A Collapse and projection-valued measures

burakumin
Messages
84
Reaction score
7
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##
 
Physics news on Phys.org
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.

burakumin said:
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.
 
  • Like
Likes bhobba
Truecrimson said:
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 definately 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
$$ \rho' = \frac{1}{\textrm{tr} (\rho \cdot \mu_A(S)) } \cdot \mu_A(S) \cdot \rho \cdot \mu_A(S) $$
 
  • Like
Likes Truecrimson
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...

Similar threads

Replies
4
Views
2K
Replies
1
Views
2K
Replies
77
Views
8K
Replies
6
Views
3K
Replies
1
Views
819
Replies
1
Views
1K
Back
Top