# 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 $$S(A|B)_{\rho}\leq S(A|B')_\sigma$$.