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The motivation for this question comes from trace distance. For any two states $\rho, \sigma$, the trace distance $T(\rho, \sigma)$ is given by

$$T(\rho, \sigma) = |\rho - \sigma|_1,$$

where $|\cdot|_1$ is the 1-norm and given by $|X|_1 = \text{Tr}(\sqrt{X^\dagger X})$. The point here is that I do not need to know $\rho$ or $\sigma$ to compute the trace distance between them. All I need to know is $\rho - \sigma$.

Can one also compute $F(\rho,\sigma)$ where $F$ is the fidelity if one is only given $\rho - \sigma$? I am aware of bounds that can be placed using the trace distance on $F(\rho,\sigma)$ but was wondering if it could be exactly computed.

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  • $\begingroup$ How many examples did you try? $\endgroup$ – Norbert Schuch Jul 3 at 14:13
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The answer is no, as the following counter-example reveals. Let $\varepsilon\in(0,1)$ and define $$ \rho_0 = \begin{pmatrix} \frac{1+\varepsilon}{2} & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & \frac{1-\varepsilon}{2} \end{pmatrix},\quad \rho_1 = \begin{pmatrix} \frac{1-\varepsilon}{2} & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & \frac{1+\varepsilon}{2} \end{pmatrix} $$ as well as $$ \sigma_0 = \begin{pmatrix} \varepsilon & 0 & 0\\ 0 & 1-\varepsilon & 0\\ 0 & 0 & 0 \end{pmatrix},\quad \sigma_1 = \begin{pmatrix} 0 & 0 & 0\\ 0 & 1-\varepsilon & 0\\ 0 & 0 & \varepsilon \end{pmatrix}. $$ The differences agree, $$ \rho_0 - \rho_1 = \begin{pmatrix} \varepsilon & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -\varepsilon \end{pmatrix} = \sigma_0 - \sigma_1, $$ but the fidelities are not equal: $$ \mathrm{F}(\rho_0,\rho_1) = \sqrt{(1 + \varepsilon)(1-\varepsilon)} \quad\text{and}\quad \mathrm{F}(\sigma_0,\sigma_1) = 1-\varepsilon. $$

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