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In chapter 5 of the book "Statistical Mechanics" by Pathria it says

Since the density matrix evolves in a unitary manner, the von Neumann entropy is time-independent

Where the von Neumann entropy is defined as the trace $$S[\rho(t)]=-\mathrm{Tr}\left(\rho(t)\ln \rho(t)\right)$$ and the evolution of the density matrix is $$\rho(t)=\exp(-iHt/\hbar)\rho(0)\exp(iHt/\hbar)$$ and $H$ is the Hamltonian operator of the system we are studying.

I couldn't prove this result, can anyone help?

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2 Answers 2

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Hint: Use the spectral decomposition to write

$$\rho(0) := \sum\limits_k \lambda_k \,|k\rangle \langle k| \tag{1} ,$$ and then find an expression for $\rho(t)$ in terms of $\lambda_k$. Especially note that $\rho(t)$ has the same eigenvalues as $\rho(0)$. Finally, again using the spectral theorem, derive that $$ S[\rho(t)] = -\mathrm{Tr} \sum\limits_k \lambda_k \ln \lambda_k \, U(t)|k\rangle\langle k| U^\dagger(t) \tag{2} \quad .$$ The cyclic properties of the trace then yield the desired result, i.e. $S[\rho(t)]=S[\rho(0)]$.

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Another neat method to prove this: write the von Neumann entropy as a limit of the Renyi entropies: $$ S[\rho] = \lim_{n \to 1} S^{(n)}[\rho] = \lim_{n \to 1} \frac{1}{1-n} \log \text{Tr} \rho^n $$ Here, the identity is manifest: $U^{\dagger} U = 1$ cancels between each of the $n$ copies of $\rho$, and then the final $U$ and $U^{\dagger}$ on the left and right ends of the string cancel using the cyclicity of the trace.

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  • $\begingroup$ That's a neat way indeed! $\endgroup$ Jun 17, 2022 at 18:14

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