# Question 14

Uhlmann’s theorem says that $$F(\rho, \sigma) = \max_{\text{all purifications}} | \langle \phi_\rho | \phi_\sigma \rangle |$$. Why is the absolute value there? Is it just to make complex numbers into reals, so that we can compare them? So technically the formula would be correct even if there was $$\text{Re} \langle \phi_\rho | \phi_\sigma \rangle$$ or $$\text{Im} \langle \phi_\rho | \phi_\sigma \rangle$$ instead? Am I right to think that we can always adjust the complex phase of $$\langle \phi_\rho | \phi_\sigma \rangle$$ by considering $$|\phi_\rho \rangle \rightarrow |\psi_\rho\rangle = |\phi_\rho \rangle |0\rangle$$ and $$|\phi_\sigma \rangle \rightarrow |\psi_\sigma\rangle = |\phi_\sigma \rangle[\cos(\theta) + i\sin(\theta)] |0\rangle$$?
This way $$| \langle \phi_\rho | \phi_\sigma \rangle | = | \langle \psi_\rho | \psi_\sigma \rangle |$$, but the complex phase of $$\langle \psi_\rho | \psi_\sigma \rangle$$ is arbitrary (depending on how we choose $$\theta$$).
I’m asking because in 2014 exam question 1 iv) (proving joint concavity of the fidelity), it would help to be able to assume $$| \langle \phi_\rho | \phi_\sigma \rangle | = \langle \phi_\rho | \phi_\sigma \rangle$$ (because of some absolute value manipulations).
Yeah, the absolute value is just to make them all real to compare them. You are right that you can find purifications so that the inner product has whatever phase you want but the same absolute value (and in fact could just maximize over the real or imaginary part). As for the joint concavity of the fidelity: that was actually Example Sheet 3, Exercise 4, and I did just as you suggested (assume $$| \langle \phi_\rho | \phi_\sigma \rangle | = \langle \phi_\rho | \phi_\sigma \rangle$$) in the solutions!