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For gapped systems, if with unique ground state, correlation function decays in exponential form. However, for gapless systems, if with unique ground state, correlation function decays as a polynomial. Why? Is there some model responsible for it? What is the physical reason?

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    $\begingroup$ Here a counter example, namely a system having exponential decay but no energy gap : Example 2 [p 596] in : The spectral gap for some quantum spin chains with discrete symmetry breaking, Commun. Math. Phys. 175, 565–606 (1996) $\endgroup$ – jjcale Jul 29 '16 at 17:36
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That gapped correlators decay exponentially can be proven from the spectral representation. Recall the two point function for a scalar is \begin{align} \langle \mathcal{O}(p)\mathcal{O}(0) \rangle=\int_0^\infty \frac{\rho(\mu^2)}{p^2+\mu^2}d\mu^2 \end{align} where $\rho(\mu^2)$ is the spectral function (the result generalises trivially for fields that aren't scalars and systems that aren't relativistic; there's still a spectral representation but the illustration will be cleaner here if we use the above). In position space, \begin{align} \langle \mathcal{O}(x)\mathcal{O}(0) \rangle=\int_0^\infty \rho(\mu^2)\frac{\exp\left(-\mu r\right)}{r}d\mu^2 \end{align}

Now if the spectrum of $\mathcal{O}$ is gapped, that means there is no spectral weight beneath an energy $\Delta$, ie $\rho (\mu^2)=0$ for $\mu<\Delta$. \begin{align} \langle \mathcal{O}(x)\mathcal{O}(0) \rangle&=\int_\Delta^\infty \rho(\mu)\frac{\exp\left(-\mu r\right)}{r}2\mu d\mu\\ &=\int_0^\infty \rho(\mu+\Delta)\frac{\exp\left(-(\mu+\Delta) r\right)}{r}2\left(\mu+\Delta\right)d\mu \end{align} where we simply shifted $\mu\rightarrow\mu-\Delta$. Pulling out a factor \begin{align} \langle \mathcal{O}(x)\mathcal{O}(0) \rangle&=\frac{\exp\left(-\Delta r\right)}{r}\int_0^\infty \rho(\mu+\Delta)\exp\left(-\mu r\right)2\left(\mu+\Delta\right)d\mu \end{align} Now consider the long distance behaviour $r\rightarrow \infty$. The integral is then dominated by the lower limit $\mu\approx 1/r\rightarrow 0$ (where the exponential can be approximated with $1$). Let's assume that in this region, i.e. close to the gap, the spectral function looks like $\rho\sim\mu^\alpha$. Then we have \begin{align} \langle \mathcal{O}(x)\mathcal{O}(0) \rangle &\rightarrow \frac{2\Delta\exp\left(-\Delta r\right)}{r}\int_0^{1/r} \mu^\alpha d\mu \\ &=\frac{2\Delta\exp\left(-\Delta r\right)}{r^{2+\alpha}} \end{align}

This shows gapped correlators decay exponentially; a comment above claims the converse is not true, and gapless systems can have exponentially decaying correlators. I admittedly only skimmed the paper, but it seemed like they only discuss gapped states. I would be interested in the details of a counter example though.

I should also add that we should always bear in mind what degrees of freedom are captured by the correlators we're talking about. If $\mathcal{O}$ is an electron operator, then we know that the electron is gapped. But this doesn't mean for instance our system is insulating: single particle Greens functions don't know about Cooper pairs for instance; that information will be encoded in the two particle Greens function$^1$.

$^1$As an aside, you can also have gapless superconductors i.e. not even the electron is fully gapped as the gap has nodes along the Fermi surface.

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