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?

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

1 Answer 1


Generally no, but they can be invertible.


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

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 $\text{rank}(A_i)<d\,,$ 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.)

  • $\begingroup$ 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. $\endgroup$
    – Rammus
    Commented Oct 30, 2023 at 10:37
  • $\begingroup$ I agree; sorry about that. I interpreted it incorrectly. I have updated my answer accordingly. $\endgroup$
    – FDGod
    Commented Oct 30, 2023 at 16:49

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