# Unitary evolution and von Neumann entropy

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?

$$\rho(0) := \sum\limits_k \lambda_k \,|k\rangle \langle k| \tag{1} ,$$ 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. $$S[\rho(t)]=S[\rho(0)]$$.
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.