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.

  • 1
    $\begingroup$ Hi ZHANG Juenjie, I removed your other subquestions, cf. this meta post. $\endgroup$
    – Qmechanic
    Oct 20, 2016 at 4:01
  • 1
    $\begingroup$ What is the physical meaning of the 1/4 spinor? $\endgroup$ Oct 20, 2016 at 10:59
  • $\begingroup$ @ 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? $\endgroup$ Oct 22, 2016 at 9:01

1 Answer 1


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.

  • $\begingroup$ 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 . $\endgroup$ Oct 20, 2016 at 5:04
  • $\begingroup$ @ZHANGJuenjie pretty much so, yes. Except for the "equivalent" part: $so(4)$ is equivalent to the two copies of $su(2)$. $\endgroup$ Oct 20, 2016 at 5:07
  • $\begingroup$ 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? $\endgroup$ Oct 20, 2016 at 5:07
  • 1
    $\begingroup$ @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. $\endgroup$ Oct 20, 2016 at 5:19
  • $\begingroup$ @Solenodon Paradoxus Just about your last note, it means that supposing Lorentz algebra (or invariance?), two dimensional case of the Dirac field vanishes? $\endgroup$ Oct 20, 2016 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.