For pure states, it is known that one can always find a unitary that relates the two i.e. for any choice of states $\vert\psi\rangle$ and $\vert\phi\rangle$, there exists a unitary $U$ such that $U\vert\psi\rangle = \vert\phi\rangle$. Hence, we have that

$$\max_U F(\vert\phi\rangle,U\vert\psi\rangle) = 1.$$

What about mixed states? Given an arbitrary pair of states $\rho,\sigma$ such that $F(\rho,\sigma) \leq \varepsilon$ for some $\varepsilon$ much smaller than $\frac{1}{d}$ where $d$ is the dimension of the Hilbert space, can we achieve a lower bound on the following?

$$\max_U F(\rho,U\sigma U^\dagger) \geq \ ?$$

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    $\begingroup$ I assume you use $U$ to give $U\sigma U^\dagger$ the same eigenbasis as $\rho$, matching up the ordering of the eigenvalues (largest to largest, smallest to smallest, etc). $\endgroup$
    – DaftWullie
    Commented Sep 21, 2021 at 10:00
  • $\begingroup$ @DaftWullie thanks, that's an interesting point, thanks! I have edited slightly to remove the arbitrary $\rho$ and $\sigma$ condition since I realized my question has a trivial answer when $\sigma$ is the identity and the unitaries are not helpful. $\endgroup$
    – JRT
    Commented Sep 21, 2021 at 10:32
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    $\begingroup$ The condition $F(\rho,\sigma)\leq \epsilon$ does not seem meaningful because one can start with orthogonal pure states $F(\psi,\phi)=0$ and end up with identical states $F(U\psi,\phi)=F(\phi,\phi)=1$. $\endgroup$ Commented Sep 21, 2021 at 14:37
  • $\begingroup$ @QuantumMechanic thanks! Have edited to make it clearer but I want a lower bound on the maximum possible fidelity. The max possible fidelity is 1 for pure states as you point out regardless of $\varepsilon$ but for mixed states, the condition helps. If not, one can consider the case where $\rho$ is pure and $\sigma$ is maximally mixed. We then obtain the max possible fidelity to be $\frac{1}{d}$. $\endgroup$
    – JRT
    Commented Sep 21, 2021 at 17:51
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    $\begingroup$ @JRT got it. Now I'm interested in how $\max_U F(\rho, U\rho U^\dagger)-\min_U F(\rho, U\rho U^\dagger)$ behaves, because otherwise $\epsilon$ may reflect some luck in your original choice of location along the unitary orbit of $\sigma$ $\endgroup$ Commented Sep 21, 2021 at 18:54


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