### Exercise 7.

Prove that the von Neumann entropy is a concave function of its inputs, i.e., given probabilities \(p_i \ge 0\), \({\displaystyle{\sum_{i=1}^r p_i =1}}\), and corresponding density operators \(\rho_i\): \[ S\left( \sum_{i=1}^r p_i \rho_i \right) \ge \sum_{i=1}^r p_i S\left(\rho_i\right). \](1)

Show that if each \(p_i > 0\) and \(\rho_i\neq \rho_j\) for \(i\neq j\), then the inequality (1) is strict. Note: means the von Neumann entropy is *strictly concave*.

*Hint: Consider the state \(\sigma_{AB} := \sum_i p_i \rho_i \otimes |i\rangle \langle i|\), its reduced states \(\sigma_A\), \(\sigma_B\) and use subadditivity.*

As the hint says, we’ll consider \[ \sigma_{AB} := \sum_i p_i \rho_i \otimes |i\rangle \langle i|. \] We notice that the support of each \(\rho_i \otimes |i\rangle \langle i|\) is orthogonal. If \(P_i\) is the projection onto the support of \(\rho_i\), then \(P_i \otimes | i \rangle\langle i |\) is the projection onto the support of \(\rho_i \otimes |i\rangle \langle i|\). But \[ (P_i \otimes | i \rangle\langle i |)(P_j \otimes | j \rangle\langle j |) = P_i P_j \otimes | i \rangle\langle i || j \rangle\langle j | = 0 \] for \(i\neq j\), so the supports are othogonal. Then \[ S(\sigma_{AB}) = \sum_i p_i S(\rho_i \otimes |i\rangle \langle i|) + H(p) = \sum_i p_i S(\rho_i) + H(p) \] using that \(S(\rho_i \otimes |i\rangle \langle i|) = S(\rho_i) + S(| i \rangle\langle i |) = S(\rho_i)\), since \(| i \rangle\langle i |\) is a pure state. But by subadditivity \[ S(\sigma_{AB}) \leq S(\sigma_A) + S(\sigma_B). \](2) Note \(\sigma_A = \sum_i p_i \rho_i\) and \(\sigma_B = \sum_i p_i| i \rangle\langle i |\). Then \(S(\sigma_B) = H(p)\), and (2) gives \[ \sum_i p_i S(\rho_i) + H(p) = S(\sigma_{AB}) \leq S(\sigma_A) + S(\sigma_B) = S \Big(\sum_i p_i \rho_i \Big) + H(p) \] which yields the result by subtracting \(H(p)\) from each side.

Now, let us assume all the \(p_i>0\) and \(\rho_i \neq \rho_j\) for \(i\neq j\). The only step in our proof which had an inequality instead of equality was in subadditivity, (2). In fact, one has equality in (2) if and only if \(\sigma_{AB} = \sigma_A\otimes \sigma_B\). One can see that by the fact that subadditivity is a rearrangement of terms of the inequality \(D(\sigma_{AB}\|\sigma_A\otimes \sigma_B)\geq 0\), which has equality if and only if \(\sigma_{AB} = \sigma_A\otimes \sigma_B\).

Therefore, we simply need to show that if \(p_i>0\) and \(\rho_i \neq \rho_j\) for \(i\neq j\), then \(\sigma_{AB}\neq \sigma_A\otimes \sigma_B = \sum_{ij} p_i p_j \rho_i \otimes | j \rangle\langle j |\). Assume otherwise for the sake of contraction: \[ \sum_i p_i \rho_i \otimes |i\rangle \langle i|. = \sum_{ij} p_i p_j \rho_i \](3) We can compare blocks (or, equivalently, act on the left of both sides by \(I\otimes \langle k |\) and on the right by \(I\otimes | k \rangle\)) to find \[ p_k \rho_k = p_k \sum_i p_i \rho_i \] which is equivalent to \(\rho_k = \sum_i p_i \rho_i = \sigma_A\). But this must be true for every \(k\) (every block), so we must have \(\sigma_A = \rho_1 = \rho_2=\dotsm = \rho_n\). This violates the assumption that the \(\rho_i\) are all different. Therefore, our assumption (3) must be wrong, and we have strict inequality in subadditivity, and therefore strict inequality in the final result.