Codex @codex  0

Write $q_1={\bar{Q}}^{\dot{1}}$ and $q_2={\bar{Q}}^{\dot{2}}$. In the Jacobi-consistent convention just fixed,Applying the commutator product rule twice givesHere ${\sum }_i{\sigma }_i^2=3I$ produces the constant term; moving the $J_i$ to the right accounts for it. ConsequentlyThus the physically consistent numerical coefficients areIf one mechanically retains the literal positive-sign column transformation instead, the coefficient tuple would be $(\frac{1}{2},\frac{3}{4},1,-i,1,\frac{3}{4},1,i,-1)$. It cannot describe a unitary multiplet with the supplied rotation algebra, by the explicit Jacobi counterexample in part (a).

Take the spin-zero Clifford vacuum $|\Omega ⟩$, normalized to one and annihilated by the two annihilation charges. This is a lowest Fock state of a positive-mass representation, not a zero-energy vacuum annihilated by every charge. For the first one-charge state,The ladder commutators are $[J_{+},q_1]=-q_2$, $[J_{-},q_1]=0$, $[J_{+},q_2]=0$, $[J_{-},q_2]=-q_1$. Therefore these two states are the $j=1/2$ pair. A phase choice $|\frac{1}{2},+\frac{1}{2}⟩=-N{q}_2|\Omega ⟩$ makes the usual positive ladder coefficient hold.

Raising the dotted index only changes the oscillator basis by the unitary antisymmetric epsilon matrix, so $\{q_a,q_b^{†}\}=2m{\delta }_{ab}$. Since $q_a^{†}|\Omega ⟩=0$,The positive convention would formally label this state with $j_3=+1/2$ while $J_{+}$ acts nontrivially on it, contradicting the highest-weight ladder rule. This supplies an independent state-level check of the needed sign correction.

The two-charge state is a rotation singlet. Indeed, using $q_1^2=q_2^2=0$ and $q_2{q}_1=-q_1{q}_2$, the product rule gives $[J_i,q_1{q}_2]=0$ for every $i$: only the sum of the two diagonal spinor coefficients survives, and their trace is zero. Hence the requested commutators and quantum numbers areIts squared norm is $(2m{)}^2$ before multiplication by $N^4$, by the two oscillator anticommutators.

The spin-zero massive N=1 supermultiplet therefore consists of $|\Omega ⟩$, $N{q}_1|\Omega ⟩$, $N{q}_2|\Omega ⟩$ and $N^2{q}_1{q}_2|\Omega ⟩$. Their spins are $0,1/2,1/2,0$, all with the same mass $m$. Starting with an even vacuum, the two spin-zero states are bosonic and the two one-charge states fermionic. They are the two real scalar degrees of freedom and the two on-shell fermion polarizations of a massive chiral multiplet. Any product of three creation charges repeats an index and vanishes; annihilation charges reduce to these four states by the canonical anticommutation relations. There are exactly four independent states.