If we take the Gibbs state $\rho _G = e^{-\beta H}/Z$, at very low tempratures $\beta \to \infty$, then we get the ground state. I understand this comes up because only the first term survives in the expansion $e^{-\beta H}=e^{-\beta E_0}|0\rangle\langle 0|$. My question is, is there a bound on how much these differ for a finite temprature $\beta$. By this I mean either in trace/hilbert distance or just bounding the difference between the average enrgy of the gibbs state with the ground state energy. I recall seeing a result at some point which connected it with the spectral gap. Now I get a very rough result by assuming only the first two terms survive \begin{align} \operatorname{Tr} (H e^{-\beta H} ) - E_0 &= \frac{e^{-\beta E_0}E_0+e^{-\beta E_1}E_1}{e^{-\beta E_0}+e^{-\beta E_1}} -E_0 \\ \\ &= 1/Z (e^{-\beta E_1}(E_1 - E_0)) \\ \\ &= \frac{e^{-\beta E_1}}{e^{-\beta E_0}+e^{-\beta E_1}}(E_1 - E_0) = \frac{1}{e^{+\beta \Delta}+1}\Delta \end{align} But this isn't how I remember it. And also I am not really satisfied with it. Does anyone have an idea? More advanced results are also welcome. Thank you for your help!
-
2$\begingroup$ Why did you expect something else? For $T \ll \Delta$ the higher energy levels are exponentially supressed, you can probably get other presentations of the same result by rewriting the expression (e.g. for $\beta \to \infty$ the +1 will not matter, so your result is asymptotically equal to $e^{-\beta \Delta} \Delta$). $\endgroup$– Sebastian RieseCommented Sep 23 at 16:57
-
$\begingroup$ @SebastianRiese Thanks for mentioning the other presentations, much more compact. I agree that the end result captures the important aspects. I am mostly disatisfied with the derivation. Specifically, even if we assume the hamiltonian is gapped that doesn't tell us anything about $E_3 - E_2$. So I don't really have any strong justification for irgnoring those terms. But I suppose the result will just get more complicated with the extra terms. As for expecting something else this is only just going by memory. But perhpas the previous result I remember was about the overlap. $\endgroup$– nikoslouCommented Sep 23 at 17:24
-
$\begingroup$ As long as $E_2$ is not degenerate, due to the nature of exponentials, for a fixed Hamiltonian if $E_3 - E_2 \ll E_1 - E_2$ for $\beta \to \infty$ the leading term will asymptotically be the same. A degeneracy will only result in a factor proportional to the grade of degeneracy. Of course things change if you have a continuous spectrum (then you will typically have a power law, but the specifics depend strongly on the spectrum). $\endgroup$– Sebastian RieseCommented Sep 23 at 17:40
1 Answer
With no loss of generality, you can assume that $E_0=0$. You are assuming a discrete spectrum, but in most cases, even if you have a spectral gap, you often have a continuum right after the gap $E_g>0$. This modifies the asymptotics, even though the logarithmic leading order is still given by the gap. Assuming that you have a density of states of the form (assuming $s>0$ in the following): $$ D(E) = D_0\delta(E)+D_g(E-E_g)_+^{s-1} $$ you get the low energy asymptotic for the mean energy: $$ \begin{align} \langle E\rangle &= \frac{\int Ee^{-\beta E}D(E)dE}{\int e^{-\beta E}D(E)dE} \\ &= \frac{e^{-\beta E_g}D_g\Gamma(s+1)\beta^{-s-1}}{D_0+e^{-\beta E_g}D_g\Gamma(s)\beta^{-s}} \\ &\sim \frac{D_g\Gamma(s+1)}{D_0}\beta^{-s-1}e^{-\beta E_g} \end{align} $$
Note that the result extends to the case where there is no gap $E_g = 0$ and $D_0 = 0$: $$ \langle E\rangle \sim s\beta^{-1} $$
-
$\begingroup$ Thank you. I think with this and the comment above from Sebastian I can consider the question answered. $\endgroup$– nikoslouCommented Sep 24 at 13:10