# Understanding proof of joint concavity of the (root) fidelity

I have some problem in understanding the proof of the concavity of root fidelity given in Chapter 9.2 of Mark M. Wilde's "Quantum Information Theory". Here, the fidelity is defined by $$F(\rho, \sigma) = ||\sqrt{\rho}\sqrt{\sigma}||_1^2$$ where $$||\cdot ||_1$$ is a Schatten 1-norm and $$\rho$$ and $$\sigma$$ are density matrices. The joint concavity of root fidelity is the following property:

$$\begin{equation*} \sqrt{F}\left( \sum_x p_X(x)\rho_x,\, \sum_x p_X(x)\sigma_x \right) \geq \sum_x p_X(x) \sqrt{F}(\rho_x, \sigma_x) \end{equation*}$$ where $$p_X(x)$$ is a probability distribution. The proof of the joint concavity is given as below in the book:

Proof. Let $$|\phi^{\rho_x}\rangle_{RA}$$ and $$|\phi^{\sigma_x}\rangle_{RA}$$ be the respective Uhlmann purifications of $$\rho_x$$ and $$\sigma_x$$, i.e, $$F(\rho_x, \sigma_x) = |\langle \phi^{\rho_x}| \phi^{\sigma_x}\rangle_{RA}|^2$$ where $$R$$ denotes the environment system.

Let $$|x\rangle$$ be the orthonormal basis of the system $$X$$. Then $$\begin{gather*} |\phi^{\rho}\rangle = \sum_x \sqrt{p_X(x)}|\phi^{\rho_x}\rangle_{RA} |x\rangle_X, & |\phi^{\sigma}\rangle = \sum_x \sqrt{p_X(x)}|\phi^{\sigma_x}\rangle_{RA} |x\rangle_X \end{gather*}$$ are respective purifications of $$\sum_x p_X(x)\rho_x$$ and $$\sum_x p_X(x)\sigma_x$$. Then \begin{align*} \sqrt{F}\left( \sum_x p_X(x)\rho_x,\, \sum_x p_X(x)\sigma_x \right) & \geq |\langle\phi^{\rho}|\phi^{\sigma}\rangle| \\ & = \Bigg| \sum_x p_X(x) \langle\phi^{\rho_x}|\phi^{\sigma_x}\rangle \Bigg| \\ & \geq \sum_x p_X(x) \,|\langle\phi^{\rho_x}|\phi^{\sigma_x}\rangle|\\ & = \sum_x p_X(x) \sqrt{F}(\rho_x, \sigma_x) \end{align*} where the first inequality is due to Uhlmann's theorem.

I don't think that the second inequality holds in this proof. Which point am I missing? Is this proof correct? If not, how can I prove the joint concavity of root fidelity? I appreciate any help.

• Note that the second inequality is an equality if $\langle\phi^{\rho_x}|\phi^{\sigma_x}\rangle \geq 0$ for all $x$. But we can enforce this by replacing $|\phi^{\sigma_x} \rangle$ with $|\hat{\phi}^{\sigma_x} \rangle = - |\phi^{\sigma_x} \rangle$ whenever it fails to be positive. This gives another purification that achieves the fidelity. So we can assume wlog that $|\langle\phi^{\rho_x}|\phi^{\sigma_x}\rangle| = \langle\phi^{\rho_x}|\phi^{\sigma_x}\rangle$ and the inequality holds with equality (which is all you need). May 12, 2020 at 10:39
• @Rammus Thank you for detailed answer!
– asdf
May 12, 2020 at 14:06

You can assign an arbitrary phase to the purifications. This way, you can make $$\langle \phi^{\rho_x}\vert\phi^{\sigma_x}\rangle\ge0$$, and thus you have equality in the 2nd inequality.