Does the dilation in Naimark's theorem produce a state?

Question

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?

Answer 1

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.

Answer 2

Looking back at this, there was no need to pass through the spectral decomposition of $\mu(a)$ to conclude that $A|\psi\rangle$ is a state. If $A$ is an isometry, then $A|\psi\rangle$ is automatically a state for any $|\psi\rangle$. Another way to put it is to observe that $$\|A|\psi\rangle\|^2= \sum_a \langle \psi|\sqrt{\mu(a)}\sqrt{\mu(a)}|\psi\rangle=\langle\psi|\psi\rangle=1.$$