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We know that (square root) Fidelity which is defined as $\text{F}(\rho,\sigma) = \| \sqrt{\rho} \sqrt{\sigma} \|_1 = \text{Tr}(\sqrt{\sqrt{\sigma} \rho \sqrt{\sigma}})$ is satisfies the property of joint concavity. That is

$$ \text{F}\left(\sum_{i=1}^{n} p_i\rho_i, \sum_{i=1}^{n} p_i\sigma_i \right) \geq \sum_{i=1}^{n} p_i \text{F}(\rho_i,\sigma_i) , \tag{1}$$

for an $n$ dimensional probability vector $p$ and ensembles of states $\{\rho_i\}_{i=1}^n$ and $\{\sigma_i\}_{i=1}^n$. It follows that $$ \text{F} \left(\sum_{i=1}^{n} p_i\rho_i,\sigma \right) \geq \sum_{i=1}^{n} p_i \text{F}(\rho_i,\sigma) \tag{2}, $$

for some state $\sigma$.

For now, assume all of the states and probabilities are distinct and all the states are full-rank. Can we show that (2) is strictly concave? That is for non-extremal probability vectors (extremal points being the $n$ dimensional standard basis vectors) the inequality of (2) becomes strict?

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  • $\begingroup$ Take all $\rho_{i}$ orthogonal to some rank $1$ $\sigma$ and it becomes an equality. $\endgroup$
    – JSdJ
    May 6 at 14:22
  • $\begingroup$ @JSdJ Thanks, that is a great example for the equality. Can we show this to be the case when all the states $\rho_i$ and $\sigma$ are full rank? $\endgroup$
    – Afham
    May 8 at 11:05
  • $\begingroup$ I'm not sure if I completely understand your question - do you mean that for $\rho_{i}$ and $\sigma$ orthogonal and full rank, it's necessarily an equality? $\endgroup$
    – JSdJ
    May 11 at 17:02
  • $\begingroup$ No, my intuition is that when all the $\rho_i$s and $\sigma$ and distinct and full rank, along with the probabilities $p_i$ being distinct, the inequality should be strict. $\endgroup$
    – Afham
    May 13 at 5:16

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