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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
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Zarathustra
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1 Answers1

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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.
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