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I suspect this is the obvious result of something in operator algebras, but that's far outside my field.

Recall that a projection-valued measure is a map $E$ from a sigma-algebra $\mathcal{F}$ on measure space $\Omega$ to the set of orthogonal projections on a Hilbert space $H$ that is countably additive and had $E(\Omega)=Id$. It is a special case of a positive operator-valued measure (POVM), in which the operators $E(F)$ need only be positive-semidefinite. (In quantum mechanics, the intuition is that a state is given by a trace-class operator $\rho$ on $H$, and the physical "probability of measuring $F$" is defined to be $tr(E(F)\rho)$.)

More generally, to describe what happens to a state after it is measured, the notion of a quantum instrument is used. It is defined as a map $\mathcal{I}$ from $\mathcal{F}$ to the set of completely positive operators on $\mathcal{L}_1(H)$ (the trace-class operators on $H$ which intuitively represent states), such that

  1. $\mathcal{I}$ is countably additive
  2. $tr(\mathcal{I}(F)(\rho))\le tr(\rho)$ for $\rho$ positive-semidefinite, $F\in\mathcal{F}$
  3. $tr(\mathcal{I}(\Omega)(\rho))=tr(\rho)$.

It is easy to see that any instrument induces a POVM (not necessarily projection-valued), given by $E_{\mathcal{I}}(F)=\mathcal{I}(F)^*(Id)$. However, the map from instruments to positive operator-valued measures is not injective, as different instruments can induce the same POVM.

My question is: Is that map surjective? Namely, given a POVM $E$ can we find an instrument $\mathcal{I}$ such that $E_{\mathcal{I}}=E$? I would even be interested in just the case when $E$ is a projection-valued measure.

Notes:

  1. When $H$ is finite-dimensional, any projection-valued measure is discrete, and given (wlog) by $E(\{x_k\})|\psi\rangle=|k\rangle\langle k|\psi\rangle$. Then a valid instrument is generated by $\mathcal{I}(\{x_k\})\rho=|k\rangle\langle k|\rho|k\rangle\langle k|.$ (Here $|k\rangle$ are the eigenvectors.)
  2. The above construction does not generalize directly to the infinite-dimensional case. Indeed there is a theorem (Theorem 4.1.2 in the book by Holevo), that says an infinite-dimensional instrument can never be repeatable in the sense that $\mathcal{I}(F_1)\mathcal{I}(F_2)=\mathcal{I}(F_1\cap F_2)$ unless the measure is discrete.

But my question doesn't need it to be repeatable (i.e. we don't care what happens to the state afterwards), we just want $\mathcal{I}(F)^*(Id)$ to give $E_{\mathcal{I}}(F)$. Can we "extend" the measure from its value on the identity?

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    $\begingroup$ Sorry, I meant to write $\mathcal{I}(f)(\rho') := \mathrm{tr}(E(F)\rho') \rho$. This amounts to simply measuring $\rho'$, discarding it, and preparing the state $\rho$ instead. $\endgroup$ Commented May 19, 2022 at 8:15
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    $\begingroup$ @YonahBorns-Weil your Note 1 is slightly wrong --- you're assuming that $E(\{x_k\})$ is a minimal projection. Nonetheless, if $\Omega = \{x_1, \ldots, x_n\}$ is finite you can still define $I(\{x_k\})\rho = \sqrt{E(\{x_k\})}\rho\sqrt{E(\{x_k\})}$, and this works for any POVM, not just projection-valued. $\endgroup$
    – Nik Weaver
    Commented May 19, 2022 at 14:37
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    $\begingroup$ Maybe there is some confusion about what the OP means by $\mathcal{I}(F)^*(Id)$. What my trivial construction achieves is the equation $\mathrm{tr}(\mathcal{I}(F)(\rho')) = \mathrm{tr}(E_{\mathcal{I}}(F) \rho')$ for all $\rho'$ and all $F$. Is this not what is being asked? $\endgroup$ Commented May 19, 2022 at 15:28
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    $\begingroup$ @TobiasFritz: Looks fine to me, for what it's worth. For example in the OP's note 1, what you're saying is that $I(x_k)(\rho)= \textrm{tr }(E(x_k)\rho) \langle \psi, \cdot\rangle \psi$ would have worked too for any fixed $\|\psi\|=1$ (one doesn't need to update to the "correct" state $k$ to have a valid instrument in the formal sense), and one can still recover $E$ from this artificial $I$ also, in the same way. $\endgroup$ Commented May 19, 2022 at 16:03
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    $\begingroup$ @TobiasFritz of course! I feel a bit stupid now; I can't believe I didn't think of just sending everything to one state! $\endgroup$ Commented May 19, 2022 at 16:54

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