0

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
  • 1
Yrogirg
  • 2,609

1 Answers1

4
  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
  • 2,324
  • Point 2 is misleading. You can have entropy increase even if the evolution is reversible! How did we explain entropy increase in classical mechanics? Remember that CM is deterministic and reversible. Entropy is a perspectival quantity. Given a closed system, which always evolves reversibly, define a split into accessible and inaccessible degrees of freedom. A given value of the inaccessible dof, corresponds to multiple configurations of the whole system. This defines an entropy in the usual way of counting microstates. Thus while the whole system is reversible, there can be entropy increase. – Andrea Feb 02 '21 at 08:17