# What can be said about the closeness of two states if the difference of their fidelity measured with respect to a fixed state is close to 0?

Suppose I have two states $$\rho$$ and $$\sigma$$. We are given that,

$$Tr((\rho - \sigma)|\psi\rangle\langle\psi|) \geq \epsilon$$ where $$|\psi\rangle$$ is a fixed state and $$\epsilon \rightarrow 0$$,

Then can we conclude anything about the closeness of two states $$\rho$$ and $$\sigma$$ in any distance measure?

In general, it would seem no. The quantity $$\mathrm{Tr}[(\rho - \sigma)|\psi\rangle\langle\psi|]$$ is only concerned with the distance between $$\rho$$ and $$\sigma$$ on the subspace $$\mathrm{span}(|\psi\rangle)$$. For example, we know we can decompose the Hilbert space as $$\mathcal{H} = \mathrm{span}(|\psi\rangle) \oplus \mathrm{span}(|\psi\rangle)^{\perp}$$. Then take $$\rho', \sigma'$$ to be operators with support only on $$\mathrm{span}(|\psi\rangle)^{\perp}$$. Then for any $$\epsilon \geq 0$$ define $$\rho_{\epsilon} = (1-\epsilon)\rho' + \epsilon |\psi \rangle \langle \psi |$$ and $$\sigma = \sigma'$$. For these states we have $$\mathrm{Tr}[(\rho_{\epsilon} - \sigma)|\psi\rangle\langle\psi|] = \epsilon.$$

However, as you mention in your question $$\epsilon$$ is small so we have (most of the time) a lot of freedom with how we can define the operators on the orthogonal subspace. If we take $$\rho' = \sigma'$$ then \begin{align} \|\rho_{\epsilon} - \sigma\| &= \|-\epsilon \rho' + \epsilon |\psi\rangle\langle\psi|\| \\ &= \epsilon \| \rho' - |\psi\rangle\langle\psi|\| \end{align} which is small if $$\epsilon$$ is small. However, in general if we use the fact that norms are continuous we have \begin{aligned} \lim_{\epsilon \rightarrow 0} \| \rho_{\epsilon} - \sigma\| &= \|\lim_{\epsilon \rightarrow 0} \rho_{\epsilon} - \sigma \| \\ &= \|\rho' - \sigma' \|. \end{aligned} So as $$\epsilon \rightarrow 0$$ the distance between $$\rho$$ and $$\sigma$$ just becomes the distance between $$\rho'$$ and $$\sigma'$$. But we were free to choose $$\rho'$$ and $$\sigma'$$ as we wished so this distance has no nontrivial a priori bound.

Caveat The case is different for qubits. There the orthogonal subspace is one-dimensional so if we tried to play the same trick we don't have any freedom in how to choose $$\rho'$$ and $$\sigma'$$. In this case we end up in the first example again where for $$\epsilon \rightarrow 0$$ we found $$\|\rho_{\epsilon} - \sigma\| \rightarrow 0$$. For qubits you can probably work out some concrete bounds on the distance.

Here's a concrete example for a single qubit.

We can always change the basis to have $$|\psi\rangle=|0\rangle$$. Let us further suppose that $$\langle0|\rho|0\rangle=0$$, so that $$\rho=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$ The requirement $$\operatorname{Tr}[(\sigma-\rho)|\psi\rangle\!\langle\psi|]=\langle\psi|\sigma-\rho|\psi\rangle=\epsilon$$ then becomes $$\sigma=\begin{pmatrix}\epsilon & a^* \\ a & 1-\epsilon\end{pmatrix}$$ for some $$a\in\mathbb C$$. To have $$\sigma\ge0$$, the coefficient $$a$$ must satisfy $$|a|^2\le \epsilon(1-\epsilon)$$ (as follows from imposing its eigenvalues to be non-negative). We then have $$\langle0|\sigma-\rho|0\rangle= \epsilon$$.

To quantify the distance between these states, we notice that the eigenvalues of $$\sigma-\rho$$ are $$\lambda_\pm=\pm\sqrt{\epsilon^2+|a|^2}$$, and therefore $$\|\rho-\sigma\|_1=|\lambda_+|=\sqrt{\epsilon^2+|a|^2}.$$ We then have the following bound on the trace distance: $$\epsilon\le\|\rho-\sigma\|_1\le\sqrt{\epsilon}$$

In the general case, suppose $$\langle0|\rho|0\rangle=p$$. Then $$\rho=\begin{pmatrix}p & b^* \\ b & 1-p\end{pmatrix}, \qquad \sigma=\begin{pmatrix}p+\epsilon & a^* \\ a & 1-(p+\epsilon)\end{pmatrix},$$ where $$a,b\in\mathbb C$$ are arbitrary complex vectors such that $$|a|^2\le p(1-p)\equiv r_{p}^2,\qquad |b|^2\le (p+\epsilon)(1-(p+\epsilon))\equiv r_{p+\epsilon}^2.$$ The trace distance then reads $$\|\sigma-\rho\|_1=\sqrt{\epsilon^2+|a-b|^2}.$$ To get maximum and minimum values of this quantity we notice that $$(r_p-r_{p+\epsilon})^2 \le |a-b|^2\le (r_p+r_{p+\epsilon})^2,$$ which immediately translates into a bound for the trace distance.