# Prove that the quantum state fidelity is upper bounded as $F(\rho,\sigma)\le1$

I am looking for proof that the quantum fidelity $$F(\rho, \sigma) = \left(\text{tr} \sqrt{\sqrt{\rho}\sigma\sqrt\rho}\right)^2$$ is bound from above by 1, i.e., $$F(\rho, \sigma) \leq 1$$.

1. I know this is a consequence of Uhlmann's theorem. However, I feel like it is far too complicated for this (seemingly) simple task. Is there more straightforward proof of the fact?

2. Also, as far as I know, the inequality is tight only for $$\rho = \sigma$$. This seems like a more difficult task to prove. Is it still possible in relatively simple manner?

You can equivalently write the (square root of the) fidelity as $$\sqrt F(\rho,\sigma) = \|\sqrt\rho\sqrt\sigma\|_1,$$ where $$\|\cdot\|_1$$ is the one norm, defined as $$\|A\|_1=\operatorname{tr}(|A|)=\operatorname{tr}[\sqrt{A^\dagger A}] = \operatorname{tr}[\sqrt{AA^\dagger}].$$ A general property of the trace norm $$\|A\|_1$$ is that it equals the max of $$\left|\langle U,A\rangle\right|$$ maximised over all unitaries $$U$$. And in particular, there is always some unitary $$U$$ such that $$\|A\|_1=\langle U,A\rangle\equiv \operatorname{tr}(U^\dagger A)$$. It's not crucial to our discussion here, but this $$U$$ is the unitary arising from the polar decomposition of $$A$$.
Suppose now $$P,Q$$ are generic Hermitian matrices. Then for some unitary $$U$$ we have $$\|\sqrt P\sqrt Q\|_1 = \langle U, \sqrt P\sqrt Q\rangle = \langle \sqrt P U,\sqrt Q\rangle,$$ where we are using the Hilbert-Schmidt inner product: $$\langle A,B\rangle\equiv \operatorname{tr}(A^\dagger B)$$. Using Cauchy-Schwarz, we get the bound $$\|\sqrt P\sqrt Q\|_1 \le \|\sqrt P U\|_2 \|\sqrt Q\|_2 = \sqrt{\operatorname{tr}(\sqrt P UU^\dagger \sqrt P)} \sqrt{\operatorname{tr}(\sqrt Q\sqrt Q)},$$ where $$\|A\|_2\equiv \sqrt{\operatorname{tr}(A^\dagger A)}$$. In conclusion $$F(P,Q) = \|\sqrt P\sqrt Q\|_1^2 \le \operatorname{tr}(P) \operatorname{tr}(Q),$$ which gives the unit upper bound for normalized density matrices.
Furthermore, the inequality is saturated when the CS inequality is, that is, when $$\sqrt P U\propto \sqrt Q$$. Being $$P,Q$$ Hermitian, this holds iff $$P$$ and $$Q$$ have the same eigenvalues and eigenvectors. In other words, iff $$P=\lambda Q$$ for some $$\lambda\in\mathbb{R}$$. When we are using density matrices, this amounts to $$\rho=\sigma$$.