Collapse and projection-valued measures

• A
Let suppose I have an observable ##A## with associated projection-valued measure ##\mu_A##
$$A = \int_{a \in \mathbb{R}} a \cdot \textrm{d}\mu_A(a)$$
for a system in the (possibly mixed) state ##\rho##. Let ##S \subset \mathbb{R}## be a measurable subset and let ##Z = \mu_A(S)## be the observable equating 1 if ##A## falls in ##S## and 0 otherwise.

Is this statement meaningful and correct:

If measuring ##A##, with probability ##\textrm{tr}( \rho \cdot \mu_A(S) )## the result will be a point in ##S## (let's call it ##a##) and the system will collapse to state ##\rho' = \mu_A(\{a\})##

Again is this statement meaningful and correct:

If measuring ##Z##, with probability ##\textrm{tr}( \rho \cdot \mu_A(S) )## the result will be 1 and the system will collapse to state $$\rho' = \frac{1}{\textrm{tr} (\rho \cdot \mu_A(S))} \cdot \int_{a \in S} \textrm{tr} ( \rho \cdot \mu_A(\{a\}) ) \cdot \textrm{d} \mu_A(a)$$ with any subsequent measurement of ##A## producing a value inside ##S##

Rethinking about that, it seems that my formulas for states after collapse are incorrect as ##\mu_A(\{a\})## is not a state at all in case ##a## is degenerate eigenvalue. But what would be the correct answer?

My question can basically be rephrased as "how do you express collapse with PVM?"

Bhobba, I think the OP just want an FAPP collapse to do calculation.

it seems that my formulas for states after collapse are incorrect as ##\mu_A(\{a\})## is not a state at all in case ##a## is degenerate eigenvalue. But what would be the correct answer?"

Typically one projects the state before the PVM measurement onto the degenerate subspace. This is called Lüders rule.

bhobba
Bhobba, I think the OP just want an FAPP collapse to do calculation.

Typically one projects the state before the PVM measurement onto the degenerate subspace. This is called Lüders rule.

Thanks Truecrimson, "Lüders rule" seems definetly the appropriate keyword. So AFAIU the correct answers would be:
$$\rho' = \frac{1}{\textrm{tr} (\rho \cdot \mu_A(\{a\})) } \cdot \mu_A(\{a\}) \cdot \rho \cdot \mu_A(\{a\})$$
and
$$\rho' = \frac{1}{\textrm{tr} (\rho \cdot \mu_A(S)) } \cdot \mu_A(S) \cdot \rho \cdot \mu_A(S)$$

Truecrimson