Setup, as I understand things so far:

One way to think about where the spin of a quantum field comes from is that it is a consequence of the ways that different types of fields transform under Lorentz transformations.

The generator of a Lorentz transformation for a Dirac spinor field $\Psi$ is $S^{a b}=\frac 1 2 [\gamma^a, \gamma^b]$ (I use roman letters for the antisymmetric indices representing the rotation direction, and greek ones for the orientation of the field. Signature is $(+,-,-,-)$.)

For a vector field like $A^{\mu}$, it is $(M^{a b})^{\nu}_{\mu}=\frac 1 4 ( \eta^{a}_{\mu} \eta^{b \nu} - \eta^{b}_{\mu} \eta^{a \nu})$

For a tensor field like $g^{\mu \nu}$, it is two copies of $M$: $(M^{ab})^{\alpha}_{\mu} \otimes I+I \otimes (M^{ab})^{\beta}_{\nu}$

As one should expect, there is clearly a structure to these representations of increasing spin.

Now, although it isn't normally done, one could use the Clifford algebra relation:

$\{ \gamma^a, \gamma^b\}=2\eta^{a b}$

to express all of these generators in terms of increasingly complex products of gamma matrices.

Okay, with all that setup, my question is stated simply:

  1. Is there a general formula that one can derive that will give the $n/2$ spin representation in terms of the appropriate gamma matrix combination?

  2. As a particular example, what does a representation for a spin-3/2 field look like, as one might find in a supersymmetric theory?


1 Answer 1


Okay, this wasn't as hard to find an answer to as I expected. However, any clarifications/ critisisms are welcome.

Weinberg basically gives the answer in section 5.6 of his QFT book:

A general tensor of rank N transforms as the direct product of N (1/2, 1/2) four-vector representations. It can therefore be decomposed (by suitable symmetrizations and antisymmetrizations and extracting traces) into irreducible terms (A,B) with A = N/2, N/2-1,... and В = N/2, N/2-1,... . In this way, we can construct any irreducible representation (A,B) for which A + В is an integer. The spin representations, for which A + В is half an odd integer, can similarly be constructed from the direct product of these tensor representations and the Dirac representation $(1/2,0)\oplus(0,1/2)$.

So it does not appear that there is any simple way to unify the representations for spinor and vector generators, but one can construct the generator for arbitrary half-spins: it has as many copies of $M$ as needed, plus one copy of $S$ if it is a half-integer.

However, the converse is not true. That is, a spin-two field must transform with two copies of $M$, but doing so does not guarantee that an object is spin-two. A counterexample is the electromagnetic tensor $F$, which is certainly spin-one. The difference lies in the ability to equate the two generators as a result of the symmetry properties of the tensor, as elaborated in this answer.

Applying this to the spin 3-2 field, we expect it to have a rotation generator that looks schematically like $S \otimes I+ I \otimes M$. And indeed this is the case- the equivalent of the Dirac equation for spin 3/2 is the Rarita-Schwinger equations:

$\gamma_a \psi^{a}_\mu=0$,

$(i \gamma^\rho \partial_\rho-m)\psi^{a}_\mu=0 $

Which transforms as $\psi'^b _\nu=(\Lambda_\nu^\mu \otimes T^b_a) \psi_\mu^a$,

whose generators are the above combination of $M$ and $S$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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