Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Quick question: In the old papers and text I occasionally see authors referring to mesonic states as having 'natural parity' or 'unnatural parity'. What was their motivation for classifying mesons this way?


share|cite|improve this question
up vote 1 down vote accepted

"Natural parity" means that the bosonic meson field behaves under reflection as 1 for even spin and -1 for odd spin. This is because this is how coordinate tensors behave under reflection. A vector changes sign under reflection of all 3 directions (three minus signs), a 2-tensor doesn't change sign (or perhaps it's better to say it changes sign twice), a 3-tensor changes sign, a 4-tensor doesn't change sign.

The reason this is called natural is because when you write down meson theories, and you expect them to preserve parity, you don't feel you should use $\gamma_5$, and you don't feel you should use $\epsilon$ tensors. Sometimes you can still define parity even when you use these objects, but then, in the parlance of the times, it becomes an unnatural parity.

For example, consider a meson coupled to a spinor $\psi$ (think nucleon) as follows:

$$ \phi(x) \bar{\psi}\gamma_5 \psi $$

This is not invariant under the natural parity, but it is invariant if you make $\phi$ parity odd. Similarly, for a vector meson coupled to a spinor:

$$ A^\mu(x) \bar{\psi}\gamma_5 \gamma_\mu \psi $$

The spinor part doesn't change sign under parity even though it would if the $\gamma_5$ weren't there, so A has to be an axial vector. The natural parity is what happens if you couple to spinor bilinears without $\gamma_5$ and this is what is expected from the example of the photon coupling in electrodynamics.

The reason people emphasize this fact is because the pions were experimentally discovered to be parity odd, this is obvious in the fact that you can't make pions singly, but in pairs, and this was a real surprise. You expected a normal Yukawa interaction a-priori, but instead, you now expect the parity odd Yukawa interaction above. Such an interaction would make a renormalizable nucleon-pion coupling.

But this is not what is going on. The dominant interaction is a gradient coupling of pions to nucleons:

$$ \partial_\mu \pi_{ij} \bar{N^i}\gamma^5\gamma^\mu N^j$$

Where the $i,j$ indices are ordinary SU(2) isospin indices, and the pion SU(2) triplet is written above in the tensor way, as a symmetric SU(2) 2-tensor. This coupling still has that "unnatural" $\gamma^5$, so the pion is parity odd (both the derivative and the nucleon bilinear would change sign under reflection, except the gamma 5 adds a minus sign).

This unnatural pion-nucleon coupling, the pseudoscalar nature of the pion, it's light mass, and the Goldberger Trieman relation, were all explained by Nambu in 1960. The vacuum is breaking a symmetry, it is known now that this is the chiral-isospin symmetry, the isospin transformation acting with a $\gamma_5$, and the pion is the Goldstone boson.

This means (assuming the symmetry is exact) that the pion field is a coordinate on the SU(2) group manifold (which happens to be a 3-sphere). Inverting the coordinates changes the sense of a chiral isospin transformation, so it changes the sign of the pion field. Having a constant pion field is just choosing a different symmetry-breaking direction, so it doesn't change the energy, so the pions must couple with derivatives, and as pseudoscalars. Further, there must be as many of them as there are broken symmetry generators, in this case, 3.

Similar ideas with similar conclusions are presented simultaneously in the sigma-model of Gell-Mann and Levy, but Nambu and Jona-Lasinio crucially add the idea that the condensate is composed of ferminic bilinears, as it is in a superconductor. This was very inspirational for future developments, since it is now known that the condensate expectation value is a quark bilinear.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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