Graduate Collapse and projection-valued measures

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SUMMARY

The discussion centers on the collapse of quantum states using projection-valued measures (PVM), specifically addressing the correct formulation of state collapse when measuring an observable A associated with a projection-valued measure μ_A. The participants clarify that the formulas initially proposed for the collapsed states are incorrect, particularly in cases of degenerate eigenvalues. The correct expressions for the post-measurement state are established as: ρ' = (1/tr(ρ · μ_A({a}))) · μ_A({a}) · ρ · μ_A({a}) and ρ' = (1/tr(ρ · μ_A(S))) · μ_A(S) · ρ · μ_A(S), with Lüders rule being the appropriate method for handling degenerate cases.

PREREQUISITES
  • Understanding of quantum mechanics and measurement theory
  • Familiarity with projection-valued measures (PVM)
  • Knowledge of Lüders rule in quantum state collapse
  • Basic proficiency in mathematical notation used in quantum mechanics
NEXT STEPS
  • Study the implications of Lüders rule in quantum mechanics
  • Explore the concept of projection-valued measures in greater detail
  • Investigate the role of degenerate eigenvalues in quantum state measurements
  • Review the Born rule and its interpretations in quantum mechanics
USEFUL FOR

Quantum physicists, graduate students in quantum mechanics, and researchers focusing on measurement theory and state collapse in quantum systems.

burakumin
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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##
 
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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.
 
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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 definitely 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
$$ \rho' = \frac{1}{\textrm{tr} (\rho \cdot \mu_A(S)) } \cdot \mu_A(S) \cdot \rho \cdot \mu_A(S) $$
 
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Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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