Since $$\{A_i\}$$ add up to identity, are hermitian and are also positive operators; generally, they will have $$\text{rank}(A_i) and hence non-invertible. But yes, $$\exists A_i$$ such that $$\det(A_i) \neq 0\,.$$

(Also, there are trivial cases where all $$\{A_i\}$$ are invertible, but those are trivial/insignificant in terms of an experiment.)

• I'm not sure this answers the question posed. The question asked whether the $A_i$ can be invertible. The answer is yes they can be, they just don't have to be. Commented Oct 30, 2023 at 10:37
• I agree; sorry about that. I interpreted it incorrectly. I have updated my answer accordingly. Commented Oct 30, 2023 at 16:49