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.

  • 1
    $\begingroup$ This also holds when only one of the density operators is pure by a similar argument (combined with the dominated convergence theorem). But the general case does indeed seem harder. $\endgroup$
    – MaoWao
    Sep 6, 2023 at 10:34
  • $\begingroup$ @LL3.14 How so? Regard the indices of the xs. One is $x_{i}$ whilst the other is $x_{j}$. $\endgroup$
    – Hldngpk
    Sep 6, 2023 at 19:55
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    $\begingroup$ @MaoWao You are correct, I have that case under control as well. Computing Fidelitites is simple when at least one of the terms in the fidelity is a pure state (i.e. a projector). This case, with both states not being pure, seems insurmountable. $\endgroup$
    – Hldngpk
    Sep 6, 2023 at 19:57
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    $\begingroup$ No, I am saying there us two times "$\lim ...$" $\endgroup$
    – LL 3.14
    Sep 6, 2023 at 20:55
  • 1
    $\begingroup$ I fixed it. Thank you @LL3.14 $\endgroup$
    – Hldngpk
    Sep 6, 2023 at 20:57


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