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Quantum physics with Green's functions

  1. Nov 13, 2007 #1
    The density of states can be expressed in terms of Green's functions and is defined as the imaginary part of the Green function (multiplied by 1/Pi).
    Can someone explain me what happens if the imaginary part is just 0? It would be the case if all Eigenvalues of the Schrödinger equation are real, which should happen very often in my understanding of quantum systems.

  2. jcsd
  3. Nov 13, 2007 #2
    The imaginary part wouldn't really be zero at the eigenvalues of the Hamiltonian, because those will be the poles of the Green's function and it will be undefined at that point.

    You get a more sensible spectral function if you add a small imaginary part to the denominator of your Green's function. Starting with a Green's function of the form:

    [tex]G(\omega) = \left[ \omega + \mu - H + i\delta \right]^{-1}[/tex]

    then the imaginary part is:

    [tex]\Im G(\omega) = \frac{-\delta}{ (\omega + \mu - H)^2 + \delta^2}[/tex]

    This pushes the poles off the real axis so you don't have singularities there, and also broadens the peaks (important for numerical studies where you are evaluating the Green's function on a regular grid). You will notice is you take the limit as [tex]\delta[/tex] goes to zero you will get a delta functions at the points where [tex]\omega + \mu - H[/tex]
  4. Nov 14, 2007 #3
    Thanks. You did help a lot.
    I didn't know that there will remain a delta function after delta went to zero. Do you know a derivation of this?

    You seem to have a background on Green's functions. Can you help me in one more question, please?
    Is it possible to get a solution of the Green's function without solving the Schrödinger equation to get the Eigenvalues?
    If it wouldn't be the case I don't really understang why the Green's function method is such an advantage. I mean, solving the Schrödinger equation for Eigenvalues and Eigenvectors, already results in having the neccessary components for the density matrix.
    Ok, there must be a reason to use Green's functions, otherwise so many people wouldn't use them. But I don't really understand the advantage.
  5. Nov 15, 2007 #4
    Off the top of my head, a derivation might go like, as [tex]\delta[/tex] goes to zero, for [tex]x = \omega + \mu - H = 0[/tex] non zero, the denominator stays non-zero and the numerator goes to zero, so you have a zero. For x = 0, then it goes like [tex]1/\delta[/tex] as [tex]\delta[/tex] goes to zero, so this is divergent. The only trick from there would be to show that the integral of [tex]\delta/(x^2 + \delta^2)[/tex] over all x is a constant value (I think it might be pi) independent of [tex]\delta[/tex].

    My experience with Green's functions is fairly limited. So the only ways I know of to evaluate Green's functions require knowledge of the full spectrum of the Hamiltonian. But I will list the advantages I can think of, all of which are condensed matter topics.

    The density of states you calculate from a non-interacting (or mean-field) theory doesn't correspond very well with experimental photo absorption/emission, because these experiments involve exciting a particle to a state with a finite lifetime that the whole system can have a response to. If the physical system were non-interacting it makes sense that the density of states would be an adequate description, because the excited particle wouldn't affect any of the other particles. But in a system with strong interactions, this isn't the case. The Green's function calculated via [tex]G(t) = \langle c(t) c^\dagger \rangle[/tex] is a direct calculation of particle excitation, so from this argument, one would expect the Green's function to give a more physically realistic picture.

    If you calculate the band structure (ie. eigenvalues of H_k) via LDA, you will have a set of energy bands which are sharp delta-function peaks at a specific energy values, all having the weight of one (ignoring degeneracy). This doesn't reflect experiment well because a variety of factors will broaden these peaks, or shift their weight to other bands. Some of these factors are: impurities, disorder, temperature, correlation effects. (In many cases these can be safely neglected, but there are certainly interesting cases where they are relevant). Impurities can be treated with the Coherent Potential Approximation, which uses a Green's function formalism (and that is the extent of my knowledge on CPA).

    My knowledge of Green's functions comes from doing Dynamical Mean Field Theory (DMFT) calculations. In the DMFT formalism, we take the Green's function at each k-value:
    [tex]G_{k}(\omega) = \frac{1}{\omega + \mu - H_k - \Sigma(\omega)}[/tex]
    where you can see a frequency dependent potential has been added to the Hamiltonian (the so-called self-energy). The spectrum of these Green's functions gives the usual k-space band structure, if the self-energy is zero. The self-energy is calculated via some other means, often with some self-consistency condition with the Green's function. The self-energy is non-Hermitian and may have a substantial imaginary part at some frequency. This can broaden the spectrum of H, create additional peaks, shift weight around, etc. to create the kinds of effects that appear in experiment. In DMFT the self-energy is used to include correlation effects. I'd imagine it's difficult / impossible to include any frequency dependent potential without using Green's functions.
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