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Greens functions and density of states

  1. Feb 22, 2010 #1
    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.

  2. jcsd
  3. Feb 23, 2010 #2
    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?
  4. Feb 24, 2010 #3


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    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 [tex] \sum (\omega - \epsilon_\nu)^{-1} [/tex]. Now using [tex] 1/x=P(1/x)+2\pi i \delta (x) [/tex] where P stands for taking the principal value, you should be able to derive your formula.
  5. Feb 24, 2010 #4


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    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.

  6. Feb 24, 2010 #5
    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.
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