Are all mass eigenstates also spin eigenstates?

Question

Is there a rigorous way to show that a mass eigenstate of a particle must also be an eigenstate of the total spin operator? If this wasn't true, you could imagine that a composite particle in a mass eigenstate could be in a superposition of two or more total spin states. I've never heard of this, so I assume it's impossible but I'm not sure why.

My question is motivated by thinking about how the 1/2 spin of the proton arises, which is currently an open question in QCD. Essentially, is there a mathematical reason for why all the intrinsic and angular momentum contributions for the valence quarks, sea quarks, and gluons must add up to total spin 1/2 if the proton is in a mass eigenstate?

Edit: to phrase my question better, I am wondering why particles are never in superpositions of eigenstates of the total spin operator $S^2$ (not the projection operator $S_z$).

Answer 1

In a non-relativistic quark model, the mass is the ground state eigenvalue of the Hamiltonian, which is rotationally invariant, and so simultaneously diagonalizable with spin. So the spectrum of the hamiltonian, hence all hadrons, can have well-defined spin, and spin is preserved in reactions, like electric charge.

Your exemplars of the quark model spectrum, with toy potentials, exhibit this, starting from quark "constituents" with also well-defined spins and masses. You may define all kinds of spin superpositions (that you have not heard of), but it's not clear what you are going to do with them, as you will always have to go through separate channels; a spin up proton will behave very differently in reaction products' spins than a spin down proton.

You may define superpositions of different charges, but, again, each charge is superselected in a different distinct sector, so you will only confuse yourself. No QM interference, so it's much unlike quantum computing, if that is what you are fantasizing about.

Going relativistic will not fundamentally change things, except now you'll have to contend with a labile "sea" of quark-antiquark pairs, to boot.

... is there a mathematical reason for why all the intrinsic and angular momentum contributions for the valence quarks, sea quarks, and gluons must add up to 1/2 if the proton is in a mass eigenstate?

In light of the above, your question here deconstructs to

Given a composition of all these contributions to a total spin 1/2, which cannot change, by conservation of spin, what determines the mass of this particular arrangement (to be what the mass of the proton ends up being)?

Nobody can compute this from first principles, but one may simulate this on a lattice approximation of QCD, the correct theory of the strong interactions. The particular operator (~arrangement of constituents) yielding the lowest baryon mass is called the proton, and it happens to be of charge +1 for obscure/recondite reasons.

The spin 3/2 hadrons turn out to have higher masses, even though there is no easy proof QCD absolutely cannot produce such a state degenerate with a proton, unlikely as this might appear from everything we know. Is this what you are asking?

By contrast, ω and φ both vector mesons, even though not degenerate, do, indeed, mix!