A limit of a particular Quantum Fidelity

Question

I have the following problem.

Let $\mathbf{\hat{\rho}}(t)$ and $\mathbf{\hat{\sigma}}(t)$ be two trace class positive operators acting on a Hilbert space of infinite dimension for all $t > 0$. More precisely assume that $$ \mathbf{\hat{\rho}}(t):= \int p_{i}(x)e^{-ixt\mathbf{\hat{B}}}\big|\psi\big\rangle \big\langle \psi\big|e^{ixt\mathbf{\hat{B}}}dx $$ $$ \mathbf{\hat{\sigma}}(t):= \int p_{j}(x)e^{-ixt\mathbf{\hat{B}}}\big|\psi\big\rangle \big\langle \psi\big|e^{ixt\mathbf{\hat{B}}}dx $$

where $\mathbf{\hat{B}}$ is a self-adjoint operator with purely absolutely continuous spectrum and $\big|\psi\big\rangle$ is any vector in the Hilbert space in question, and $p_{i}$ and $p_{j}$ are probability distributions with compact support which is nonoverlapping. I am trying to prove that $$\lim_{t\rightarrow \infty}\big\|\sqrt{\mathbf{\hat{\rho}}(t)}\sqrt{\mathbf{\hat{\sigma}}(t)}\big\|_{1}= 0 \;\; (quantum\; fidelity)$$

However, this has proven to be quite a challenge since there are no good upper bounds for the quantum fidelity in the general case were both of the operators in question are not pure. I have tried using the following celebrated bound.

$$ \big\|\sqrt{\mathbf{\hat{\rho}}(t)}\sqrt{\mathbf{\hat{\sigma}}(t)}\big\|_{1}\leq\sqrt{1-\big\|\mathbf{\hat{\rho}}(t)-\mathbf{\hat{\sigma}}(t)\big\|_{1}^{2}} $$ but this just replaces a very difficult problem with one of equal complexity.

For the simpler version of this problem where

$$\mathbf{\hat{\rho}}(t):=e^{-ix_{i}t\mathbf{\hat{B}}}\big|\psi\big\rangle \big\langle \psi\big|e^{ix_{i}t\mathbf{\hat{B}}} $$

and

$$ \mathbf{\hat{\sigma}}(t):=e^{-ix_{j}t\mathbf{\hat{B}}}\big|\psi\big\rangle \big\langle \psi\big|e^{ix_{j}t\mathbf{\hat{B}}} $$

with $x_{i}\neq x_{j}$ and all of the other assumptions preserved I can easily show the analogous hypothesis.

Here

$$ \lim_{t\rightarrow \infty}\big\|\sqrt{\mathbf{\hat{\rho}}(t)}\sqrt{\mathbf{\hat{\sigma}}(t)}\big\|_{1} = \big|\langle \psi\big|e^{-t(x_{i}-x_{j})\mathbf{\hat{B}}}\big|\psi\big\rangle\big| = \int e^{-t(x_{i}-x_{j})\lambda}d\mu_{\psi}(\lambda) $$ where $d\mu_{\psi}(\lambda)$ is the absolutely continnuous spectral measure afforded by $\big|\psi\rangle$. Owing to the Riemann Lebegues lemma indeed $\lim_{t\rightarrow \infty}\int e^{-t(x_{i}-x_{j})\lambda}d\mu_{\psi}(\lambda) = 0$. Due to this result, I am led to believe that the more general case where $\mathbf{\hat{\rho}}(t)$ and $\mathbf{\hat{\sigma}}(t)$ are uncountable mixtures as presented above, we should have the same sort of behavior as $t\rightarrow \infty$. However, the quantum fidelity is unwieldy. Any help tackling this problem would be greatly appreciated.

Answer 1

Neat question! Here is an almost complete answer; I encourage someone to edit my answer to tighten up the math at the end, but all of the steps are clear.

Inspired by another inequality for fidelity, let's resolve the identity using states that are Fourier transforms of eigenstates $\hat{\mathbf{B}}$; i.e., using states that are displaced by $e^{-i x t \hat{\mathbf{B}}}$: $$\mathbb{I}=\int dx |x\rangle\langle x|;\quad e^{-i y t \hat{\mathbf{B}}}|x\rangle=|x+ y t\rangle; \quad \langle x|y\rangle=\delta(x-y).$$ These can always be found by using $|x\rangle=\int dp e^{-ixp}|p\rangle$ with $|p\rangle$ the delta-orthogonal eigenstates of $\hat{\mathbf{B}}$.

We use that the trace norm can always be rewritten using $\mathrm{Tr}(|A|)=\mathrm{Tr}(A U)$ for some unitary $U$ and the usual $|A|=\sqrt{A A^\dagger}$. The fidelity $F(\rho,\sigma;t)=||\sqrt{\rho(t)}\sqrt{\sigma(t)}||_1=\mathrm{Tr}(|\sqrt{\rho(t)}\sqrt{\sigma(t)}|)$ can then be written as $$F(\rho,\sigma;t)=\mathrm{Tr}(\sqrt{\rho(t)}\sqrt{\sigma(t)}U)$$ for some $U$. Inserting the resolution of identity and then using a Cauchy-Schwarz inequality $|\mathrm{Tr}( A^\dagger B)|\leq \sqrt{\mathrm{Tr}(A^\dagger A)\mathrm{Tr}(B^\dagger B)}$ yields \begin{aligned} F(\rho,\sigma;t)=&\int dx \mathrm{Tr}(\sqrt{\rho(t)}|x\rangle\langle x|\sqrt{\sigma(t)}U)\\ =&\int dx \mathrm{Tr}(\sqrt{\rho(t)}\sqrt{|x\rangle\langle x|}\sqrt{|x\rangle\langle x|}\sqrt{\sigma(t)}U)\\ &\leq \int dx \mathrm{Tr}(\sqrt{\rho(t)}\sqrt{|x\rangle\langle x|}\sqrt{|x\rangle\langle x|}\sqrt{\rho(t)})\mathrm{Tr}(\sqrt{|x\rangle\langle x|}\sqrt{\sigma(t)}UU^\dagger \sqrt{\sigma(t)}\sqrt{|x\rangle\langle x|})\\ &= \int dx \langle x|\rho(t)|x\rangle\langle x|\sigma(t)|x\rangle. \end{aligned} For this we simply needed that $\rho$, $\sigma$, and $|x\rangle\langle x|$ are Hermitian and that $\mathrm{Tr}(\sqrt{\rho}|x\rangle\langle x|\sqrt{\sigma}U)\leq |\mathrm{Tr}(\sqrt{\rho}|x\rangle\langle x|\sqrt{\sigma}U)|$.

Next, we expand $|\psi\rangle=\int dx \psi(x)|x\rangle$ in our chosen basis and use $e^{it \hat{\mathbf{B}}}|x\rangle=|x-t\rangle$ to compute $$\langle x|\rho(t)|x\rangle=\int dy p_i (y) \langle x-yt|\psi\rangle\langle \psi|x-yt\rangle=\int dy p_i(y) |\psi(x-yt)|^2$$ and $$\langle x|\sigma(t)|x\rangle=\int dz p_j(z) |\psi(x-zt)|^2.$$

Putting these together, our inequality becomes $$F(\rho,\sigma;t)\leq \int dx \int dy\int dz p_i(y) p_j(z) |\psi(x-yt)|^2 |\psi(x-zt)|^2$$ and the goal is to prove that the integral on the right-hand side vanishes in the limit of $t\to \infty$ when $p_i(x)$ and $p_j(x)$ have nonoverlapping compact support.

Now is where the hand-waving begins, but I'm sure someone can make this rigorous. Define the support of $p_i(x)$, $p_j(x)$, and $|\psi(x)|^2$ respectively as $S_i$, $S_j$, and $S_\psi$. The integrand is only nonzero for regions where $y\in S_i$, $z\in S_j$, $x-yt\in S_\psi$, and $x-zt\in S_\psi$. Since $S_i$ and $S_j$ are disjoint, we must have $|y-z|>0$ for the integrand to be nonzero (ignoring regions of zero measure). This means that the the integrand can only be nonzero if $S_\psi$ has two "spaced apart" regions with $x^\prime\equiv x-yt\in S_\psi$ and $x^\prime+(y-z)t\in S_\psi$. In the limit of $t\to\infty$, the integrand is only nonzero if there are two "infinitely spaced apart" regions $S_\psi$.

So, for any state $|\psi\rangle$ whose support in the Fourier transform basis of $\hat{\mathbf{B}}$ does not have two regions that are infinitely far apart, the relationship $$\lim_{t\to\infty} F(\rho,\sigma;t)=0$$ holds.