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

Question

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 1

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.

Answer 2

I would like to point out that the accepted answer is not correct, there is no need to halve the physical angles.

In the language of representation theory, the Lie algebras $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$ of infinitesimal rotations are isomorphic and have the same representations indexed by an integer $m=2j\in\mathbb{Z}$, where $j$ is the spin. The problem is that, as $\text{SO}(3)$ is not simply connected, not all representations of its Lie algebra give rise to representations of itself.

For even $m$ (integer spins), these exponentiate to representations of both $\text{SU}(2)$ and $\text{SO}(3)$. However, for odd $m$ (half-integer spins), there is no such common representation, so we must stick with representation of the former group. So in this case we are forced to exponentiate the representations of $\mathfrak{su}(2)$ and call them "rotations".

However, since $\text{SU}(2)$ is a double cover of $\text{SO}(3)$, $\pm e\in \text{SU}(2)$ both correspond under projection to $e\in \text{SO}(3)$, so roughly speaking a rotation by $\pi$ via $\text{SU}(2)$ (which results in a phase $\exp i\pi=-1$ on the vectors of the representation) is a physical rotation of $1=\exp i2\pi\in \text{SO}(3)$. So the moral is that despite this identification we must remember that it is not $\text{SO}(3)$ doing the rotations, but rather its double cover $\text{SU}(2)$, and that is why the angles halve, regardless of whether we use an odd $m$ or a half-integer $j$ to identify spin.

You may wonder what happens for even $m$, as this argument should apply as well. In this case, the representations are not faithful, i.e., not one-to-one, so $-1\in \text{SU}(2)$ acting on a vector space $V$ is actually given by $1\in\text{End}(V)$ and this is why they are also representations of $\text{SO}(3)$: the identification of each group's action is automatic. The whole crux of the matter is that for odd $m$, the representations of $\text{SU}(2)$ are faithful.

Answer 3

I would say it's because at its core quantum mechanics is the study of atomic transitions, and the knowledge of the internal states must be inferred. What we see from that perspective is that all angular momentum transitions are integer multiples of $\hbar$, and are quite frequently exactly $\hbar$. There are sets of related states in some atoms which are grouped in singlets, triplets, quintuplets, etc., and related states in other atoms which are grouped in doublets, quadruplets, sextuplets, etc. The fundamental bit of insight is that the odd-plet atoms may sometimes have zero angular momentum, while the even-plet atoms can never have zero angular momentum. Integer steps between half-integer values is a very clear way to achieve this; I think that dividing $\hbar$ again so that the smallest transition was "two spin quanta" would make the separation between these two systems much more confusing.

There's a lovely book by Tomonaga called The Story of Spin where he takes three competing rules for assigning quantum numbers to atomic angular momentum states from the early days of quantum mechanics, by Sommerfeld, Landé, and Pauli. Tomonaga presents all three models at face value. Each of them has its own internal consistency. We use Sommerfeld's model, where a state with multiplicity $n$ has angular momentum $j=\frac{n-1}{2}$. Landé used a supplementary quantum number $R=\frac n2$, which switches the places of integer and half-integer spin from the statistical standpoint, but by means of some different selection rules could recover the same physics. Deciding for myself which of the three models I preferred, and then convincing myself it wasn't just that I already had a lot invested in Sommerfeld's approach, was a very valuable intellectual exercise. None of the three has only integer quantum numbers for systems with spin; I suppose that the integer quantum numbers for systems with orbital angular momentum were already too well-established by the time that spin was discovered.

Answer 4

This is required in 3D world only, 2D world is richer!

Particle's spin has relation to its statistics (spin-statistics theorem). Namely, the quantum evolution of exchanging two identical particles gives idea about the intrinsic spin of that particles. This action of exchanging depends solely on the topology of exchanging paths (see Leinaas & Myrheim 1977). Interestingly, the topology of exchanging paths of two indistinguishable particles in 3D is equivalent to permutation group which implies only fermionic or bosonic statistics (spin n/2 and n). On the other hand, in 2D, the topology of exchanging two identical particles does not require any constraint on the statistics of particles, which implies arbitrary spins (not only $1/2$ but also $1/3$, $\sqrt{2}$..)

Answer 5

As can be read in the first comment, you can assign the value one to spin half particles (as the value is in the current agreement). You have to change all elementary particle spins accordingly though. I mean, spin-one must be spin-two in that case, and spin-two has to be spin-four.

I think that assigning the value $\frac{1}{2}$, makes the distinction between matter fields and force fields more clear. Matter fields inherently carry a spin $\frac{1}{2}$ (or $\frac{1}{2}+n$, though particles for which $n$ is different from zero aren't known to exist), while force fields carry a spin of $!$ or $2$ (again, in theory, any spin with an integer value, $n$, but only the spin $1$ and spin $2$ are known). As soon as you see that a particle's spin is an integer you know that it is a force-carrying particle. And the same holds for spin $\frac{1}{2}+n$-particles.

On top of that, you would have to make quite some adaptations in measuring devices (rewrite values).