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 \(\sigma_i\)