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2

Okay, this is a rather subtle situation, but I think I've figured it out. The key is to be very careful about which mathematical results about Hermitian operators do and do not hold for generic operators. Let $H$ represent an arbitrary Hermitian matrix, $N$ an arbitrary normal one, $D$ be a generic diagonalizable matrix, and $M$ an arbitrary matrix, all ...

2

Square root is not differentiable at 0, so that cyclic property cannot be applied While $\rho\sigma$ has the same non-negative eigenvalues as $\sqrt\rho\sigma\sqrt\rho$, it's not self-adjoint. Non self-adjoint matrices are not diagonalizable in general, so the square root $\sqrt{\rho\sigma}$ can be not well-defined (see edit below). Anyway, $\text{Tr}(\... 2 You could probably reach the same conclusions by identifying that$tr(\rho \sigma)$is just the expectation value of$\rho$under the mixed state$\sigma$, but let's do it explicitly: for$\rho = \sum_i p_i |\psi_i\rangle \langle \psi_i|$and$\sigma = \sum_i q_i |\phi_i\rangle \langle \phi_i|$, we have:$tr(\rho \sigma)$=$\sum_{ij} p_i q_j tr(|\psi_i\...

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If these are qubit states, the formula in your question simplifies dramatically to $$F'(\rho,\pi)=\sqrt{\text{tr}(\rho \pi) + 2 \sqrt{\text{det}(\rho) \text{det}(\pi)}}.$$ If you consider the components of the vectors, $\vec s = (s_1, s_2, s_3)$ and $\vec r = (r_1, r_2, r_3)$, this can be expressed simply as $$F'(\rho, \pi) = \frac{1}{\sqrt{2}} \left[1+ \... -2 Assuming you are working with qubits, according to Nielsen and Chuang,$$1 - F(\rho, \pi) \le D(\rho, \pi) = \frac{|\vec{s}^{\,} - \vec{r}^{\,}|}{2} Where $F(\cdot, \cdot)$ and $D(\cdot, \cdot)$ are fidelity and trace distance respectively.

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