8
$\begingroup$

In his QFT vol. 1 (paragraph 3.5), Weinberg discusses the so-called old-fashioned perturbation theory (OFPT), i.e. the one based on the perturbative expansion of the Hamiltonian. As a result, in this theory time is dedicated, and there is no explicit Lorentz covariance. He claims though that its advantage as compared to the Feynman's perturbation theory is that the OFPT allows for indicating the parameter space for which different intermediate states produce singularities in S-matrix, such that the perturbation theory becomes inapplicable.

As an example, at the beginning of chapter 14, he discusses the low-energy electron-proton scattering. In Feynman's language of the perturbation theory, the Hydrogen atom pole is invisible independently of the order of the perturbation. However, the OFPT shows the divergence of the perturbative expansion for the CM momentum lower than $|\mathbf q|\sim m_{e}e^{4}$, which is of the order of the binding energy of the electron in the Hydrogen atom.

My question is why the main feature of OFPT - that time is dedicated - allows us to see these divergencies, whereas Feynman's theory does not, although the second one is just the resummation of the OFPT in the explicitly Lorentz-covariant form (such that virtual particles in the intermediate states are replaced by propagators). I would be grateful for the help.

An edit

The OFPT series from the Lippman-Schwinger equation may be rearranged in the covariant perturbation theory series. The procedure is to arrange the OFPT series summands at the given order of the coupling in pairs, in which the contribution of the intermediate states divided by the energy difference may be replaced by the propagator. In this sense, the covariant perturbation theory solves the LS equation as well as OFPT. Given this statement, I conclude that the main role for determining the effect of the resonance is played by the intermediate states in OFPT.

For instance, in the example with the electron-proton scattering, the IR enhancement is given by the energy difference between the intermediate and initial states (see details in Weinberg's QFT Vol 1, chapter 14). Is it correct, and if yes, why the intermediate state here is so important, whereas the propagator does not allow to "see" the bound state?

$\endgroup$
1
  • $\begingroup$ what do you mean by saying ' time is dedicated'? $\endgroup$
    – S. Kohn
    Feb 28 at 5:47

1 Answer 1

4
$\begingroup$

I think what is going on here is that "old-fashioned" perturbation theory can make the power counting of higher-order terms manifest. This is similar to the use of effective field theories like NRQED instead of QED to make the power counting manifest.

The appearance of bound states at weak coupling (as in the Coulomb problem) means that the suppression of higher-order diagrams by $\alpha$ must be compensated by an IR enhancement involving inverse powers of $m_e/|\mathbf{q}|$. In order to see this we would like to have an a-prior estimate ("power counting") of diagrams in terms of $m_e/|\mathbf{q}|$.

Some guidance is provided by the fact that we already know that these diagrams are summed by the Schrodinger equation. If we can find a perturbative scheme that corresponds to solving the Schroedinger (Lippmann-Schwinger) equation by iterating diagrams, then the power counting should be manifest. This is what old-fashioned perturbation theory does. By isolating the pieces that correspond to the non-relativistic limit, we can isolate the terms that are summed by the Schroedinger equation, and indeed, we can estimate the magnitude of diagrams without explicitly computing them, and find the $m_e/|\mathbf{q}|$ enhancement.

Of course, fully covariant perturbation theory can be used to compute the same diagrams, but the $m_e/|\mathbf{q}|$ is not manifest (you can't see it without an explicit computation), and there is no straightforward way to sum the $(e^2m_e/|\mathbf{q}|)^n$ terms.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer! I still do not understand something though and would be grateful if you will clarify it to me. I have added a paragraph to my question. $\endgroup$
    – Name YYY
    Aug 26, 2021 at 8:30

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.