# Exercise 4

## from Example Sheet 3

### Exercise 4.

Show that the fidelity is jointly concave. That is, given any finite probability distribution $$\{p_i\}_{i=1}^n$$ and quantum states $$\rho_i$$ and $$\sigma_i$$ for $$i=1,\ldots,n$$, $F\left(\sum_{i=1}^n p_i \rho_i ,\sum_{i=1}^n p_i \sigma_i\right) \geq \sum_{i=1}^n p_i F(\rho_i,\sigma_i).$

Hint: Choose purifications $$\phi_i$$ of $$\rho_i$$ and $$\psi_i$$ of $$\sigma_i$$ such that $$F(\rho_i,\sigma_i) = \langle \phi_i|\psi_i \rangle$$, using Uhlmann’s theorem.

As the hint suggests, we will take purifications $$\phi_i$$ of $$\rho_i$$ and $$\psi_i$$ of $$\sigma_i$$ such that $$F(\rho_i,\sigma_i) = \langle \phi_i|\psi_i \rangle$$. Uhlmann’s theorem gives that there exist purifications with $$F(\rho_i,\sigma_i) = |\langle \phi_i|\psi_i \rangle|$$, but we may always multiply one of them by the required phase to their inner product is real1.

Then consider the states $| \Psi \rangle = \sum_i \sqrt{p_i} | \psi_i \rangle \otimes | i \rangle, \qquad | \Phi \rangle = \sum_i \sqrt{p_i} | \psi_i \rangle\otimes | i \rangle.$ We notice $\langle \Phi|\Psi \rangle = \sum_i p_i \langle \phi_i | \psi_i \rangle = \sum_i p_i F(\rho_i,\sigma_i).$ On the other hand, $\Phi = \sum_{i,j} \sqrt{p_i p_j} | \psi_i \rangle\langle \psi_j | \otimes | i \rangle\langle j |,$ and if we trace out the last system, we have $\sum_{i} p_i | \psi_i \rangle\langle \psi_i |.$ Since each $$\psi_i$$ is a purification of $$\rho_i$$, we can trace out again to find $\sum_i p_i \rho_i.$ Therefore, $$\Phi$$ is a purification of $$\sum_{i=1}^n p_i \rho_i$$ (where the purifying reference is the tensor product of the purifying space for the $$\phi_i$$’s with the additional space spanned by $$| i \rangle$$ we added to make $$\Phi$$). Similarly, $$\Psi$$ is a purification of $$\sum_{i=1}^n p_i \sigma_i$$. Thus, $\sum_i p_i F(\rho_i,\sigma_i) = \langle \Phi,\Psi \rangle \leq |\langle \Phi,\Psi \rangle| \leq \max_{\phi,\psi} |\langle \phi,\psi \rangle| = F\left(\sum_{i=1}^n p_i \rho_i ,\sum_{i=1}^n p_i \sigma_i\right),$ where the maximum is over all purifications $$\phi$$ of $$\sum_{i=1}^n p_i \rho_i$$ and $$\psi$$ of $$\sum_{i=1}^n p_i \sigma_i$$, and the final equality is by Uhlmann’s theorem, again.

1. I.e. if $$\langle \phi_i|\psi_i \rangle = e^{i \theta}|\langle \phi_i|\psi_i \rangle|$$ then we take $$| \tilde \psi_i \rangle = e^{-i \theta} | \psi_i \rangle$$ instead of $$| \psi_i \rangle$$, which is still a purification of $$\sigma_i$$