The Walecka or $\sigma$/$\omega$-model is an effective theory describing nucleon-nucleon interaction by an exchange of $\sigma$/$\omega$-mesons. Why does it not include interactions by pions?

  • $\begingroup$ Er ... isn't the $\sigma$ a stand in for all the light scalar mesons? I certainly don't recall him addressing this question when I took his class at W&M back in the late 90's (he generously threw it open to all JLAB grad-students and even intervened with the W&M parking office to get us permits). $\endgroup$ Jul 15, 2014 at 0:11
  • $\begingroup$ No. I see that this does not answer the question as the pion is a pseudoscalar. So I have nothing to offer. $\endgroup$ Jul 15, 2014 at 0:13
  • 1
    $\begingroup$ I haven't read it fully (not much of a particle physicist), but this review of QHD (co-authored by Walecka) seems to introduce the pion alongside the sigma and omega mesons. $\endgroup$
    – Kyle Kanos
    Jul 15, 2014 at 2:48
  • $\begingroup$ To clarify the situation, QHD was an attempt to go beyond the mean field approximation in this effective interaction model. There are no symmetry arguments for excluding the pion from these higher order terms. $\endgroup$ Dec 6, 2015 at 14:23

2 Answers 2


I want to cite the original paper of Walecka where he awnsers your question:

From Annals of Physics 83/2 (1974) p. 491 "A theory of highly condensed matter

"The reader might object to the fact that there is no one-pion exchange tail in this interaction; however, the strong spin and isospin dependence of the potential arising from the exchange of an isovector, pseudoscalar pion implies that the contribution of the one-pion exchange potential to the bulk proerties of nuclear matter largely averages to zero"

Hope this helps


The answer given in Walecka's 1974 paper is mostly correct. The one pion exchange contribution to the Hartree energy vanishes in balanced nuclear matter. The same point was made in the 1972 paper of Miller and Green (Phys Rev C5 241) where the same type model was used for doubly magic (finite) nuclei. If exchange is included (Hartree Fock as opposed to Hartree) then the pion would contribute.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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