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A Dyson's equation and Green's functions

  1. Aug 14, 2016 #1
    Hi,
    Is the Dyson's equation basis independent (for instance, I construct the basis set where the elements are atomic orbitals and those orbitals are non-orthogonal) ?

    What is the unperturbed retarded Green's function for one-particle case in matrix notation if the basis functions are not orthogonal?

    Thank you!
     
  2. jcsd
  3. Feb 23, 2018 #2

    TeethWhitener

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    I'm pretty sure the general form of Dyson's equation is basis-independent (there is no reference to a basis).
    $$G(\mathbf{k},\omega) = G_0(\mathbf{k},\omega)+G_0(\mathbf{k},\omega)\Sigma(\mathbf{k},\omega)G(\mathbf{k},\omega)$$
    But the Green's functions are typically defined in the position basis, so you have to use completeness relations to go to the atomic orbital basis.

    Edit: So if you have
    $$i\hbar G^+ (\mathbf{x}_2,t_2;\mathbf{x}_1,t_1) = \theta(t_2-t_1)\langle\mathbf{x}_2|U(t_2,t_1)|\mathbf{x}_1\rangle$$
    and you want to work in the energy eigenbasis ##\{\phi_n\}##, you need to insert the completeness relations in the appropriate places:
    $$i\hbar G^+ (\mathbf{x}_2,t_2;\mathbf{x}_1,t_1) = \theta(t_2-t_1)\sum_{m,n}\langle\mathbf{x}_2|\phi_m\rangle\langle \phi_m|U(t_2,t_1)|\phi_n\rangle\langle \phi_n|\mathbf{x}_1\rangle$$
    and since ##\langle\mathbf{x}_2|\phi_m\rangle = \phi_m(\mathbf{x}_2)##, we can rewrite this as:
    $$i\hbar G^+ (\mathbf{x}_2,t_2;\mathbf{x}_1,t_1) = \theta(t_2-t_1)\sum_{m,n}\phi_m(\mathbf{x}_2)\phi_n^*(\mathbf{x}_1)\langle \phi_m|U(t_2,t_1)|\phi_n\rangle$$

    Further edit (based on post #6 below): you don't have to use an orthogonal basis if you don't want to, but you can't use the completeness relations as I illustrated above. You would have to orthogonalize the non-orthogonal basis for use in the completeness relations.
     
    Last edited: Feb 23, 2018
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