 Solution

The irreducible massive N=1 vector multiplet consists of one massive vector, one real scalar, and two massive spin-$\frac{1}{2}$ fermions. Thus its on-shell boson and fermion counts areThe 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 $\{Q_{\alpha },Q_{\beta }^{†}\}=2M{\delta }_{\alpha \beta }$. The two normalized fermionic creation operators form a spin-$\frac{1}{2}$ doublet. Start with a spin-$\frac{1}{2}$ fermionic Clifford vacuum annihilated by the two lowering operators. The zero- and two-creation-operator levels each have spin $\frac{1}{2}$, whereas the one-creation-operator level hasIt 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 $\frac{1}{2}$.