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A QM system (defined by a Hamiltonian $H$), at temperature $T$, is described by a density matrix $\rho(t)$, which has an associated entropy $S(t)=-tr(\rho(t) \ln \rho(t)$.

Now the question is:

Write down the time evolution for the density matrix and calculate $dS/dt$.

The solution in the SM is:

The time evolution of the density operator is given by the Heisenberg equation: $\partial \rho(t)/\partial t = i \hbar [ \rho , H]$, leading to: $$dS/dt = i\hbar tr((1+\rho)[\rho, H]) = i\hbar\{ tr(\rho H) -tr(H\rho)+tr(\rho^{\epsilon} H)-tr(\rho H \rho)\} = 0 $$

Now, I did my calculation and got something different here: $$dS/dt = -tr(\partial \rho / \partial t \ln \rho + \partial \rho /\partial t) = -i\hbar tr( (\ln \rho +1)[\rho, H])$$

Where does the mistake lie here?

The solution manual can be found on library genesis, on page 166 there's the problem and its solution.

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  • $\begingroup$ Why is there a mistake? Your answer is also zero. $\endgroup$
    – ComptonScattering
    Commented Jan 31, 2017 at 12:25
  • $\begingroup$ He got a different term for $dS/dt$, but I understand that the end result that it's either way zero what's matter in the end. $\endgroup$
    – MathematicalPhysicist
    Commented Jan 31, 2017 at 12:29
  • $\begingroup$ There must be a mistake in the SM or the transcription. But, as already appreciated, any function of ρ inside the trace will provide the same vanishing of time evolution. This is a generic "unitary evolution" result of the trace. $\endgroup$
    – Cosmas Zachos
    Commented Jan 31, 2017 at 15:51
  • $\begingroup$ I mean, there are already two transcription errors in the SM quote. The exponent in the penultimate term should be 2 instead of ε. Moreover, from dimensional analysis, you need 1/ħ wherever you have ħ. But the crucial point is that the time-evolution of ρ is a similarity transform of the initial ρ(0). So the time-evolute of a function of ρ is the same similarity transform of the same function of ρ(0). The similarity transform of the argument of a trace is the argument itself by the cyclicity of the trace. The trace is independent of the time. You need no time derivatives anywhere. $\endgroup$
    – Cosmas Zachos
    Commented Jul 20, 2017 at 0:35

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