Quantum Computing Asked on June 29, 2021
A POVM is typically defined as a collection of operators ${mu(a)}_{ainSigma}$ with $mu(a)inmathrm{Pos}(mathcal X)$ positive operators such that $sum_{ainSigma}mu(a)=I$, where I take here $Sigma$ to be some finite set and $mathcal X$ some vector space (using the notation from Watrous’ TQI).
On the other hand, when discussing observables in a measure-theoretic framework, given a nonempty set $Omega$, a $sigma$-algebra $mathcal F$ on $Omega$ (in other words, given a measurable space $(Omega,mathcal F)$), and denoting with $mathcal E(mathcal X)$ the set of effects on the space $mathcal X$, that is, the set of Hermitian operators $E$ such that $0le Ele I$, we say that a mapping $mathrm A:mathcal Ftomathcal E(mathcal X)$ is an observable if, for any state $psiinmathcal X$, the function
$$mathcal Fni Xmapsto langle psi|mathrm A(X)psirangleinmathbb R$$
is a probability measure. Here I’m taking definition and notation from Heinosaari et al. (2008).
In other words, $mathrm A$ is an observable iff, for all $psiinmathcal X$, the triple $(Omega,mathcal F,mathrm A_psi)$ is a probability space, with $mathrm A_psi$ defined as $mathrm A_psi(X)=langle psi|mathrm A(X)psirangle$.
I’m trying to get a better understanding of how these two different formalisms match.
In particular, an observable as thus defined is closer to a POVM than an "observable" as usually defined in physics (which is just a Hermitian operator), right?
Are these observables equivalent to POVMs? That is, does any such observable correspond to a POVM, and vice versa?
I can see that a POVM can be thought of as/is the map $mu:Sigmatomathrm{Pos}(mathcal X)$, which then extends to a map $tildemu:2^{Sigma}tomathrm{Pos}(mathcal X)$ such that $tildemu(2^Sigma)=1$, which is then an observable.
However, I’m not sure whether any observable also corresponds to such a POVM.
These two definitions define the same concept: the POVM measurement. The observable definition is how POVM is defined for use in the case of infinite index set and dimension (see e.g. POVM) and POVM definition in the question is how it is simplified for use in the finite case. If you are working in finite dimensions, the two constructions are equivalent.
POVM from observable
We will construct a POVM ${mu(a)}_{ainSigma}$ from the observable $A: mathcal{F} to mathcal{E}(mathcal{X})$. Begin by choosing a finite $Sigma subset mathcal{F}$ such that $cup_{ainSigma}a = Omega$ and $a cap b = emptyset$ for every $a, binSigma$. This serves as the index set for the POVM and establishes the connection between the two definitions. Define
$$ mu(a) := A(a). $$
Note that this defines valid POVM elements because $0 le A(X) le I$ for every $Xinmathcal{F}$ and
$$ sum_{ainSigma} mu(a) = sum_{ainSigma} A(a) = Aleft(bigcup_{ainSigma} aright) = A(Omega) = I $$
where in the second equality we used $sigma$-additivity of $A_psi$ for every $|psirangle$ and in the last step we used the fact that $A_psi$ is a probability measure for every $|psirangle$.
Observable from POVM
(This merely fills in some details in the construction you sketched in the question.)
We will construct an observable from the POVM ${mu(a)}_{ainSigma}$. Let $Omega = Sigma$ and $mathcal{F} = mathcal{P}(Sigma)$. For $Xinmathcal{F}$ define
$$ A(X) := sum_{ain X}mu(a) $$
where we are using the fact that $X$ is finite. Note that the right-hand side is a valid effect. We need to show that for every $|psirangle$, $A_psi$ is a probability measure. It is clear that for every $Xinmathcal{F}$, $A_psi(X) in [0, 1]$ and $A_psi(emptyset) = 0$. Now, $mathcal{F}$ is finite, because $Omega$ is finite. Thus, to show $sigma$-additivity we just need to show additivity. Let $X_{i=1,dots,k}$ be a collection of disjoint subsets of $Omega$, then
$$ A_psileft(bigcup_{i=1}^k X_iright) = sum_{aincup_{i=1}^k X_i} mu_psi(a) = sum_{i=1}^k sum_{ain X_i} mu_psi(a) = sum_{i=1}^k A_psi(X_i) $$
where $mu_psi(a) = langlepsi|mu(a)psirangle$ and in the second equality we used the fact that $X_i$ are disjoint.
Remark: Note that the definition of observable appears to be more flexible: it doesn't just give rise to a single POVM. Instead, every partitioning of $Omega$ using the elements of the $sigma$-algebra gives rise to a potentially different POVM. However, this fact corresponds to the trivial observation that you can obtain a new POVM by grouping and adding elements of a given POVM.
Correct answer by Adam Zalcman on June 29, 2021
Upon some more reflection, the answer is probably as follows.
Let $mathrm A$ be an observable according to the definition in the question, and assume $Omega$ is finite. Then any $Xinmathcal F$ is also some finite subset of $Omega$. By definition of observable, we require the mapping $mathrm A_psi$ to be additive and non-negative, and therefore $$langle psi|mathrm A(X)psirangle = sum_{omegain X}langle psi|mathrm A({omega})psirangle ge0.$$ It follows that, for any $omegainOmega$, the operator $mathrm A({omega})$ is positive semi-definite: $mathrm A({omega})ge0$.
Furthermore, asking $mathrm A_psi$ to be a probability measure also means that $mathrm A_psi(Omega)=1$ for all $psi$, that is, $sum_{omegainOmega}A({omega})$ to be an operator whose expectation value is $1$ on any state $psi$. This implies $sum_{omegainOmega}A({omega})=I$.
It follows that, defining $mu(omega)equiv A({omega})$, we get the definition of POVM as collection of positive semi-definite operators summing to the identity.
Answered by glS on June 29, 2021
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