If $\rho,\sigma$ are classical-quantum states, can the fidelity $F(\rho,\sigma)$ be expressed in terms of $F(\rho_i,\sigma_i)$?

Question

Let $\rho = \sum_i \vert i\rangle\langle i\vert \otimes \rho_i$ and $\sigma = \sum_i\vert i\rangle\langle i\vert\otimes\sigma_i$ where we are using the same orthonormal basis indexed by $\vert i\rangle$ for both states.

The quantum fidelity is defined as $F(A,B) = \|\sqrt{A}\sqrt{B}\|_1$ (one can also define it as the square of this quantity if that helps). Can one express

$$F(\rho,\sigma)$$

in terms of the various $F(\rho_i,\sigma_i)$? For the trace distance, one has indeed that $\|\rho - \sigma\|_1 = \sum_i \|\rho_i - \sigma_i\|_1$ so does the fidelity also have a similar property?

Answer 1

I assume for $\rho$ and $\sigma$, you meant to write $\rho = \sum_i p(i) \vert i\rangle\langle i\vert \otimes \rho_i$ and $\sigma = \sum_i q(i)\vert i\rangle\langle i\vert\otimes\sigma_i$. From QIT by Mark Wilde, pg 250, $$\sqrt{F}(\rho,\sigma)=\sum_{i}\sqrt{p(i)q(i)}\sqrt{F}(\rho_{i},\sigma_{i})$$ for classical-quantum states like the one above.

Answer 2

Observe that, for any collection of matrices $A_i$, we have $$\sqrt{\sum_i |i\rangle\!\langle i|\otimes A_i} = \sum_i |i\rangle\!\langle i|\otimes \sqrt{A_i}, \\ {\rm Tr}\left(\sum_i |i\rangle\!\langle i|\otimes A_i\right) = \sum_i {\rm Tr}(A_i).$$ It immediately follows that $\|\sqrt\rho\sqrt\sigma\|_1\equiv {\rm Tr}|\sqrt\rho\sqrt\sigma|$ can be written as a weighted sum over the fidelities $\|\sqrt{\rho_i}\sqrt{\sigma_i}\|_1$.