Exercise 10

from Example Sheet 3

Exercise 10.

Let \(\mathcal{H}_A, \mathcal{H}_B\) and \(\mathcal{H}_{B'}\) be Hilbert spaces and \(\rho\in\mathcal{D}(\mathcal{H}_{AB})\) be a state. Further, let \(\Lambda:\mathcal{H}_B\rightarrow\mathcal{H}_{B'}\) be a CPTP map and define \[ \sigma_{AB'}= ( \operatorname{id}_A \otimes \Lambda)\rho_{AB}. \] Prove that \[ S(A|B)_{\rho_{AB}} \leq S(A|B')_{\sigma_{AB'}}. \] Hint: Use strong subadditivity.

To use subadditivity, we need three systems. We’ll introduce a third: by Stinespring’s dilation theorem, there is a Hilbert space \(\mathcal{H}_C\), an isometry \(U : \mathcal{H}_B\otimes \mathcal{H}_C \to \mathcal{H}_{B'}\otimes \mathcal{H}_{C}\) and a pure state \(\varphi\in\mathcal{D}(\mathcal{H}_C)\) such that \(\Lambda(\rho)=\operatorname{tr}_C(U(\rho\otimes\varphi)U^\dagger)\) for all \(\rho\in\mathcal{D}(\mathcal{H}_B)\). Now let \[\omega_{AB'C} = (I_A\otimes U)(\rho_{AB}\otimes \varphi)(I_A\otimes U^\dagger)\] such that \(\operatorname{tr}_C\omega_{AB'C} = \sigma_{AB'}\). Strong subadditivity of the von Neumann entropy implies \[ S(\omega_{AB'C})+S(\omega_{B'})\leq S(\sigma_{AB'}) + S(\omega_{B'C}). \] We have \(\omega_{B'} = \operatorname{tr}_A(\operatorname{tr}_{C}\omega_{AB'C}) = \operatorname{tr}_A\sigma_{AB'} = \sigma_{B'}\) and \[S(\omega_{AB'C}) = S(\rho_{AB}\otimes\varphi) = S(\rho_{AB}) + S(\varphi) = S(\rho_{AB})\] by invariance under unitaries of the von Neumann entropy and the fact that \(S(\varphi)=0\) for pure states \(\varphi\). Furthermore, \[\omega_{B'C} = U(\rho_B\otimes\varphi)U^\dagger\] and hence \(S(\omega_{B'C}) = S(\rho_B)\) by the same arguments.

In summary, we have \[S(\rho_{AB})-S(\rho_B)\leq S(\sigma_{AB'})-S(\sigma_{B'})\] which is equivalent to \(S(A|B)_{\rho}\leq S(A|B')_\sigma\).