From Greens function to DOS

Math  DFT 

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

G(E)=innn(EEn+iη)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)=nnn(EEn+iη).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:

j<jG(E)j>=jnjnnj(EEn+iη)=j1(EEj+iη)\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 $G$:

Tr[G(E)]=j1(EEj+iη)(1)\mathrm{Tr}[G( E)] = \sum_j \frac{1}{(E-E_j+i\eta)} \tag{1}

According to :link: the Sokhotski-Plemelj formula:

Tr[G(E)]=j1(EEj+iη)=j[P1EEjiπδ(EEj)](3)\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 :

ρ(E)=jδ(EEj)(4)\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:

ImTr[G(E)]=πjδ(EEj)=πρ(E)(5)\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 :link: this post) of the DOS since Eq. 3 only works when $\eta \rightarrow 0^+$.



Author | Chengcheng Xiao

Currently a postdoctoral research associate at Imperial College London. Predicting electron behaviour since 2016.