We know that we can calculate the Green function $G(\tau;\lambda)= -\langle \mathcal{T}c(\tau)c^*\rangle$ of an interacting Hamiltonian $H=H_0 + \lambda V$ using connected Feynman diagrams. Of course, this is technically not true since the summation of such diagrams do not necessarily converge. The best we can usually do (correct me if I'm mistaken) is find that $G(\tau)$ is smooth with respect to $\lambda$ and that the derivatives $$ \frac{d^n}{d\lambda^n} G(\tau;\lambda) $$ is a summation over connected Feynman diagrams with $n$ vertices, and hope that the first few terms of the Taylor series expansion gives us a good approximation.

With that in mind, the self-energy is then defined by rearranging/grouping the connected Feynman diagrams in such a way so that $$ G=\frac{1}{G_0^{-1}-\Sigma} $$ However, I can't understand why we can do this mathematically, since even a conditionally convergent summation cannot have its terms arbitrarily rearranged (Riemanns rearrangement theorem).

My guess would be that even though the motivation comes from the above, the actual rigorous logic would be to define $\Sigma(i\omega)=G_0^{-1}(i\omega)-G^{-1}(i\omega)$ and find that $\Sigma$ is smooth with respect to $\lambda$. We can then repeat the logic for $G$ and compute the first few derivatives of $\Sigma$, which just so happen to be exactly equal to the definition of $\Sigma$ by regrouping the Feynman diagrams. Is this the case? If so, now the difficulty here is actually calculating $G_0^{-1},G^{-1}$ since these are inverses of operators and not numbers.

  1. Briefly, the full (connected) propagator $$G_c~=~G_0 + {\cal O}(\lambda_i)$$ is a formal power series in coupling constants $\lambda_i$.

  2. If we assume that the zero-order term $G_0$ is invertible, then we can uniquely construct the inverse formal power series $G_c^{-1}$ order by order in perturbation theory, cf. a multi-variable generalization of Lagrange inversion theorem.

  3. Then the self-energy $$ \Sigma~=~G_0^{-1}-G_c^{-1}$$ is also well-defined as a formal power series.

  4. For more information including renormalization, see e.g. this, this & this related Phys.SE posts.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.