Why don't we construct a spin 1/4 spinor?

I am learning quantum field theory. I understand that the solution of Dirac equation has four states and each corresponds to a spinor. These four states are exactly the eigenstates of the spin operator and their eigenvalues are +1/2 or -1/2. But it seems that I can construct another operator (matrix), may be 8 by 8 or some other dimensions. The corresponding eigenvalues of this matrix can of course be 1/4, 1/2, 3/4, etc. With these eigenstates I can also construct a formula and make the eigenstates the solution of this formula. Then I can have a spin 1/4 spinor. I don't understand why Dirac equation can be so special.

• Hi ZHANG Juenjie, I removed your other subquestions, cf. this meta post. – Qmechanic Oct 20 '16 at 4:01
• What is the physical meaning of the 1/4 spinor? – Suzu Hirose Oct 20 '16 at 10:59
• @ Suzu Hirose No physical meaning. But Should the spin have to have a meaning mathematically? I mean are there any thing deeper behind Dirac Equation that force the equation to look like this or this is simply a guess work with trial and error? – ZHANG Juenjie Oct 22 '16 at 9:01

There is no such thing as spin $1/4$ in 4-dimensional spacetime.

Spin has to do with representations of the Lorentz Lie algebra. So let me shed some light on how these are classified.

First, the complex-valued Lorentz algebra $so(1,3)$ is equivalent to $so(4)$. The $so(4)$ algebra is equal to the direct sum of two copies of $su(2)$, which means that irreducible representations of $so(1,3)$ are labeled by ordered pairs of the irreducibles of $su(2)$.

The $su(2)$ representation theory can be found in any textbook on Lie groups or even in some QFT textbooks. One of the crucial facts is that irreducibles of $su(2)$ are labeled by nonnegative half-integers called spins:

$$j = 0, \; 1/2, \; 1, \; 3/2, \; 2, \; 5/2, \; \dots$$

This is enough to start constructing irreducibles of the Lorentz algebra. The basic blocks of the representation theory are the two fundamental representations $(1/2, 0)$ and $(0, 1/2)$ which are called the left and right Weyl spinors respectively. Both are two-dimensional.

Dirac spinors actually belong to the 4-dimensional reducible representation $$(1/2, 0) \oplus (0, 1/2).$$

Another 4-dimensional representation is the irreducible $(1/2, 1/2)$ to which the 4-vectors belong.

As you can see, this is all just representation theory and there is no spin-$1/4$ representation in 4 spacetime dimensions.

However, in $2$ spacetime dimensions this is no longer valid and representations with fractional spins exist.

• It seems that the keys are 1. SO(4) is equivalent to SU(2); 2. Irreducibles of SU(2) are labeled by nonnegative half-integers . – ZHANG Juenjie Oct 20 '16 at 5:04
• @ZHANGJuenjie pretty much so, yes. Except for the "equivalent" part: $so(4)$ is equivalent to the two copies of $su(2)$. – Solenodon Paradoxus Oct 20 '16 at 5:07
• What do you mean by saying "irreducible"? Could please suggest any references? Even if I understand that algebraically spin 1/4 is not allowed, I still cannot get an intuitive sense about how this thing relates to spin. Do you have any suggestions or references? – ZHANG Juenjie Oct 20 '16 at 5:07
• @ZHANGJuenjie irreducible representations are the ones which don't contain invariant subspaces. I would suggest taking a full course on Lie groups and algebras, though it might take a considerable amount of time. Unfortunately I am no expert in which literature is better, so I would suggest posting a resource recommendation question. Or you could try this notes: solenodonus.com/file/lie-theory.pdf – Solenodon Paradoxus Oct 20 '16 at 5:11
• @ZHANGJuenjie also you probably misunderstand the meaning of spin-$1/2$. See, this is just a convention. I could instead call "spin" what others call "spin divided by two" which would make Dirac spinors spin-$1/4$ in my terminology. The important fact is that there is a discrete set of irreps, and one can not achieve a smaller yet nonzero spin than that of Dirac spinors. – Solenodon Paradoxus Oct 20 '16 at 5:19