1
$\begingroup$

Consider the Chiral Perturbation Lagrangian to first order(quark masses set to zero): $$L = L^{(2)} = \frac{f^2}{4}tr[\partial_{\mu} U \partial^{\mu} U^{\dagger}] ,$$ where U is a $2 \times 2 $ exponential matrix collecting the pion-goldstone fields and the terms of the Lagrangian come from expanding the exponential- this term is similar to the non-linear sigma model for pions. Consider now the relativistic inclusion of nucleons so that: $$L =L_{\pi N}^{(1)} + L_{\pi \pi}^{(2)} ,$$ and where the first term is similar to the non-linear sigma model with the inclusion of nucleon-pion interaction- it's of the form $N D(x) \bar{N} $, where $D(x)$ is a differential operator containing the pion fields and couplings with the nucleon. The nucleon transforms in a highly non-linear way under the chiral group and the same for the pion fieds.

QUESTION: From all I have found there is no clear phrasing of using this Lagrangian for the calculation of NN-scattering amplitudes, instead, higher order terms must be included- the first such term is of the form: $$L_{NN} \simeq \bar{N}N \bar{N}N + (\bar{N}\sigma N)(\bar{N}\sigma N) ,$$ in the heavy baryon formalism.

But what I though I could do, without considering yet the heavy baryon formalism, was to take the $L_{\pi N}$ Lagrangian and use it the usual QFT way, that is in second order perturbation theory: then I wold have something of the form $S =\simeq \int d^4 x H_{\pi N}^2 $, a term that contains four N fields and pion fields as exchange terms. I did the same think using a phenomenological Yukawa type Lagrangian for the exchange of pions between nucleons as described by Erkelenz or Machleidt in papers about pion exchange nuclear forces- a Lagrangian with interaction of the type $$L_{phenomenology}= g \bar{N} \tau \cdot \pi N $$ with the inclusion of gamma matrices for pseudo-scalar terms and other kind of couplings.

So, if the $L_{\pi N} $ is of the same form just for chiral symmetry, is there a theoretical reason not to use it in second order for the calculation of the scattering amplitude between two nucleons- or such a use is valid and I just haven' t found any references on this?

$\endgroup$
1
$\begingroup$

1) The basic reference is Weinberg's work on effective lagrangians for the $NN$ interaction, see https://inspirehep.net/record/29549?ln=en .

2) Weinberg shows that in $NN$ scattering the one-pion exchange interaction and the $NN$ contact terms (with no derivatives) appear at the same order.

3) Note that in the $NN$ sector you cannot simply do perturbation theory. This is because there are $NN$ bound states and resonances at very low energy. Weinberg proposes to compute the $NN$ potential in perturbation theory, and then iterate the potential by solving the Schroedinger (Lippmann-Schwinger) equation.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for this answer, it' s very helpful and so is the reference. Some secondary, follow-up, questions that if you like can clarify. 1) By your second argument on the post, do you mean that to calculate the NN scattering at first order in the chiral perturbation scheme we must calculate the contribution from both terms or you mean there is another problem from the fact that they appear both at the same order(in contrast I imagine with the Yukawa potential I presented for usual QFT perturbation theory)? $\endgroup$ – Constantine Black Sep 3 '17 at 16:10
  • $\begingroup$ 2) Do you have any references on the iteration process you mention in your third argument? Why isn' t the old-fashioned style of perturbation theory not problematic? Anyway, thanks for the help and the reference again. $\endgroup$ – Constantine Black Sep 3 '17 at 16:12
  • $\begingroup$ I should also note that at a first point, although I see the arguments made by Weinberg on the problems of NN scattering from a χPT point of view(the fact that ordinary perturbation theory must not work because of the presence of bound states and resonances and that the cure of infrared divergences in the non-relativistic limit presents problems due to the assumption made, I not see the motivation to switching to old-fashiones perturbation theory- what makes us believe it should work). $\endgroup$ – Constantine Black Sep 3 '17 at 18:04
  • 1
    $\begingroup$ There are some cases where we know how to do this. An example is NRQED. At leading order we just get the Schroedinger equation with the Coulomb interaction, and at higher order we perturbatively include relativistic corrections (retardation, radiative corrections, etc). Weinberg makes a proposal for how to do this in the case of NN. This is not necessarily unique, or the best way to do it, or even correct. $\endgroup$ – Thomas Sep 3 '17 at 19:09
  • 1
    $\begingroup$ Also, the range of convergence of strict perturbation theory is extremely small, restricted to energies less than 1 MeV (because the scattering length is very large, and there are nearby bound states). $\endgroup$ – Thomas Sep 8 '17 at 14:03

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.