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?

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    $\begingroup$ Did this answer your question? $\endgroup$ – Jakob Mar 1 at 16:57
  • $\begingroup$ yess, it definitely helped understanding the issue $\endgroup$ – 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 $S_{\mathrm{N}} (\rho) = S_{\mathrm{N}} (\rho^\prime)$.


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