固定状態に関して測定された忠実度の差が0に近い場合、2つの状態の近さについて何が言えるでしょうか?


2

2つの州 $ \ rho $ $ \ sigma $ があるとします。それが与えられます、

$$ Tr((\ rho-\ sigma)| \ psi \ rangle \ langle \ psi |)\ geq \ epsilon $$ ここで、 $ | \ psi \ rangle $ は固定状態であり、 $ \ epsilon \ rightarrow 0 $

次に、2つの州 $ \ rho $ $ \ sigma $ の近さについて何でも結論付けられますか>距離の目安は?

4

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.


4

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.