6
$\begingroup$

Let $\mu:\Sigma\to\mathrm{Pos}(\mathcal X)$ be some POVM, with $\Sigma$ the finite set of possible outcomes, and $\mathrm{Pos}(\mathcal X)$ the set of positive semidefinite operators on a finite-dimensional space $\mathcal X$. Write the components of the measurement with $\mu(b), b\in\Sigma$.

From Naimark's dilation theorem, we know that $\mu$ can be understood as a projective measurement on an isometrically enlarged space. More precisely, that there is an isometry $U_\mu:\mathcal X\to\mathcal X\otimes\mathcal Y$, for some $\mathcal Y$, such that $\mu(a)=U_\mu^\dagger (I\otimes P_a) U_\mu$ for all $a\in\Sigma$, with $P_a\equiv |a\rangle\!\langle a|$. Explicitly, we can write this isometry as $$U_\mu =\sum_{a\in\Sigma}\sqrt{\mu(a)}\otimes |a\rangle =\begin{pmatrix}\sqrt{\mu(1)}\\ \vdots \\ \sqrt{\mu(|\Sigma|)}\end{pmatrix}.$$

Consider now the notion of extremality of a POVM $\mu$. We say that $\mu$ is extremal if it cannot be written as a (nontrivial) convex combination of other POVMs. Physically, this is interesting because a non-extremal POVM amounts to some degree of "classical uncertainty" added on top of whatever is being measured about the state.

How does the extremality of a POVM $\mu$ reflect on its "Naimark isometry" $U_\mu$? If $\mu$ is extremal, that would mean that $U_\mu$ is an isometry such that each of its $\dim(\mathcal X)\times \dim(\mathcal X)$ blocks, call these $U_\mu(a)\equiv\sqrt{\mu(a)}$, cannot be written as $$U_\mu(a)^2=\frac12(U_{\mu_1}(a)^2+U_{\mu_2}(a)^2).$$ In other words, convex combinations of measurements, in terms of the associated isometries, translate into these weird sums taken after squaring the different blocks building the individual isometries. What kind of structure is this? Is there some known notion for isometries/unitaries corresponding to the impossibility of performing such a type of decomposition?

$\endgroup$
3
  • $\begingroup$ The weird sum formulations uses additional assumptions on the form of the U's, I'd say (basically you don't have to take a square root for U, but can take any X with $X^\dagger X = \mu$). $\endgroup$ Jul 23 at 16:36
  • $\begingroup$ Also note that the same POVM measurement (in your formulation) can have vastly different post-measurement states, and thus vastly different U's. (E.g., a filtering operation, from the pure POVM point, can equally be built by a projetive measurement + classical mixing.) $\endgroup$ Jul 23 at 16:41
  • $\begingroup$ ... Overall, this makes me suspect that, in order to get a meaningful answer, you need to add some quantifier (most likely: Does there exist a U with a special structure). $\endgroup$ Jul 23 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.