# Greens functions and density of states

1. Feb 22, 2010

### Niles

Hi guys

Here they say that whenever we write the Hamiltonian for a non-interacting system in its eigenbasis, then we have that

$$A(\nu, \omega) = 2\pi \delta(\omega-\varepsilon_\nu).$$

How can this statement be proven? Do you have any hints for this?

Any help will be greatly appreciated.

Sincerely,
Niles.

2. Feb 23, 2010

### Niles

Ok, perhaps I should reformulate my question: I keep reading that the imaginary part of the retarded Greens function is the density of states, and I am quite sure it has got something to do with

$$A(\nu, \omega) = 2\pi \delta(\omega-\varepsilon_\nu).$$

since if we sum over ν we get something proportional to the density of states on the RHS. SO I guess my question is: How do we show that in the eigenbasis of the Hamiltonian for a non-interacting system, the imaginary part of the Green's function is equal to the density of states?

3. Feb 24, 2010

### DrDu

I don't know where the eigenbasis should enter as the Greensfunctions do not depend on a basis. But for a free particle, the greens function is basically $$\sum (\omega - \epsilon_\nu)^{-1}$$. Now using $$1/x=P(1/x)+2\pi i \delta (x)$$ where P stands for taking the principal value, you should be able to derive your formula.

4. Feb 24, 2010

### ZapperZ

Staff Emeritus
You need to consider this as the spectral function of the single-particle propagator. To get the density of states, you sum up over all momentum. So you show that the result of the DOS matches the density of states for a free non-interacting electron gas.

Mattuck actually did this in his book "A Guide to Feynman Diagrams in the Many-Body Problem" for this very case. It is a Dover book, so it's cheap. It is also extremely useful.

Zz.

5. Feb 24, 2010

### Niles

In my book the Greens functions in real-space and momentum-space look different; so how it is that they cannot depend on the chosen basis?

The point is that in the book they only do it in momentum-space, and add as a side-note that "this in fact is true for all non-interacting fermions in any bases you like" - that is the statement I think needs more explanation.