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Let $F\equiv\{F^a\}_a$ be a POVM in some finite-dimensional Hilbert space $\mathcal X$. It is well-known that one can always understand $F$ as a projective measurement (PVM) in an isometrically enlarged space. More precisely, defining $V_F:\mathcal X\to\mathcal Y\otimes \mathcal X$ as $$V_F u \equiv \sum_a |a\rangle\otimes (\sqrt{F^a}\,u), \qquad u \in\mathcal X$$ one verifies that $V_F$ is an isometry and $F^a=V_F^\dagger (\mathbb{P}_a\otimes I)V_F$ with $\mathbb{P}_a\equiv|a\rangle\!\langle a|$.

Let me now consider a slightly different scenario: let $W:\mathcal X\to\mathcal Y\otimes\mathcal X$ be some isometry (not necessarily tied to any specific POVM). We can now build a corresponding POVM describing evolution through this isometry followed by a PVM on the ancillary space. This reads $$(F_W)^a \equiv W^\dagger (\mathbb{P}_a\otimes I)W.$$ Any isometry can be decomposed wrt a basis for $\mathcal Y$ as $$W u= \sum_a |a\rangle\otimes (W_a u), \qquad u\in\mathcal X,$$ but in general, the operators $W_a\equiv (\langle a|\otimes I)W:\mathcal X\to\mathcal X$ are not positive semidefinite operators. The only thing we can say about them is that, due to the isometric constraint $W^\dagger W=I$, they must satisfy $\sum_a W_a^\dagger W_a=I$ (i.e. they can always be understood as the Kraus operators for some channel).

But then again, given $F_W$, I could follow the construction in Naimark's theorem to obtain another isometry, call it $V_{F_W}$, whish has the form $$V_{F_W} = \sum_a |a\rangle\otimes \sqrt{(F_W)^a}.$$

I'm not entirely sure what to make of this: $F_W$ is clearly a POVM for any isometry $W$, but then it doesn't seem like $V_{F_W}$ equals $W$ in general (as $W_a$ won't necessarily be positive semidefinite, while the corresponding blocks of $V_{F_W}$ are). But still, both isometries feel like they are performing a similar kind of operation. Is there a way to see whether they are indeed "equivalent", at least as far as the measurement at hand is concerned?

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  • $\begingroup$ physics.stackexchange.com/questions/226477/… I think this might be relevant, although the answer is not fully clear to me. If you get it could you please help me with it? $\endgroup$ Apr 19, 2022 at 10:50
  • $\begingroup$ Please don't use Watrous notation, reading this is torture. $\endgroup$ Apr 19, 2022 at 14:48
  • $\begingroup$ @MateusAraújo I mean, that's pretty subjective, but which part are you referring to exactly? Using $\mathcal X$ rather than $\mathcal H$ for spaces? Or $\mu(a)$ instead of $\mu_a$? The latter I use mostly because $\sqrt{\mu_a}$ doesn't render very nicely in latex imo, and it's easier to adapt when I have to use an index to denote POVMs as in here $\endgroup$
    – glS
    Apr 19, 2022 at 16:25
  • $\begingroup$ In general that this notation gratuitously breaks the conventions used in hundreds of papers, so we have to look at the definition of each symbol to know what it means, instead of just knowing. Using lowercase Latin letters for operators, for example. These are vectors in pretty much any paper you'll find on the arXiv. Also, POVM elements are $E^a$ or $F^a$, projectors are $P^a$ or $\Pi^a$. Now a real-valued measured is written as $\mu(a)$. Using this notation for operator-valued measures is just evil. $\endgroup$ Apr 20, 2022 at 6:02
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    $\begingroup$ A ket is just a vector, it doesn't need to be normalised. You're just making sure nobody can understand what you write. $\endgroup$ Apr 21, 2022 at 6:38

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You could define $$ V_F^S u \equiv \sum_a |a\rangle\otimes (S^a\sqrt{F^a}\,u), \qquad u \in\mathcal X, $$ where $S^a$ is any unitary. Then $V_F^S$ is also an isometry and $$ F^a=(V_F^S)^\dagger (\mathbb{P}_a\otimes I)V_F^S. $$

In fact, for any $W$ with the property $F^a=W^\dagger (\mathbb{P}_a\otimes I)W$ there exists a set of $S^a$ such that $W = V_F^S$. It follows from the polar decomposition of $W_a$.

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  • $\begingroup$ ah, brilliant, makes perfect sense. Given any POVM $F_W$ corresponding to measurement through a given isometry $W$, we can find some set of unitaries $S$ such that the associated "Naimark's dilation isometry" $V_F^S$ satisfies $W=V_F^S$. $\endgroup$
    – glS
    Apr 21, 2022 at 9:10

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