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I am trying to find out the degrees of freedom of the vector-spinor field ($s=3/2$). The degrees of freedom are given by $N=\frac{1}{2}\left(N_{F}-N_{C}\right)$ for this spin where $N_F$ is the number of (real) field components and $N_{C}$ the number of (real) constraints on the field that follow from the field equation.

I have derived the two constrains it has, thus $N_C=2$. I have to figure out what $N_F$ is for this kind of field to see what $N$ will be, but I am not able to reason how many components will it have although it was kind of intuitive to figure out for lower spin fields.

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  • $\begingroup$ did you consider spinor components of the two constraints? Each should have 4 independent components, thus 8 constraints. If $N_F$ is off-shell d.o.f., it should be 12. Then $N=(12-8)/2=2$ on-shell d.o.f. (assuming the field is massless, and 4 spacetime dimenions) $\endgroup$
    – Kosm
    Commented Mar 20, 2022 at 11:11
  • $\begingroup$ Because of the derivations I followed in order to get the constrains, I am pretty sure I am considering a massive field. The equation of motion I started with is $i \gamma^{\mu \nu \rho} \partial_{\nu} \psi_{\rho}+m \gamma^{\mu \nu} \psi_{\nu}=0$ and using the two constrains I derived, I was able to write it as a Dirac equation for a vector-spinor, $\left(\mathrm{i} \gamma^{\mu} \partial_{\mu}-m\right) \psi_{\nu}=0$. Thus I think $m\neq 0$. What would be $N_F$ then? Could you also elaborate why $N_F=12$ if we are considering it to be off-shell? @Kosm $\endgroup$
    – god_operator
    Commented Mar 20, 2022 at 11:22

2 Answers 2

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Massless Rarita-Schwinger (=spin-vector) field in $4$ spacetime dimensions has $4$ gauge degrees of freedom (this symmetry becomes local supersymmetry once graviton is included), since the gauge parameter is a Weyl spinor. You subtract those from $16$ real components to get $12$ off-shell DOF.

Equations of motion then show that the time-component of the spin-vector $\psi_\mu$ is non-propagating, thus $-4$ DOF. There is another constraint from the equations of motion: $\partial^\mu\psi_\mu=0$ which further subtracts $4$ DOF. So far we have $12-8=4$ DOF which should be divided by two due to the field equations for the remaining spatial components of $\psi_\mu$ (Weyl/Dirac equations reduce the number of independent components of spinors by half).

Now for massive spin-vector the gauge freedom is lost, so off shell we have $16$ DOF, while on-shell $(16-4-4)/2=4$ independent DOF. The two extra degrees of freedom are longitudinal modes, similarly to a massive vector field having one longitudinal DOF. In the context of supersymmetry breaking, the two extra degrees of freedom of a massive gravitino are from a Goldstone (Weyl) fermion.

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  • $\begingroup$ I have only been doing quantum field theory for 5 weeks and I was looking for a simpler answer as I do not have knowledge about some of the concepts you are talking about but thank you for the explanation, it might be of help to someone $\endgroup$
    – god_operator
    Commented Mar 20, 2022 at 12:23
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I might have solved my question after giving it a thought but please correct me if I am wrong. I am considering a vector-spinor field and thus from $\psi_{\mu \alpha}$, we will have 16 complex components as $\mu=0,1,2,3 ; \alpha=1,2,3,4$ and therefore, it will have 32 real components meaning that $N_F=32$.

Now for the case of a vector-field, we have 2 constrains, each having 4 complex components and thus 8 real entries meaning that in total, $N_C=2\cdot 8=16$.

Thus, we have that $$ N=\frac{1}{2}\left(N_{F}-N_{C}\right)=\frac{1}{2}\left(32-16\right)=8 $$

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  • $\begingroup$ $\psi_\mu$ is a Weyl fermion for each $\mu$, having 4 real components. Thus $4\times 4=16$ components in total. $\endgroup$
    – Kosm
    Commented Mar 20, 2022 at 12:27

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