= Solution
The irreducible <massive N=1 vector multiplet> consists of \b[one massive vector, one real scalar, and two massive spin-$\tfrac12$ fermions]. Thus its on-shell <boson> and <fermion> counts are
$$
n_B=3+1=4,\qquad n_F=2+2=4.
$$
The two <fermions> can be written as two <Majorana spinors> or one <Dirac spinor>. All these states have the same mass when <supersymmetry> is unbroken; <auxiliary fields> are not extra propagating states.
To obtain rather than assume this spectrum, use the rest-frame <supersymmetry algebra> $\{Q_\alpha,Q_\beta^\dagger\}=2M\delta_{\alpha\beta}$. The two normalized <fermionic creation operators> form a spin-$\tfrac12$ doublet. Start with a spin-$\tfrac12$ <fermionic> Clifford vacuum annihilated by the two lowering operators. The zero- and two-creation-operator levels each have spin $\tfrac12$, whereas the one-creation-operator level has
$$
\tfrac12\otimes\tfrac12=0\oplus1.
$$
It is <bosonic>, giving exactly the scalar and vector above. A higher-spin Clifford vacuum would introduce spin above $1$, while a spin-zero Clifford vacuum gives maximum spin only $\tfrac12$.
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