It feels like the transition from a sea of probability of a droplet of precision (or wave, depending what you're looking for) would be a loss of entropy, if so wouldn't this violate the Second Law? Or does Heisenberg's insistence that we cannot ask questions about what happens before observation exempt it from such considerations?
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$\begingroup$ A measurement converts an eigenstate of one observable to an eigenstate of some other observable. In what sense, and for what reason, would you expect to associate a higher value of entropy with one than with the other? $\endgroup$– WillOCommented Oct 30, 2023 at 21:52
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This is an excellent question and is the subject of ongoing research in the fields of quantum thermodynamics and quantum information theory.
Projective measurements appear to contradict the laws of thermodynamics in three ways:
- They do not conserve energy
- They allow for an arbitrary decrease in entropy (Ideal Projective Measurements Have Infinite Resource Costs)
- They permit cooling to absolute zero using finite resources, which violates Nernst’s formulation of the 3rd law (Landauer vs. Nernst: What is the True Cost of Cooling a Quantum System?).
Recently, a potential resolution to this apparent contradiction was proposed in the paper Quantum measurements and equilibration: the emergence of objective reality via entropy maximisation, where the authors models the process of measurement as the system of interest and its environment undergoing an equilibration process, thus being driven by an increase in entropy.
