angular momentum - Is there a reason why the spin of particles is integer or half integer instead of, say, even and odd?

It seems to me that we could change all the current spin values of particles by multiplying them by two. Then we could describe Bosons as even spin particles and Fermions as odd spin particles. Is there some consequence or reason why we can't simply change 1/2 as the smallest unit of spin into 1?

It just seems prettier this way. In fact, there's almost an interesting relation to even and odd functions in that if you switch any two bosons around the wave function stays the same while if you switch any two fermions around the wave function becomes negative.

Answer

The "spin" tells us how the wavefunction changes when we rotate space (or spacetime). Just because I double all charges by convention, the behaviour of the wavefunction will not be any different. What will happen is that the "doubling" or charges will lead to the "halving" of your definition of angles such that the physical results (which depends on angle multiplied by spin) remain the same.

Wrt. the observation on "odd" and "even" functions -- that is not an accident and it works quite like you think it does.

The crux of the matter is that a "full rotation" corresponds to $2 \pi$ so the phase picked up by a spin $\frac{1}{2}$ wavefunction is $e^{i \pi} = -1$.

Recall that (even in classical mechanics) "angular momentum" is the generator of rotations. So if I start using different units, eg: $\tau \equiv 2 \pi$ to represent a half rotation (instead of $\pi$) then the values of charge will halve to maintain the value of $e^{i q \theta}$

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If you understand some representation theory, here goes:

1. 
   

   Representations of $SO(3)$ have integer charges. Since we're referring to the group of rotations, we call that charge as "angular momentum" or "spin". The representations correspond to scalars (spin 0), vectors (spin 1) and tensors (in general of spin 2 or higher).

   
   


2. 
   

   $SU(2)$ is a "double cover" of $SO(3)$ so representations of $SU(2)$ can have the "charges" as $SO(3)$ representations. Thus we also get half-integer spin. The new representations correspond to spinors.

   
   
   


3. 
   

   When we consider quantum relativistic physics (aka QFT), all physical fields/particles must form kosher reps of the Lorentz algebra, which happens to be $so(3,1) \sim so(4) = su(2) \oplus su(2)$. So (up to the "unitary trick") reps of the Lorentz group can be written as a tensor product of reps of the left and right-handed $SU(2)$ algebras. Based on #2 above, these continue to have integer or half-integer spin.