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  Nuclear physics from perturbative QFT

+ 9 like - 0 dislike

Is there a renormalizable QFT that can produce a reasonably accurate description of nuclear physics in perturbation theory? Obviously the Standard Model cannot since QCD is strongly coupled at nuclear energies. Even the proton mass cannot be computed from perturbation theory, as far as I understand.

The strong force sector should probably be comprised of a nucleon spinor field Yukawa-coupled to a scalar pion field, the later with quartic self-coupling. The electroweak sector should probably be the Glashow-Weinberg-Salam model with hadrons replacing quarks in some manner.

Some key parameters of what a "reasonably accurate description" should be are:

  1. Reproducing the entire spectrum of nuclei and predicting which nuclei are stable
  2. Estimating nuclear masses with precision of ~ 0.1%
  3. Estimating nuclear decay rates with precision ~ 10%

Assuming this is feasible:

How the precision of the results depends on computation loop order?

It is possible to consider the nonrelativistic limit of the QFT, in which nuclei are described by quantum mechanics of nucleons coupled by a (multibody) interaction potential. The potential can be computed in perturbation theory by considering the nonrelativistic limit of nucleon scattering. Obviously this cannot give the right masses, since mass is additive in non-relativistic physics, but does it give reasonably precise stability criteria and binding energies?

What if we remain with the 2-body potential only?

What happens if we extract the potential from the nonrelativistic limit of the full nonperturbative Standard Model (theoretically, I'm not sure it's manageable in practice)?


  1. Let me explain my motivation to think such a renormalizable QFT exists. The (sufficiently) fundamental description of nucleons and pions is QCD, but in low energies they can be described by an effective field theory. Such effective field theories are normally nonrenormalizable but the nonrenormalizable interactions are suppressed by powers of E / Lambda, where E is the energy scale of interest and Lambda is the fundamental energy scale: in this case, the QCD scale. Since nuclear binding energies are significantly lower than the QCD scale (highest binding energy per nucleon is about 9 MeV and the QCD scale is about 200 MeV so the ratio is < 5%), it doesn't seem absurd to use renormalizable interactions only
  2. It is not really possible to compute bound state energies and lifetimes at finite order of perturbation theory. This is because they appear as poles in the S-matrix but these poles don't appear at any finite order. However, maybe it's possible to introduce some meaningul infinite (but not complete) sum which will converge (at least after some resummation technique) and exhibit the required poles? It is certainly possible for bound states in external fields (in this case the infinite sum amounts to including the external field effect in the propagator), maybe there's a way to do it for bound states of dynamic particles as well? Now that I think about it, this issue should have been a separate question... In any case, we can still use perturbation theory to extract the many-body potentials

EDIT: I now realize issue 2 above can be tackled using the Bethe-Salpeter equation. However, I found no good discussion of it so far. Any recommendations? I'd prefer something mathematically-minded

EDIT: Thomas' answer below raised some doubts as to the possibility of describing nuclear physics using a renormalizable QFT. Therefore I wish to expand the question to include nonrenormalizable QFTs. As long as we stick to finite loop order there is a finite number of parameters so the approach makes sense. The question is then: Which QFTs (renormalizable or not) can produce nuclear physics from perturbation theory + Bethe-Salpeter equation? What is the required loop order?

This post has been migrated from (A51.SE)
asked Dec 26, 2011 in Theoretical Physics by Squark (1,725 points) [ no revision ]
Not an expert, but I know that people work on this. E.g., see: http://en.wikipedia.org/wiki/Chiral_perturbation_theory

This post has been migrated from (A51.SE)

2 Answers

+ 7 like - 0 dislike

The theory your asking about is an effective field theory (in this case the nuclear eft developed by Weinberg), so it is not renormalizable. QCD is the only renormalizable field theory that can account for nuclear physics (quantum hadrodynamics, mentioned above, is a renormalizable model field theory, with no real predictive power). Also note that nuclear bound states are non-perturbative, so nuclear eft is a powerful tool to organize calculations, but even leading order calculations are not perturbative (you must solve the Schroedinger equation, or an equivalent Dyson-Schwinger equation). Because of the efimov effect, calculations may require three body forces at leading order, but there is a hierarchy of many body forces. Finally asking for 0.1% accuracy on masses (that is roughly 10% for binding energies) is challenging, but not crazy.

A few more comments: 1) It is not true in general that the leading order terms in an EFT correspond to renormalizable interactions. 2) Solving a Bethe-Salpeter equation that sums two-body ladders with an interaction that is local in time is equivalent to solving the Schroedinger equation (this is used not only in nuclear EFT, but also in NRQED or NRQCD calculations). 3) The issue with the Efimov effect is that according to Weinberg's power counting three-body operators should be suppressed relative to two-body operators, but Weinberg counting can fail non-perturbatively (if there is an Efimov effect).

This post has been migrated from (A51.SE)
answered Dec 27, 2011 by tmschaefer (720 points) [ no revision ]
Most voted comments show all comments
1. The pion is a Goldstone boson, so in an EFT it appears derivatively coupled. 2. Yes, but a low energy EFT for massive particles (nucleons) automatically leads to a non-rel setup. 3. See, for example, nucl-th/9809025.

This post has been migrated from (A51.SE)
1. Interesting. Does it mean non-derivative coupling is forbidden by chiral symmetry?

This post has been migrated from (A51.SE)
2. Relativistic effects are less significant but I'm not sure it means we can take the completely nonrelativistic limit

This post has been migrated from (A51.SE)
1. You may consider linear representations (the linear sigma model), where the leading terms are indeed renormalizable couplings to sigma an pi fields, but as an EFT you should integrate out the sigma, and you end up with a non-linear representation, and a derivatively coupled pion (this is explained in Weinberg's QFT book).

This post has been migrated from (A51.SE)
2. The leading term in the EFT for nucleons is the free Schroedinger lagrangian, but higher order terms contain relativistic corrections (kinetic energy corrections, spin-orbit forces, etc). This is essentially the Foldy-Wouhuysen reduction. The same technology is used in NRQED to do order-by-order to corrections to Coulomb bound states.

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Most recent comments show all comments
1. Well, it might be that no renormalizable interactions are allowed by the symmetries but it doesn't seem to be the case since we have the Nucleon-pion Yukawa coupling and pion-pion quartic self-coupling

This post has been migrated from (A51.SE)
2. Bethe-Salpeter is only equivalent to Shrodinger in the nonrelativistic limit. It can be used in a fully relativistic context as well (in which the interaction is not instantaneous)

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+ 3 like - 0 dislike

You're looking for quantum hadrodynamics. See, e.g. Serot and Walecka.

This post has been migrated from (A51.SE)
answered Dec 26, 2011 by S Huntsman (405 points) [ no revision ]

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