1
$\begingroup$

According to the Noether theorem, we only have the conserved quantity $$J+S,$$ where $J$ is the orbital angular momentum and $S$ is the spin angular momentum. But I am always impressed that the spin is a conserved quantity such as in the case of EPR paradox. One may think that in the EPR paradox, we only deal with non-relativistic system and spin $S$ decouples from angular momentum $J$. But even in QFT, spin is always used to characterize the particle state and so is also treated as conserved quantity.

Could you please point out what did I miss in the analysis?

$\endgroup$
  • $\begingroup$ Spin is not always treated as conserved! Consider, for instance, the case of an electron in an atom absorbing a photon. The photon has spin-1, but the spin of the electron is not going to change. $\endgroup$ – ACuriousMind Jan 6 '16 at 20:42
  • $\begingroup$ @ACuriousMind, yes you are right, but my motivation for asking this question is actually reflected by Timaeus's answer. $\endgroup$ – Wein Eld Jan 6 '16 at 21:09
2
$\begingroup$

One sense that is co served is that a photon is always spin 1 and an electronic always spin 1/2 and a Higgs is always spin 0 and those don't ever ever change.

But basically the spin state of a system is just a linear combination of tensor products of a bunch of 1d vectors for every spin 0 particle in the system, a bunch of 2d vectors for every spin 1/2 particle in the system, a bunch of 3d vectors for every spin 1 particle in the system, a bunch of a bunch of 4d vectors for every spin 3/2 particle in the system, and a bunch of 5d vectors for every spin 2 particle in the system.

But if a particle is created or destroyed, then the spin is gone. As for whether $J=L+S$ is conserved, it's not even an observable. As for observables just as $J^2$ or $J_z$ whether they are conserved just depends on whether they commute with the Hamiltonian.

$\endgroup$
  • $\begingroup$ Your answer is very enlightening. Regarding the first point ( the sense of conservation of the intrinsic property of quantum field), is there anything hide there? I mean, is there something kind of operator representing conserved quantity analogous the angular momentum in the representation of $SO(3)$ in the QM case. The quantity should not simply be the spin, because it should correspond the field rather the one-particle state. $\endgroup$ – Wein Eld Jan 6 '16 at 21:07
  • $\begingroup$ @WeinEld I'm not sure what you are asking about. If the spin were different it would be a different particle. Even a neutrino that flavor oscillates is still described as one neutrino being destroyed and another being created. A particle is what a particle is. That way you can talk about creating it or destroying it and you know what you got or what you lost. $\endgroup$ – Timaeus Jan 6 '16 at 21:11
  • $\begingroup$ It has the similar taste of the supersymmetry in superspace. $\endgroup$ – Wein Eld Jan 6 '16 at 21:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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