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A POVM, as defined for example in (Peres and Wooters 1991), is defined by a set of positive operators $\mu(a)$ satisfying $\sum_a \mu(a)=\mathbb 1$. We do not require the $\mu(a)$ to be projectors, nor commuting operators, like is done in the case of projective measurements. The probability of getting the $a$-th outcome when the state is $\lvert\psi\rangle$ is then $\langle\psi\rvert \mu(a)\rvert\psi\rangle$.

One way to understand the significance of such a general kind of measurement operation is via Naimark's theorem. Following the notation (sorta) used in Watrous' book (chapter 5), we can write $$\mu(a)=A^\dagger(\mathbb 1_{\mathcal X}\otimes E_{a,a})A,$$ where $A$ is an isometry, satisfying $A^\dagger A=\mathbb 1$, whose action we can write as $$AX=\sum_a\sqrt{\mu(a)}X\otimes\lvert a\rangle.$$ Should I understand this to be saying that the measurement $(\mu(a))_a$ corresponds to correlating $\lvert\psi\rangle$ with a bigger system, and then "looking at the enlarged system" (that is, performing a standard projecting measurement)?

If so, it should be sensible to write the state in the enlarged system where the projective measurement actually occurs as $$A\lvert\psi\rangle=\sum_a[\sqrt{\mu(a)}\lvert\psi\rangle]\otimes\lvert a\rangle.$$ However, writing the spectral decomposition of $\mu(a)$ as $\mu(a)=\sum_k\lambda^a_k\lvert\lambda^a_k\rangle\!\langle\lambda^a_k\rvert$, we would have $$A\lvert\psi\rangle=\sum_{a,k}\sqrt{\lambda^a_k}\lvert\lambda^a_k\rangle\!\langle\lambda^a_k\rvert\psi\rangle\otimes\lvert a\rangle.$$ But is such an object a proper state? The normalisation reads $$\|A\lvert\psi\rangle\|^2=\sum_a \|\sqrt{\lambda^a_k}\lvert\lambda^a_k\rangle\!\langle\lambda^a_k\rvert\psi\rangle\|^2=\sum_{a,k} \lambda^a_k \lvert\langle\lambda^a_k\rvert\psi\rangle\rvert^2,$$ and it does not seem so obvious whether this always sums to one (and intuitively I would say it doesn't, due to its dependence on $\lvert\psi\rangle$).

Is it wrong to think of $A\lvert\psi\rangle$ as a proper state? Or if not, how can we see that it is always normalised?

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First of all, $\|A|\psi\rangle\|^2 = (A|\psi\rangle)^\dagger(A|\psi\rangle) = \langle \psi | A^\dagger A|\psi\rangle = 1$.
To see why your last expression equals to $1$ note that $$ \langle\psi\rvert \mu(a)\rvert\psi\rangle = \sum_{k} \lambda^a_k \lvert\langle\lambda^a_k\rvert\psi\rangle\rvert^2 ~~\text{and}~~ \sum_a \mu(a) = \mathbb 1 $$

So, it's perfectly legal state from mathematical point of view. Just like in the case of purification.

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  • $\begingroup$ so I guess the answer to the question's title would be yes, good to know =). I still find not obvious the fact that indeed that sum gives one (although it is clear that it must be the case mathematically), but it is probably better to ask about that specific point in a separate question $\endgroup$ – glS May 19 at 12:04

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