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Put $N_a=a^{†}a$ and $N_b=b^{†}b$. The canonical commutation relations give $[N_a,a^{†}]=a^{†}$, $[N_a,a]=-a$, and the analogous relations for $b$, with all cross-mode operators commuting. Consequently $[K_3,K_{+}]=K_{+}$ and $[K_3,K_{-}]=-K_{-}$. Alsoso $[K_{+},K_{-}]=N_a-N_b=2K_3$. This is the Schwinger boson representation of the angular momentum algebra, with $\hslash =1$ in the stated commutators.

The vacuum has $N_a=N_b=0$, and $(a^{†}{)}^2|0⟩$ has occupation numbers $(2,0)$, hence $K_3$ eigenvalue $1$. With $N=N_a+N_b$,Thus $\boxed{K^2(a^{†}{)}^2|0⟩=2(a^{†}{)}^2|0⟩}$. It is the highest-weight state of the spin-one irreducible representation. There are two other states in this multiplet, $a^{†}{b}^{†}|0⟩$ and $(b^{†}{)}^2|0⟩$, with $K_3$ eigenvalues $0,-1$. A normalized basis is $(a^{†}{)}^2|0⟩/\sqrt{2}$, $a^{†}{b}^{†}|0⟩$, $(b^{†}{)}^2|0⟩/\sqrt{2}$. This count assumes the usual two-mode bosonic Fock space; an extra unmentioned vacuum multiplicity could replicate the multiplet.

Contracting the result of part (c) with the Pauli matrix givesThus the $T^a$ form the Schwinger boson representation of the $\mathfrak{su}(2)$ commutation relations. The second Pauli matrix givesand therefore

On the energy level $E=N+1$, the total occupation is $N$. The $N+1$ states $|n_x,N-n_x⟩$ form the spin-$j=N/2$ irreducible representation. Every component of its angular momentum generator has eigenvaluesSince $L=2T^2$, the possible orbital angular momenta areThere are $N+1$ distinct values, exactly matching the degeneracy found in part (a); hence each occurs once. Restoring $\hslash$ multiplies every $\ell$ by $\hslash$.