# Exercise 7

## from Example Sheet 3

### 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.