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

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


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

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


$$ \lim_{t\rightarrow \infty}\big\|\sqrt{\mathbf{\hat{\rho}}(t)}\sqrt{\mathbf{\hat{\sigma}}(t)}\big\|_{1} = \big|\langle \psi\big|e^{-t(x_{i}-x_{j})\mathbf{\hat{B}}}\big|\psi\big\rangle\big| = \int e^{-t(x_{i}-x_{j})\lambda}d\mu_{\psi}(\lambda) $$ 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.

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    $\begingroup$ If B=0, aren't rho and sigma equal (and thus your statement incorrect)? $\endgroup$ Sep 18, 2023 at 16:11
  • $\begingroup$ re: Norbert, the distributions are non-overlapping and hence $\rho,\sigma$ are different. $\endgroup$
    – hulsey
    Sep 18, 2023 at 19:07
  • $\begingroup$ @NorbertSchuch thank you for your response. I am actually only interested in the case where $\mathbf{\hat{B}}\neq \mathbf{0}$; my apologies for not being specific. Indeed, if $\mathbf{\hat{B}}=\mathbf{0}$ then both states in question would be the same since the probability densities would integrate out to 1. $\endgroup$
    – Hldngpk
    Sep 19, 2023 at 7:51
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    $\begingroup$ @hulsey This is incorrect for B=0. -- The point seems that the questions demands a purely absolutely continuous spectrum, which rules out such cases. $\endgroup$ Sep 19, 2023 at 8:33
  • $\begingroup$ @NorbertSchuch Good point! Thank you. $\endgroup$
    – Hldngpk
    Sep 19, 2023 at 11:43

1 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.

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    $\begingroup$ @Hldngpk I have nothing to add! Your comments are all spot-on: 1) is well taken, I have only proven for a certain class of operators; 2) is helpful, I appreciate you tightening that up; 3) yes since this method proves your relation for more than just pure states I feel that it should be general enough. If not, and you need another method to prove the general case, this is at least stronger evidence that your relation is valid $\endgroup$ Sep 23, 2023 at 22:00
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    $\begingroup$ @Hldngpk does it help that I did not actually define $|x\rangle$ as eigenstates of anything? I simply formed them from a continuous superposition of eigenstates of $\hat{\mathbf{B}}$, and that state will obey the shift property and orthonormality property required so long as the eigenstates of $\hat{\mathbf{B}}$ are delta-orthogonal $\endgroup$ Sep 24, 2023 at 3:02
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    $\begingroup$ @Hldngpk all I need is an infinite set of eigenstates of $\hat{\mathbf{B}}$ such that $\langle p|p^\prime\rangle=\delta(p-p^\prime)$ and $\int dp |p\rangle\langle p|=\mathbb{I}$, then I just define $|x\rangle=\int dp e^{-ixp}|p\rangle$. Now you might say $\hat{\mathbf{B}}$ has compact support so this is not true, and then my suggestion would be to instead look at the eigenstates of $t\hat{\mathbf{B}}$ and those will have eigenvalues spanning $\mathbb{R}$ and so the Fourier transform should be nice $\endgroup$ Sep 24, 2023 at 13:41
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    $\begingroup$ @Hldngpk say we want $|x;t\rangle$ such that $e^{-iytB}|x;t\rangle=|x+y;t\rangle$. Just define $|x;t\rangle=\int dp e^{-iptx}|pt\rangle$ where $tB|pt\rangle=tB|pt\rangle$. Then $\langle x;t|x^\prime;t\rangle=\int dp \exp[i pt(x-x^\prime)]=t^{-1}\int d(pt) \exp[i (pt)(x-x^\prime)]$ where the latter integral takes the domain of $pt$ which is infinite even if $B$ has compact support $\endgroup$ Sep 24, 2023 at 13:46
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    $\begingroup$ very interesting response. I have to process it. By the way, you wrote $tB|pt\rangle=tB|pt\rangle$ above? Is this a typo? Did you mean to write $ tB|pt\rangle=tp|pt\rangle$ ? $\endgroup$
    – Hldngpk
    Sep 25, 2023 at 1:53

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