From Greens function to DOS

Remembering that in this post we have derived that the greens function for non-interacting time-independent is (expressed in eigenvalue basis):

$$G(E) = i \sum_n \frac{\ket{n} \bra{n}}{(E-E_n+i\eta)}$$

For historical reason (I’m not actually sure why but that’s what everytext book says…), we ditch the imaginary sign:

$$G(E) = \sum_n \frac{\ket{n} \bra{n}}{(E-E_n+i\eta)}.$$

Now, if we sandwich it in between two eigenvectors (of the Hamiltonian) and take the sum over these eigenvectors, we get:

$$\begin{aligned} \sum_j<j|G(E)|j> &= \sum_j \sum_n \frac{\braket{j \vert n} \braket{n \vert j}}{(E-E_n+i\eta)}\\ &=\sum_j \frac{1}{(E-E_j+i\eta)} \end{aligned}$$

This can also be written as the trace of the matrix representation of $G$:

$$\mathrm{Tr}[G( E)] = \sum_j \frac{1}{(E-E_j+i\eta)} \tag{1}$$

According to the Sokhotski-Plemelj formula:

$$\mathrm{Tr}[G( E)] = \sum_j \frac{1}{(E-E_j+i\eta)} = \sum_j \left [ \mathcal{P} \frac{1}{E-E_j} - i \pi \delta(E-E_j) \right ] \tag{3}$$

Remembering that the definition of the density of state (DOS) is :

$$\rho(E) = \sum_j \delta(E-E_j) \tag{4}$$

and since we have $\delta(E-E_j)$ as the imaginary part of the Eq. 3, we can express it as:

$$\mathrm{Im} \mathrm{Tr}[G(E)] = -\pi \sum_j \delta(E-E_j) = - \pi \rho(E) \tag{5}$$

$\eta$ acts as an smearing factor (because it is a manually added finite life time to the time evolution propagator of states. See Eq. 10 in this post) of the DOS since Eq. 3 only works when $\eta \rightarrow 0^+$.