Greens functions and density of states

[Niles]

Hi guys

I found this on Google Books: http://books.google.dk/books?id=v5v...resnum=5&ved=0CCwQ6AEwBA#v=onepage&q=&f=false

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.Niles.

 

[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?

 

[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.

 

[ZapperZ]

Hi guys

I found this on Google Books: http://books.google.dk/books?id=v5v...resnum=5&ved=0CCwQ6AEwBA#v=onepage&q=&f=false

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.


Niles.

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.

 

[Niles]

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.

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?

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.

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.