I am learning about parity of nucleus and (elementary) particles. For example, proton has a intrinsic parity of $+1$. However, a more general term for parity of proton is $(-1)^l$ where $l$ is orbital quantum number. So proton parity is $+1$ only when $l=0$ which implies that proton is in ground state (in some nucleus I guess, not sure "ground state" is applicable for free particles?).

My question is can free particle, like proton have orbital quantum number? (eg. proton that has just been released in beta+ decay so he flies away as free particle)

I ask this because there is this nuclear reaction: $$ \rm p^+ + {^{7}_{3}Li} → {^{8}_{4}Be} → {^{4}_{2}He}+ {^{4}_{2}He}+ 17.2\,MeV $$

If I understood this reaction correctly, we have free flying proton that hits nucleus of Li and then Be is created. This Be is unstable so it decays to two He nuclei. Given the law of conservation of parity (that applies for EM and strong interaction) we must have equal parity on far left and far right side.

Parity for proton is +1 while for $\rm{^{7}_{3}Li}$ is -1 which gives total parity -1 on the left side (they are multiplied). On the right side we have two He nuclei which both have +1 so we have total parity of +1 on the right side.

So this reaction will not happen (because we have odd parity on the left and even on the right) unless proton has $l=1$ which will make it parity to $(-1)^l = (-1)^1 = -1$. Then on the left side we will have parity of $(-1)(-1) = +1$ which will make this process possible. In my notes from the class it says "this reaction will not happen if proton is in ground state, meaning it has $l=0$, because then parity is not conserved". I am wondering how can proton be in the ground state, or how can it have orbital angular momentum, if it is a free particle like in this process?

  • $\begingroup$ A ground state only applies to bound states, that is electrons in a potential well, such as the nucleus. A free state particle can have any energy value, but a bound state particle has restrictions on its angular momentum and energy levels $\endgroup$
    – user108787
    Commented Jul 14, 2016 at 19:29
  • $\begingroup$ Ok... Can free particle have some angular momentum other than spin? Like some kind of orbital angular momentum? $\endgroup$
    – matori82
    Commented Jul 14, 2016 at 19:33
  • $\begingroup$ The angular momentum of the system must be conserved, so after a reaction, a free particle may have angular momentum, but when you say orbital angular momentum, I think that would generally be taken to mean the electron is not free, and has acquired angular momentum because it is in a potential, produced by say an atomic nucleus. What would a free particle orbit around? When you see the LHC particle pictures, the paths of the particles are curved but that is not orbital a.m. $\endgroup$
    – user108787
    Commented Jul 14, 2016 at 19:47
  • $\begingroup$ I think, what you meant to ask is whether it is possible for a free particle to have a CONSERVED OAM. The answer is no. Thanks for your question! It answered my question=) $\endgroup$
    – MsTais
    Commented Dec 15, 2017 at 16:29

1 Answer 1


Recall that the value and sign of $\mathbf{r} \times \mathbf{p}$ angular momentum depends on the point around which you chose to measure it.

A free particle can (indeed, does) have angular momentum around any and all points relative which it is moving and has a non-zero impact parameters.

But, frankly, that's not a very interesting statement.

  • $\begingroup$ ok but in that case angular momentum is not quantized, right? $\endgroup$
    – matori82
    Commented Jul 14, 2016 at 19:52
  • 1
    $\begingroup$ The application of the angular momentum operator to those states would generate a quantized result, but the system need not be in a eigenstate of angular momentum if you don't measure it. $\endgroup$ Commented Jul 14, 2016 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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