# What property ensures that von Neumann entropy is conserved?

So I always had this idea in my mind that unitary evolution in quantum mechanics conserves information (or in other words von Neumann entropy) because unitary evolution preserves the trace. But this cannot be true because general evolution with a superoperator ($$\\rho \rightarrow \rho'$$) of a mixed state does not preserve information and is also trace preserving!

So what property of unitary evolution preserves the von Neumann entropy? As it is defined as $$S = -tr(\rho \log \rho)$$ and we can always bring this in diagonal form $$S= -\sum_i \lambda_i \log \lambda_i$$, I would expect unitary evolution to preserve the eigenvalues? But I don't find any thing related to this online... Does unitarity imply that "eigenvalues" are conserved or what property makes sure that the entropy is conserved?

Yes, unitarity preserves eigenvalues. This is because the definition of eigenvalues is that any Hermitian matrix $$H$$ can be brought into diagonal form by a unitary $$V$$, $$VHV^\dagger=D,$$ and the diagonal elements of $$D$$ are the eigenvalues. So, now consider $$UHU^\dagger$$. This can be diagonalised by a unitary $$VU^\dagger$$ into the same matrix $$D$$: $$(VU^\dagger)UHU^\dagger(VU^\dagger)^\dagger=VU^\dagger UHU^\dagger UV^\dagger=VHV^\dagger=D.$$ Hence the eigenvalues are the same, and the von Neumann entropy is the same.
Having a constrant trace is a necessary condition for preserving the eigenvalues, but is not sufficient - it only ensures that $$\sum_i\lambda_i$$ is constant.