I have to answer the following question:

Consider the decay $\rho^0\to\pi^+\pi^-$. The rho meson has angular momentum 1, what must the orbital angular momenta of the pions be, given that they are both spin-zero?

I know this question has been asked before, but I am only marginally interested in the numerical result itself (which I believe to be 1).

I believe angular momentum must be defined with respect to something -- e.g. a reference point, another particle, and so on. Therefore, 1) in "the rho meson has angular momentum 1", am I correct in assuming that it's entirely spin angular momentum? My reasoning is that it would not make sense to talk about the angular momentum of a single particle without a reference point, so I assume it's all spin. Following-up: every time I hear "particle X has angular momentum y", 2) can I assume it is the spin that is being mentioned?

3) is "the orbital angular momenta of the pions" the orbital angular momentum shared by the pions? In this case, with two particles, I believe it makes sense to talk about orbital angular momentum, if we implicitly refer to the angular momentum between them.

UPDATE Actually, in a later question I just read "Both the ground state $D^0$ meson and the excited state $D^{0*}$ have zero orbital angular momentum". I interpret this as saying that the orbital angular momentum is a well-defined quantity -- in this case it is zero, but it needn't be. How can you talk about the orbital angular momentum of a single particle? Without mentioning a reference point? For example, when in atomic physics I used to read about the orbital angular momentum of the electron, I always assumed that implicitly they meant the orbital angular momentum with respect to the nucleus.

  • 1
    $\begingroup$ Indeed, without contemplating a space wavefunction of sorts, you must talk about spin (intrinsic angular momentum). But the space-wavefunction of two pions, or a quark-antiquark pair has "geography" and a meaningful action by the orbital angular momentum operator. $\endgroup$ Commented Mar 30, 2020 at 18:22
  • $\begingroup$ @CosmasZachos thanks! $\endgroup$
    – martin
    Commented Mar 30, 2020 at 18:47

2 Answers 2


What is spin? Spin is the amount of angular momentum necessary so that in the particle interactions we have studied, angular momentum would be conserved, so that angular momentum conservation would still be a strong law. It works, because there has been no falsification of this hypothesis in the study of particle interactions at present. That is how the spin of the particles and resonances has been assigned.

For particles and their composites, when one is talking of angular momentum, one means angular momentum about the center of mass ( where the nucleus sits is in your example). In the center of mass of the $\rho^0\to\pi^+\pi^-$ system , the two pions have to go in equal and opposite directions,because of momentum conservation. Their spin is zero and if they are not imagined in an "orbit", they cannot build up the spin 1 of the rho. The decay would be forbidden by angular momentum conservation. Giving an angular momentum of 1 in the center of mass system, restores conservation of angular momentum and allows the decay.


Single composite particles can have orbital angular momentum $L$ as well as spin $S$. In a simplified & intuitive picture, you can think of $L$ as resulting from the motion of the constituent particles and $S$ as resulting from their alignment (e.g. aligned quarks in a meson make $S=1$, anti-aligned make $S=0$).

You may be familiar with this from electrons in an atom or nucleons in a nucleus... it's the same principle for quarks in a hadron.

  • $\begingroup$ This approach works in (some) atoms where the $LS$ coupling is small. But it fails in strongly-interacting systems. Even the deuteron ground state has a substantial $D$-wave admixture. In a strongly-interacting system, $L$ and $S$ for the constituents are not good quantum numbers; the total angular momentum is referred to as the “spin” of the system. $\endgroup$
    – rob
    Commented Dec 11, 2021 at 12:53
  • $\begingroup$ Fair enough, but what about heavy quarkonium systems? You have states named like $\Upsilon(4S)$. Surely $L$ and $S$ are well-defined in that case. $\endgroup$
    – dukwon
    Commented Dec 12, 2021 at 15:18
  • $\begingroup$ Hmmm, heavy quarks are not my area. I see those term symbols in bottomonium like the $\Upsilon$, but not in mixed-generation particles like the $B$ mesons. It could be that the third generation is so massive that it has its own rules about $LS$ coupling; consider that the top quark doesn't even bother to hadronize. $\endgroup$
    – rob
    Commented Dec 12, 2021 at 17:49

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