# Why is the relative angular momentum of quark+antiquark in a pion equal to zero?

In a set of lecture notes provided by my lecturer, it says that a pion consists of a quark ($$q$$) and an antiquark ($$\bar{q_2}$$) with relative $$L=0$$. The negative parity and the fact that $$S=0$$ makes the pion a state with $$J=0$$ and odd parity, which is called a pseudoscalar.

Why is it that $$L=0$$? Why not $$1$$? or $$2$$?..

• Of course, the relative angular momentum of the electron and proton in plain 'ole hydrogen (in the ground state) is also zero... – dmckee --- ex-moderator kitten Dec 19 '18 at 14:45
• "Why not 1? or 2?" They exist but they are not called pions. – Lewis Miller Dec 19 '18 at 14:57
• What are they called then? – Luismi98 Dec 19 '18 at 15:01
• $a_1$ and $\pi_2(1670)$, respectively, of course. Why don't you simply study the PDG listings? They are all there (mostly). – Cosmas Zachos Dec 19 '18 at 19:51

Pions are the lowest mass meson. There are two reason for this:

1. Their valence quark content is only up and down quarks, which are the lowest mass quarks.
2. They represent the lowest energy state of those quarks, part of which comes from having no angular kinetic energy (which you will recall from intro mechanics can be expressed as $$L^2/(2I)$$). Other angular momentum states would simple be more massive (and indeed we see those states in the form of $$\rho$$ and $$b$$ mesons).

This is really the key: orbital angular momentum implies energy (which means mass).

• Thank you for your answer. From my understanding, a rho meson has the same quark content as a pi meson but the bound state has spin 1 instead of 0. And the b meson involves a bottom quark. – Luismi98 Dec 19 '18 at 19:30
• For question like that you should be looking in Particle Data Book. Go to pdg.lbl.gov (that is the Particle data Group's homepage). In addition to on-line access you can order some of there products. The Booklet is plenty for the sorts of question you've just asked. – dmckee --- ex-moderator kitten Dec 19 '18 at 19:46
• "And the b meson involves a bottom quark." There is also an excited light quark state called $b$ (and one called $a$ for that matter) which is what i was referring to there. Again, see the PDG's listings. – dmckee --- ex-moderator kitten Dec 19 '18 at 19:48
• @Lusmi98 To be pedantic, actually there is a 3rd reason that trumps the rest... Pseudoscalars are "abnormally", "exceptionally" light, so, much lighter than their vector cousins, the ρ s, because they are the "pseudogoldstone bosons of dynamically broken chiral symmetry". Don't worry about it yet, but there is a theorem protecting their mass from being large.... – Cosmas Zachos Dec 19 '18 at 19:55
• @Dan Indeed, parton distributions of all mesons are not even available, and their wave functions are spectacularly simple than those of baryons. All mesons are virtually trivial combinations of S and L, with the odd admixture of glueballs. Pseudoscalars are special, in that one should think of them as quark bound states second, and as pseudogoldstons first, but this is a much deeper story, adequately discussed in other questions... This one addresses their quark picture which is simple. – Cosmas Zachos Dec 20 '18 at 1:13

It doesn’t have to be! L=0 is simply the lowest-mass one.

By definition, a pion is a $$J^{PC}=\mathrm{even}^{-+}$$ light unflavored meson with I=1. With naive quark content that decomposes to S=0, L=even, and anything matching it is defined as a pion. Like pi(2)(1670). Its actual quark content is a matter of debate, and it could be an exotic excitation.