# Gap system's correlation function

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?

• 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) – jjcale Jul 29 '16 at 17:36

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}
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.