Saturating the Fuchs-van de Graaf inequality

Question

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.

keisuke.akira · Answer

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

Saturating the Fuchs-van de Graaf inequality

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