2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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

1
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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

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.