Change of Mutual Information in Isolated Quantum Systems

Question

I was reading some publications regarding correlation and mutual information for composite quantum systems.
I noticed that most papers give the expression for the mutual information to be:
$$Delta I(A:B) = Delta S(A) + Delta S(B)$$
Stating that the term $Delta S(AB)$ must be equal to zero for isolated composite systems undergoing $rho_{AB}^0 rightarrow Urho_{AB}^0U^dagger = rho_{AB}^t$.
Why can we say that the entropy for composite systems does not change under unitary transformations?

Jakob · Accepted Answer

I don't know exactly how to understand your notation, but if you ask why the von Neumann entropy of a density matrix does not change under a unitary transformation, then this is due to the fact that a unitary transformation does not change the eigenvalues of the density matrix. Consider a density matrix $rhoequiv sumlimits_k lambda_k P_k$ and the corresponding von Neumann entropy:
$$S_{mathrm{N}} (rho) = - sumlimits_k lambda_k , ln(lambda_k) quad .$$
For a unitary $U$, the operator $rho^prime equiv U,rho,U^dagger$ has the same eigenvalues as $rho$ and hence $S_{mathrm{N}} (rho) = S_{mathrm{N}} (rho^prime)$.

Change of Mutual Information in Isolated Quantum Systems

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