# Saturating the Fuchs-van de Graaf inequality

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.

• 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 $\rho = U \sigma U^\dagger$. Jun 26, 2020 at 13:35

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, $$1 - F(\rho,\sigma) = d(\rho,\sigma)$$