 Solution

Use the unsquared convention for quantum fidelity. For density operators on a common finite-dimensional Hilbert space,Here $\Vert A{\Vert }_1=\mathrm{Tr}\sqrt{A^{†}A}$ is the trace norm, and all square roots are the positive operator square roots. The two displayed expressions agree because $\sqrt{\rho }\sqrt{\sigma }$ and its adjoint have the same singular values. This convention has $0\le F\le 1$; some literature squares this quantity, but that convention is not used here.

For normalized pure states, the rank-one operator $\sqrt{\rho }\sqrt{\sigma }=⟨\phi |\psi ⟩|\phi ⟩⟨\psi |$ has a single nonzero singular value. Therefore