According to Fetter's book, the Feynman diagrams contribute to the proper self-energy are:

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where the first and second orders here refer to the perturbation expansion of the interacting Green's function.

The question is: Why do only two figures (${\rm{(a)'}}$ and ${\rm{(e)}}$) contribute to the proper self-energy of the Anderson mode? (see for example Fig. 6.20 in here)


1 Answer 1


The Fetter&Walecka's diagrams are written for a general coulomb interaction, something like $$ V=\frac{1}{2}\sum_{k,k',q, q'}\sum_{\sigma,\sigma'}v_{k,k';q,q'}c_{k,\sigma}^\dagger c_{k',\sigma'}^\dagger c_{q',\sigma'}c_{q,\sigma}, $$ whereas the interaction in the Anderson model is $$ V=Un_\uparrow n_\downarrow=\frac{1}{2}\sum_{\sigma}n_\sigma n_{\bar{\sigma}}= \frac{1}{2}\sum_{\sigma}d_\sigma^\dagger d_\sigma d_{\bar{\sigma}}^\dagger d_{\bar{\sigma}}= \frac{1}{2}\sum_{\sigma,\sigma'}d_\sigma^\dagger d_{\sigma'}^\dagger d_{\sigma'}d_\sigma $$ There are here important differences in comparison to the full Coulomb interaction

  • there is only exchange term present, i.e., there is no term with equal spin projections ($\sigma'=\bar{\sigma}$ but no $\sigma'=\sigma)
  • There is only one orbital state for electrons, that is we cannot have more than one spin of each spin projection, and we cannot have more than 2 electrons.

The following should be checked by derivation (which I strongly recommend to do oneself, e.g., at zero temperature), but I would claim the following:

  • b, c, d are really included in the self-energy by dressing the inner lines
  • f is actually the Kondo diagram and should not be neglected (although one usually arrives to Kondo effect by expansion in tunneling)

See also: Higher-order perturbation in Kondo problem

Anderson model without tunneling
It might be however that the reference cited considers Anderson model without tunneling. In this case most of the diagrams would vanish due to the restricted phase space. In this case the Green's function can be calculated exactly in respect to any state: $$ |\psi\rangle=p_0|0\rangle+\sum_\sigma p_\sigma d_\sigma^\dagger |0\rangle + p_2d_\uparrow^\dagger d_\downarrow^\dagger |0\rangle= p_0|0\rangle+\sum_\sigma p_\sigma |\sigma\rangle + p_2|\uparrow\downarrow\rangle, $$ with the result that it contains only terms $$\frac{1}{\omega-\epsilon_\sigma\pm i0}\text{ and }\frac{1}{\omega-\epsilon_\sigma-U\pm i0},$$ the latter corresponding to the sum of the ladder diagrams, i.e. a' and e.

TL; DR: An easy way to see which diagrams give zero contribution is by assigning spin to every electron line. Electron spin does not change between two vertices, whereas an interaction line necessarily connects two electron lines of opposite spin.


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