# Can the fidelity $F(\rho,\sigma)$ be computed knowing only $\rho - \sigma$?

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.

• How many examples did you try? Jul 3 '20 at 14:13

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: $$\mathrm{F}(\rho_0,\rho_1) = \sqrt{(1 + \varepsilon)(1-\varepsilon)} \quad\text{and}\quad \mathrm{F}(\sigma_0,\sigma_1) = 1-\varepsilon.$$