How composite particles have definite spin?

[ShayanJ]

Consider a composite particle. Its spin is determined by the spins of its constituent particles. But the constituent particles are in a superposition of different spin states and so don't have a definite spin. So it seems it shouldn't be possible to ascribe a definite spin to the composite particle. I know, one particular state may be the stable state but that only means this state is the most probable one and the composite particle is only mostly in that state, not always!

 

[Vanadium 50]

A composite particle may or may not be in an eigenstate of spin. If the total spin commutes with the Hamiltonian, a particle in an eigenstate of energy will also be in an eigenstate of spin.

 

[ShayanJ]

But what is that which is considered to be the partcile's spin?
I mean...there is only one spin eigenstate?
Or you're talking about the state with the least energy?

 

[Vanadium 50]

I don't understand anything you wrote.

 

[atyy]

A simple example is worked out in http://www.physics.udel.edu/~msafrono/424/Lecture 25.pdf

There are more examples in http://www.physics.utoronto.ca/~krieger/Phy489_Lecture6_2013.pdf

 

[ShayanJ]

Ok, Thanks to atyy, I can ask my question more clearly. In the first of atyy's files, the spin states of the hydrogen atom are mentioned. There are three states of spin 1 and one state of spin 0. So Hydrogen should have spins 0 and 1. You can't say its spin is 1 or 0. Its spin is both.

 

[DrDu]

Consider a composite particle. Its spin is determined by the spins of its constituent particles.

That's not true. The spin is the angular momentum in the rest frame of the center of mass (or energy in relativistic mechanics) of the particle. Therefore, also the orbital angular momentum of the particles contributes to it's spin. In contrast to an elementary particle, a compound particle may have several spin eigenvalues or be in a superposition of them. E.g. for an hydrogen atom (neglecting nuclear spin), the total spin J is the sum of the orbital momentum L and the spin of the electron S.

 

[ShayanJ]

a compound particle may have several spin eigenvalues or be in a superposition of them.

That's exactly my point. My quesion is, while that is known, how is that e.g. spin of the $\alpha$ particle is considered to be 0? Why people ascribe a definite spin to it?

 

[DrDu]

I suppose there are no, even metastable, excited states of the He nucleus with higher spin.

 

[Vanadium 50]

It's not "considered to be 0". It's measured to be 0.

In any event, an alpha particle is a state of definite binding energy, and since total spin commutes with the total Hamiltonian, it's also in a state of definite total spin.