My question concerns the self-energy of a diagonal propagator for a single-particle lattice problem. The context is Anderson Localisation, but really it's a problem of complex analysis. I would like to show that the imaginary part of the self-energy corresponds to a rate of loss of probability amplitude from a site $j$ to overlapping eigenstates at an energy $\omega$.

Let the site-diagonal propagator in the time and frequency domains be \begin{equation} G_{jj}(t) = -i\Theta(t)\langle j|e^{-i\hat{H}t}|j\rangle\longleftrightarrow G_{jj}(\omega) = \frac{1}{\omega+i\eta-\epsilon_j-S_j}, \end{equation} where $\omega$ is frequency/energy, $\epsilon_j$ is the site energy of site $j$, $\eta\equiv 0^+$ and $S_j$ is the self-energy. Now, the self-energy is complex, \begin{equation} S_j = X_j(\omega) - i\Delta_j(\omega). \end{equation} In the time domain, therefore, \begin{equation} G_{jj}(t) = \frac{1}{2\pi}\int_{-\infty}^\infty\mathrm{d}\omega\frac{e^{-i\omega t}}{\omega-\epsilon_j-X_j(\omega)+i(\eta+\Delta_j(\omega))}. \end{equation} I can show that with the assumption of $\omega$-independence for the self-energy, $S_j(\omega)\equiv S_j$ $\forall\omega$, $\Delta_j$ corresponds to the rate of loss of probability amplitude for site $j$. That is, the reverse Fourier transform gives a factor of $\exp(-\Delta_jt)$. $X_j$ gives an overall oscillatory phase factor.

However, how does one show this connection for a general, $\omega$-dependent self-energy? I have seen stated in the literature that "$\Delta_j(\omega)$ is physically the rate of loss of probability amplitude from site $j$ to overlapping eigenstates at energy $\omega$". This I would like to show.

If it helps for background knowledge, the Hamiltonian will be of form \begin{equation} \hat{H} = \sum_j\epsilon_j|j\rangle\langle j| + t\sum_j |j\rangle\langle j+1| + \text{h.c} \end{equation} In $1d$, e.g., and with the site energies $\{\epsilon_i\}$ independent random variables, this is an Anderson problem and all states are localised.


1 Answer 1


In general case one cannot show that that imaginary part of the self-energy is the decay rate (inverse lifetime). This interpretation is valid only in those cases where the Green's function has distinct poles (rather than branch cuts), and the lifetime is sufficiently long to speak meaninffully about quasiparticles (which is a way of saying that the self-energy is slowly changing with frequency - making it close to the case of constant self-energy, treated in the OP).

This approximation works rather well for certain problems, such as Landau Fermi liquid, but breaks in many notable cases - e.g., one-dimensional electronic liquid (Luttinger liquid). One could also easily concoct artificial examples where this is not the case: E.g., let us consider a central site coupled by tunneling to a Fermi sea and another similar site (with possibly different energy). If we write the Green's function for the site in question, the self-energy will contain broadening due to the continuum of stated in the Fermi sea, but also the information about the other discrete states that it is coupled to, corresponding to the tunneling back and fourth between the sites. Tunneling back and fourth is clearly not a lifetime, but an extreme case of collapse-and-revival.

  • $\begingroup$ This is interesting. Do you happen to have links/references to some pedagogical resources related to what you write? I'm afraid my general knowledge of Green functions is rather limited. $\endgroup$
    – dsfkgjn
    Commented Jun 28, 2021 at 13:57
  • 1
    $\begingroup$ @dsfkgjn I think it is a good idea to start with carefully reading the part where they discuss the meaning of poles, which is found in most man-body texts. I am a condensed-matter person, so my suggestions would be AGD or Fetter&Walecka. Then my example is done with the quantum dots in mind, which could be a rather simple calculation , if one takes broad-band approximation. Something along the lines of this answer but for two dots. $\endgroup$
    – Roger V.
    Commented Jun 28, 2021 at 14:02
  • $\begingroup$ @dsfkgjn regarding the answer linked: with no bias, one can do the calculation within the zero temperature formalism, without bothering with Keldysh. $\endgroup$
    – Roger V.
    Commented Jun 28, 2021 at 14:05

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