= Solution
In a massive <four-dimensional N=1 supersymmetry> representation, the two normalized raising <supercharges> transform as a spin-half doublet. Start with a <Clifford vacuum> of <superspin> $j=1/2$. At occupation levels zero, one and two, the ordinary <spins> are
$$
\frac12,\qquad \frac12\otimes\frac12=0\oplus1,\qquad \frac12.
$$
The two-creation state has a spin-singlet oscillator factor because the creation operators anticommute. Thus the <massive N=1 vector multiplet> contains
$$
\boxed{1\ \oplus\ \frac12\ \oplus\ \frac12\ \oplus\ 0.}
$$
There are four bosonic polarizations, $3+1$, and four fermionic polarizations, $2+2$, as required by <boson-fermion degeneracy in a supermultiplet>.
In the massless limit the <helicity> multiplicities become $+1,-1$, two copies each of $+1/2,-1/2$, and two zero-helicity states. They separate into the four irreducible <helicity> pairs
$$
\boxed{(1,\tfrac12),\quad(\tfrac12,0),\quad(0,-\tfrac12),\quad(-\tfrac12,-1).}
$$
With the lower-helicity convention for <superhelicity>, a pair $(\kappa,\kappa+1/2)$ has labels $\kappa=1/2,0,-1/2,-1$, respectively. A midpoint convention shifts each label by $1/4$; the explicit <helicity> pairs remove any naming ambiguity.
Pairing under the <CPT theorem>, the first and last pairs form a massless <vector multiplet>, and the middle two form a massless <chiral multiplet>. This is the <massless limit of a massive N=1 vector multiplet>. The longitudinal vector polarization supplies one of the <chiral multiplet>'s two real scalar modes; the original spin-zero state supplies the other. Simply discarding the longitudinal polarization would lose a physical degree of freedom and give the wrong decomposition.
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