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:

enter image description here

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:enter image description here

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?

  • 2
    $\begingroup$ 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). $\endgroup$ Commented Jan 28, 2020 at 13:35

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.