Two definitions:

  1. $\boxed{\Delta}$ is defined to be the energy gap above the ground state.
  2. $\boxed{\xi_t}$ is defined to be the temporal correlation length, i.e. it is the number such that for a generic observable $\mathcal O$ (with a zero expectation value in the ground state) we have that $\langle \mathcal O(t=T) \mathcal O(t=0) \rangle_\textrm{gs} \sim \exp(-T/\xi_t)$ as $T \to \infty$.

I would now like to derive that $\boxed{\xi_t= \frac{1}{\Delta}}$.

It is natural to rewrite the following (setting $E_\textrm{gs} = 0$):

$$ \langle \mathcal O(t=T) \mathcal O(t=0) \rangle_\textrm{gs} = \langle \mathcal O e^{-iHT} \mathcal O \rangle_\textrm{gs} = \color{red}{\sum_\alpha |\langle 0 | \mathcal O | \alpha \rangle |^2 \; \exp(-iE_\alpha T)}.$$

The goal is to show that (under generic conditions), the latter expression has the asymptotic behaviour $\color{red}{\sim \exp(- E_1 T)}$ (where I have implicitly ordered the eigenstates $|\alpha\rangle$ ($\alpha = 1,2,\cdots$) in ascending energy). Indeed, since $\Delta = E_1$, this would prove the above.

However, I am not sure how to justify that asymptotic expression. I get the intuitive idea (coming from cancellations of phase factors) and I can imagine deriving it from some version of the method of steepest descent / saddle-point approximation, but I am not sure. The basic issue is that I am struggling to see how the imaginary phase would become a real exponential factor.

EDIT: It is probably useful to rewrite the above red expression as thus: $$ \langle \mathcal O(t=T) \mathcal O(t=0) \rangle_\textrm{gs} = \int_\Delta^\infty g(\omega) e^{-i\omega T} \mathrm d \omega $$ with $$ g(\omega) := \sum_\alpha |\langle 0 | \mathcal O | \alpha \rangle |^2 \delta(\omega - E_\alpha). $$ Moreover, we might need to presume that the spectrum is a continuum (above the gap), with a density of states $\nu(\omega)$. which is smooth for $\omega > \Delta$. Moreover, if we define $\mu(\omega)$ as the weight density of the state $\mathcal O |0\rangle$ in the part of Hilbert space with energy $\omega$, then we the above sum becomes $$g(\omega) = \left(\int \mathrm d E \; \nu(E)\right) \mu(E) \; \delta(\omega - E) = \nu(\omega) \mu(\omega).$$ The main point is then that we can consider $g(\omega)$ to be a positive function which is smooth, aside from possible issues at the onset $\omega\approx \Delta$. We are now trying to understand the asymptotics of the Fourier-transform of $g(\omega)$.


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