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It is well-known that one side of the Fuchs-van de Graaf inequality is saturated for pure states, i.e. $F(\rho,\sigma)^2 = 1-d(\rho,\sigma)^2$ when $\rho$ and $\sigma$ are pure (here we are using the definition $F(\rho, \sigma) := \|\sqrt{\rho}\sqrt{\sigma}\|_1$ for fidelity). However, are there other situations where it is known that this equality holds? How far has this been characterized?

As a starting point, I am aware that when the states are qubits, it can be shown that $F(\rho,\sigma)^2 = 1-d(\rho,\sigma)^2$ holds if and only if the states have the same eigenvalues. (This is not too difficult to prove using specialized qubit expressions for the fidelity, but as far as I am aware, it does not seem to be well-known.) The "have the same eigenvalues" condition does not generalize even to qutrits, however, and hence it may perhaps not be the best approach to characterizing the conditions.

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  • $\begingroup$ Interesting question. Not an answer but your 'have the same eigenvalues' condition can be more cleanly stated as they are unitarily equivalent, i.e., there exists some unitary $U$ such that $\rho = U \sigma U^\dagger$. $\endgroup$
    – Rammus
    Jun 26, 2020 at 13:35

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This is not a saturation of the the Fuchs-van de Graaf upper bound, $F(\rho,\sigma)^2 = 1-d(\rho,\sigma)^2$, but rather the lower bound, $1 - \sqrt{F(\rho,\sigma)} \leq d(\rho,\sigma)$ (see, for example, here).

Consider, $\rho = | \psi \rangle \langle \psi |$, a pure state and $\sigma = \frac{\mathbb{I}}{d}$ is the maximally mixed state (for a $d$-dimensional Hilbert space). Then, the fidelity reduces to $F(| \psi \rangle \langle \psi | , \frac{\mathbb{I}}{d}) = \langle \psi | \frac{\mathbb{I}}{d} | \psi \rangle = \frac{1}{d}$ (see, for example, Wikipedia).

And the trace norm distance is $\left\Vert | \psi \rangle \langle \psi | - \frac{\mathbb{I}}{d} \right\Vert_{1} = \left( 1-\frac{1}{d} \right) + \frac{d-1}{d} = 2(1-\frac{1}{d})$. Therefore, using the normalized trace norm, $d(\rho, \sigma) \equiv \frac{1}{2} \left\Vert \rho - \sigma \right\Vert_{1}$, we have, $1 - F(\rho,\sigma) = d(\rho,\sigma)$

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  • $\begingroup$ Thank you! Indeed, I'm familiar with some saturating examples for this side of the bound (though I think there is a typo in its statement in the first sentence). Though it might perhaps also be interesting to find necessary conditions for saturating this side of the bound. Ironically, there's a qutrit example where both states have the same eigenvalues (equivalently, are unitarily equivalent) that saturates this side of the bound, indicating that the qubit condition in the main post does not generalize. $\endgroup$
    – helloworld
    Jul 24, 2020 at 11:52

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