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If we stick with Heisenberg picture where density matrix $\rho$ is constant, how do we account for entropy increase?

I've read the answer to State collapse in the Heisenberg picture but I don't see how the explanation can be used to incorporate the increase of entropy, if it is defined like $$ S = \operatorname{Tr} (\rho \ln \rho). $$

Or is Shroedinger picture preferable for irreversible evolution?

Community
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Yrogirg
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1 Answers1

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  1. You're missing a minus in the entropy definition - $S=-Tr(\rho\ln\rho)$
  2. Entropy of a unitarily evolving system (doesn't matter in which picture) is conserved (The entropy is a trace of a function of the density matrix "operator" thus it depend solely on the eigenvalues of it's input operator, but the eigenvalues of the density matrix don't change under unitary transformation).
  3. After measuring the system, you have the system in a pure state - thus the entropy of the system after measurement is identically 0.
  4. So how can a system increase it's entropy? The generalization of unitary evolution operators are Kraus maps, which don't preserve density matrix eigenvalues.. See https://en.wikipedia.org/wiki/Quantum_operation#Kraus_operators
Alexander
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