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.

  • $\begingroup$ 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). $\endgroup$
    – Rammus
    May 12, 2020 at 10:39
  • $\begingroup$ @Rammus Thank you for detailed answer! $\endgroup$
    – asdf
    May 12, 2020 at 14:06

1 Answer 1


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.

  • $\begingroup$ I got it. Thank you! $\endgroup$
    – asdf
    May 12, 2020 at 14:05

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