Remembering that in this post we have derived that the greens function for non-interacting time-independent is (expressed in eigenvalue basis):
G(E)=in∑(E−En+iη)∣n⟩⟨n∣
For historical reason (I’m not actually sure why but that’s what everytext book says…), we ditch the imaginary sign:
G(E)=n∑(E−En+iη)∣n⟩⟨n∣.
Now, if we sandwich it in between two eigenvectors (of the Hamiltonian) and take the sum over these eigenvectors, we get:
j∑<j∣G(E)∣j>=j∑n∑(E−En+iη)⟨j∣n⟩⟨n∣j⟩=j∑(E−Ej+iη)1
This can also be written as the trace of the matrix representation of $G$:
Tr[G(E)]=j∑(E−Ej+iη)1(1)
According to the Sokhotski-Plemelj formula:
Tr[G(E)]=j∑(E−Ej+iη)1=j∑[PE−Ej1−iπδ(E−Ej)](3)
Remembering that the definition of the density of state (DOS) is :
ρ(E)=j∑δ(E−Ej)(4)
and since we have $\delta(E-E_j)$ as the imaginary part of the Eq. 3, we can express it as:
ImTr[G(E)]=−πj∑δ(E−Ej)=−πρ(E)(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 $\eta \rightarrow 0^+$.