What's the corresponding energy dispersion of Green function?

I want write a "toy" Green function witch can describe the electrons in the band with a width of $$±W$$ with uniform density of states (DOS). The reference gives an explicit expression of imaginary-time Green function: $$G(i\omega)=\frac{1}{2\pi}\ln \frac{i\omega+W}{i\omega+W}$$ with uniform DOS like this:

Thus, I am confused of the origin for above form of Green function?

In addition, I have made some attempts: from the start point of tight-binding model, (half-filling one dimension for simplicity), the Hamiltonian and Green function are: $$H=-W\sum_{i,j}c_{i}^\dagger c_j+h.c.=-W \cos k c_k^\dagger c_k\\G(i\omega,k)=\frac{1}{i\omega+W\cos k}$$ to obtain the similar expression of initial form for Green function, I integral in term of momentum: $$G(i\omega)=\int_{-\pi}^\pi G(i\omega,k)dk=\frac{-2i(-1)^{Floor[\frac{\pi-2Arg[i+\omega]+Arg[1+\omega^2]}{2\pi}]}}{\sqrt{1+\omega^2}}$$ the result is very tedious and the DOS is like this:

which is not similar to the initial form. Thus, I am also confused the explicit expression of band structure(energy dispersion), or model, corresponding to the initial form of Green function?

• The density of states $g \propto \frac{\partial K}{\partial E} \propto \frac{1}{\frac{\partial E}{ \partial k}}$. So for a constant density of states you need $v = \frac{\partial E}{\partial k} = const$, that is you need a linear dispersion of the form $E = v k$ (inside the band). Commented Jan 28, 2020 at 13:35

To get this Green function you should take a wide-band limit. That is, a continuum limit where the energy of the electrons is $$\epsilon_k = v_F k$$ and then $$k$$ has limits $$\pm W/v_F$$. This is what you get when you linearize the spectrum about the Fermi energy. From the tight binding model you will get if you add a chemical potential $$\mu$$, and then linearize and take the continuum limit. You get something like $$H = \sum_k v_F k c^{\dagger}_k c_k$$, and $$k=2\pi n/L$$ and it is measured from $$k_F$$.
The single-particle Matsubara Green function is then $$g_{\epsilon_k}(i\omega) = \left( i\omega-\epsilon_k\right)^{-1}$$ and you sum over it to get the GF at a certain point [note that the factor of $$1/L$$ is added because we are looking at the correlation function of $$\psi(x)$$] $$G(i\omega) = \frac{1}{L} \sum_{k} g_{\epsilon_k}(i\omega) = \frac{1}{2\pi}\int_{-W/v_F}^{W/v_F}\! dk g_{\epsilon_k}(i\omega) = \frac{1}{2\pi v_F}\int_{-W}^{W}\frac{d\epsilon}{i\omega-\epsilon}$$ where we used $$2\pi/L = dk$$ in taking the continuum limit. This integral will result in what you wrote.