I am working on calculating the Green's function for a Hamiltonian $H = H_0 + V$ numerically, where I'm specifically interested in $G(\omega) = \frac{1}{\omega - H + i\epsilon}$ at $\omega = 0$.

A challenge arises due to the eigenvalues of $H_0$ containing several zeros. This causes $G_0(\omega) = \frac{1}{\omega - H_0 + i\epsilon}$ to diverge or "explode" at $\omega = 0$, when $\epsilon$ is small.

The Hamiltonians $H$ and $H_0$ are discrete and single-particle. Therefore, the zero eigenvalues are not part of a continuum.

I am considering the use of the Dyson series, which expresses the Green's function $G$ in terms of a perturbative expansion:

$$G = G_0 + G_0 V G_0 + G_0 V G_0 V G_0 + \ldots = G_0 \frac{1}{1-VG_0}$$

where $G_0$ is the Green's function of the unperturbed Hamiltonian $H_0$. Since the zero eigenvalues of $H_0$ casue $G_0$ to diverge, the Dyson series, or the geometric series $\frac{1}{1-VG_0}=1+VG_0+VG_0VG_0+\ldots$, doesn't converge.

My question is: Given the zero eigenvalues in $H_0$ and the associated divergence in $G_0(\omega)$ at $\omega = 0$, how should I approach the calculation of $G(\omega = 0)$ using the Dyson series? Are there specific techniques or modifications to the series expansion that can be applied in this scenario to manage the divergence caused by the zero eigenvalues?

Any guidance or suggestions on how to effectively handle this situation in the calculation would be greatly appreciated.

Thank you!

  • $\begingroup$ Is it an isolated (degenerate?, non-degenerate?) eigenvalue, or rather part of a continuum? $\endgroup$
    – Qmechanic
    Jan 9 at 9:21
  • $\begingroup$ @Qmechanic The Hamiltonians are discrete. Therefore, the zero eigenvalues are not part of a continuum, although I'd appreciate any way of handling the zero eigenvalues in a continuum, too. $\endgroup$
    – Frank
    Jan 9 at 10:18
  • $\begingroup$ Do you mean numerically or what are you interested in? $\endgroup$ Jan 9 at 10:42
  • 1
    $\begingroup$ @TobiasFünke I mean numerically. I will make that clear in the question. $\endgroup$
    – Frank
    Jan 9 at 10:52
  • 2
    $\begingroup$ There is nothing special about eigenvalue being zero - the denominator always vanishes (has a pole) near an eigenvalue, i.e., when $\omega \approx \epsilon_n$. One could shift the energy origin and make it non-zero, and it wouldn't change. The problem lies elsewhere - possibly in degeneracy, but it is hard to say without more details about the Hamiltonian. $\endgroup$
    – Roger V.
    Jan 9 at 10:53


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