# Why is spin number half integer, especially +1/2,-1/2 for electrons?

1. Jul 16, 2014

### Nick Jackson

Ok guys, I know this must be pretty basic for but I am new to this section of physics. Anyway, my question is a two-part one, I guess:
1) Why does the spin number get only half integer values in fermions and integer values in bosons, mesons, etc.?
2) How do we conclude that the spin number is taking the values +1/2 and -1/2 in electrons (I mean mathematically)?

2. Jul 16, 2014

### Avodyne

1) The spin-statistics theorem of relativistic quantum field theory states that particles with integer spin must be bosons, and that particles with half-odd-integer spin must be fermions.

2) For a particle with position operator $\vec x$ and momentum operator $\vec p$, we can define an orbital angular momentum operator $\vec L = \vec x \times \vec p$. From the canonical commutation relation $[x_i,p_j]=i\hbar\delta_{ij}$, we can derive the commutation relation $[L_i,L_j]=i\hbar\varepsilon_{ijk}L_k$. We then postulate that, in addition to orbital angular momentum, a particle may have spin angular momentum. This involves an additional operator $\vec S$ with commutation relations $[S_i,S_j]=i\hbar\varepsilon_{ijk}S_k$. It is then possible to prove mathematically that the eigenvalues of ${\vec S}{}^2$ are $s(s+1)\hbar^2$ where the allowed values of $s$ are $0,1/2,1,3/2,2,\ldots$. It is then an empirical fact that $s=1/2$ for electrons.

In relativistic quantum field theory, each type of particle arises as an excitation of a particular field. How the fields transform under the Lorentz group determines the spin of the corresponding particle.

Last edited: Jul 16, 2014