# Change of Mutual Information in Isolated Quantum Systems

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 $$\rho_{AB}^0 \rightarrow U\rho_{AB}^0U^\dagger = \rho_{AB}^t$$.

Why can we say that the entropy for composite systems does not change under unitary transformations?

• Did this answer your question? – Jakob Mar 1 at 16:57
• yess, it definitely helped understanding the issue – Oti Dioti Mar 11 at 9:55

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 $$\rho\equiv \sum\limits_k \lambda_k P_k$$ and the corresponding von Neumann entropy: $$S_{\mathrm{N}} (\rho) = - \sum\limits_k \lambda_k \, \ln(\lambda_k) \quad .$$
For a unitary $$U$$, the operator $$\rho^\prime \equiv U\,\rho\,U^\dagger$$ has the same eigenvalues as $$\rho$$ and hence $$S_{\mathrm{N}} (\rho) = S_{\mathrm{N}} (\rho^\prime)$$.