1

I have been reading about the Yukawa interaction https://en.wikipedia.org/wiki/Yukawa_interaction, which apparently is also used to model the nuclear force between nucleons. The fermions are considered to be relativistic, since they are Dirac fermions satisfying the Dirac equation. However, I have also read that usually the nucleons can be considered to be "at rest", since their rest mass is way bigger than their kinetic energy term, so usually they can be assumed to be non-relativistic. How does this match with the action of the Yukawa interaction? Shouldn't then fermions be just described by the simpler Schrödinger equation? The fermions I refer to here are the protons and neutrons of the nucleus, whose interaction is mediated by the scalar field $\phi$.

Qmechanic
  • 201,751
Zarathustra
  • 234
  • 2
    For any massive particle you can choose a reference frame where that particle is at rest. Does that mean you don’t have to treat it relativistically? – Oбжорoв Oct 23 '21 at 13:52
  • 1
    You can treat it relativistically of course, the question is whether the relativistic corrections would matter in this specific setup. You can treat any body in mechanics using relativistic theory, yet not all relativistic corrections prove to be important in all dynamical systems. – Zarathustra Oct 23 '21 at 13:56
  • the purpose of my comment was to point out that the mass of the nucleon has nothing to do whether you can use the Schrödinger equation or not. What matters is what you want to describe and whether you are in an energy regime where such corrections matter. – Oбжорoв Oct 23 '21 at 19:21
  • Related: https://physics.stackexchange.com/questions/292913/why-is-the-density-of-the-fermi-gas-in-a-neutron-star-not-changing-the-potential/293523#293523 – Lewis Miller Oct 23 '21 at 21:24

1 Answers1

0

I'll set $c=\hbar=1$ throughout.

How does this match with the action of the Yukawa interaction?

From Lorentz-invariant terms in a Lagrangian density, we can obtain amplitudes that become potentials in the non-relativistic limit. For example, the Feynman rules for the scalar interaction $g\overline{\psi}\phi\psi$ imply the interaction between two fermions has first-order amplitude $\frac{-4\pi g^2}{k^2+m^2}$ in $3$-momentum space, hence $\frac{-g^2}{r}e^{-mr}$ in $3$-position space. A similar analysis applies to the pseudoscalar option, $g\overline{\psi}i\gamma^5\phi\psi$.

Shouldn't then fermions be just described by the simpler Schrödinger equation?

The prescription$$E\mapsto i\partial_t,\,p_j\mapsto-i\partial_j,$$together with a factor of the wavefunction $\Psi$ on the RHS, converts in general potential-dependent energy-momentum relations into results such as the Schrödinger equation. Insofar as special relativity can be neglected (which technically means we still have Galilean relativity), we can get away with the Schrödinger equation in particular, but we have to regard it as a consequence of the relativistic result so the latter can be used more generally. In special relativity, we generalize the free-particle relation $E^2=m^2+p^2$ to $(E-V)^2=m^2+(p-Q)^2$, so the aforementioned potential-deriving technique works just fine.

So putting it all together:

  • The action tells us a potential;
  • We can use that potential in either a Galilean or special relativistic energy-momentum relation, with the aforementioned prescription, to get a differential equation;
  • The Galilean limit is fine for low-energy cases such as typical nuclear physics, but not in e.g. a particle accelerator. In particular, protons' interactions with (anti)protons require a full Yukawa interaction treatment.
J.G.
  • 24,837