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With the Schwinger boson representation and $a_n=L^{-1/2}{\sum }_q{e}^{-iqn}{A}_q$, the Hamiltonian operator of a periodic quantum XY chain takes the displayed form. The four inverse discrete Fourier transforms and site sum give $1/L$ and lattice momentum conservation. The forward and reverse nearest-neighbour exchange phases are $e^{i(q-t)}$ and $e^{-i(q-t)}$, giving the cosine. All momenta are defined modulo $2\pi$; neighbouring sites must be distinct for the stated normal ordering.

The Schwinger boson representation satisfies $S_n^2=(N_n/2)(N_n/2+1)$, where $N_n$ is the local total boson number operator. Thus a prescribed local spin $s$ requires the displayed operator constraint at every site. A condition on $⟨N_n⟩$ alone is insufficient: for $⟨N_n⟩=2s$, $⟨S_n^2⟩=s(s+1)+\mathrm{Var}(N_n)/4$. Projecting onto the required eigenspace at each site removes the unwanted local irreducible spin representations.

Use $\hslash =1$ throughout. Let $N_a=a^{†}a$, $N_b=b^{†}b$ and $N=N_a+N_b$. Operators belonging to different boson species commute. The proposed Schwinger boson representation givesConsequently the first of the spin commutation relations isThe number operator commutators $[N_a,a^{†}]=a^{†}$, $[N_a,a]=-a$ and their $b$ counterparts giveThus all the spin commutation relations hold. Also $[N,S^{\pm }]=[N,S^z]=0$: each fixed-$N$ sector is invariant. To identify its spin, calculate the Casimir operator:The irreducible spin representation with prescribed $s$ therefore requires $N=2s$, and its normalized Fock basis correspondence isConversely, $s=(n_a+n_b)/2$ and $m=(n_a-n_b)/2$. The nonnegative integer occupation numbers give precisely $m=-s,-s+1,\dots ,s$, and there are $2s+1$ such states. As a check, the spin raising operator acts aswhich is the usual spin ladder operator coefficient $\sqrt{s(s+1)-m(m+1)}$.

For a lattice of fixed spins, the local spin constraint in the Schwinger boson representation is essential:The full two-species bosonic Fock space also contains other local irreducible spin representations; fixing only the total boson number does not remove them.

Take $L\ge 2$, so neighbouring operators act at distinct sites. The identityexpresses the Hamiltonian operator of this quantum XY chain in spin ladder operators. Substituting the Schwinger boson representation gives the requested quartic expression:Operators at different sites commute, allowing normal ordering without a correction here. For the first quartic term, the inverse discrete Fourier transforms giveHere ${\delta }^{(L)}$ is one when its two momenta agree modulo $2\pi$ and zero otherwise; it expresses lattice momentum conservation. The other quartic term has the same ordered operators with phase $e^{i(p-r)}$. On the support of this Kronecker delta, $p-r=t-q$ modulo $2\pi$, so the two phases sum to $2\cos (q-t)$. Thus the Fourier vertex of a Schwinger boson chain isAn equivalent three-sum expression sets $t=p+q-r$ modulo $2\pi$, so its coefficient is $\cos (r-p)$. The factor $1/L$ comes from four factors $L^{-1/2}$ and one site sum, while the factor of two from the two exchange terms cancels the original $1/2$. The formula is Hermitian: conjugation exchanges $p$ with $t$ and $q$ with $r$, and the coefficient is unchanged after imposing lattice momentum conservation. It also preserves each local total boson number in its real-space form, hence preserves the physical constrained subspace.

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 matrices 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 $\hslash$.

The canonical commutation relations are $[x,p_x]=[y,p_y]=i\hslash$, with $[x,y]=[p_x,p_y]=[x,p_y]=[y,p_x]=0$. Operators from different Cartesian modes commute, so $(p_x{p}_y{)}^{†}=p_y{p}_x=p_x{p}_y$ and $(xy{)}^{†}=yx=xy$. Thus $J_1$ is Hermitian and represents an observable on the usual oscillator domain.

Introduce annihilation operators $a_x=\sqrt{m\omega /(2\hslash )}x+i{p}_x/\sqrt{2m\hslash \omega }$ and similarly $a_y$. Set $N_x=a_x^{†}{a}_x$, $N_y=a_y^{†}{a}_y$ and $N=N_x+N_y$. The Schwinger boson representation givesFor $A=a_x^{†}{a}_y$ and $B=a_y^{†}{a}_x$, $AB=N_x(N_y+1)$ and $BA=N_y(N_x+1)$. ThereforeThe two-mode oscillator Casimir identity isSince $[N,A]=[N,B]=0$, $[H,J_1]=0$, and $J_1$ is conserved. However, $[J_1,J_2]=i\hslash J_3$ is not the zero operator, so there is no general joint eigenbasis. The exception is important: a common eigenstate of two angular-momentum components must satisfy $J_3\psi =0$, then the other commutators force $J_1\psi =J_2\psi =0$. Thus only the $j=0$ sector admits a common state; here that is the oscillator ground state.

Using $J^2={\hslash }^{2}j(j+1)$, with $j=0,1/2,1,\dots$, and selecting the positive energy branch givesEquivalently $N=2j=0,1,\dots$, so $E_N=\hslash \omega (N+1)$ and the degeneracy is $N+1$, corresponding to $N_x=0,\dots ,N$.

Use $[a_i,a_j^{†}]={\delta }_{ij}$ and define $E_{ij}=a_i^{†}{a}_j$. A direct multiplication givesWith $S_{+}=\hslash a_{\uparrow }^{†}{a}_{\downarrow }$, $S_{-}=\hslash a_{\downarrow }^{†}{a}_{\uparrow }$, this implies $[S_{+},S_{-}]=2\hslash S_z$ and $[S_z,S_{\pm }]=\pm \hslash S_{\pm }$. Resolving the raising and lowering combinations givesThese operators realize the Schwinger boson representation of angular momentum. Each commutes with $\hat{N}=n_{\uparrow }+n_{\downarrow }$, so fixed total occupation is an invariant subspace.

Set $P=a^{†}{b}^{†}$ and $Q=ab$. Independent bosonic canonical commutation relations give $[Q,P]=N_a+N_b+1=2K_3$, while $[K_3,P]=P$ and $[K_3,Q]=-Q$. Substituting $K_1=(P+Q)/2$ and $K_2=(P-Q)/(2i)$ yieldsThereforewhich matches the matrix Lie algebra. This boson-pair realization of SU(1,1) differs from the number-conserving Schwinger boson representation of $SU(2)$.

For the two-dimensional isotropic harmonic oscillator, $H=\hslash \omega (N+1)$ and the Schwinger boson representation gives $J^2={\hslash }^{2}N(N+2)/4$. Substitution yields the displayed relation, including its quantum constant term. The fixed-total-occupation sector is a spin-$j=N/2$ representation, so energy is $\hslash \omega (2j+1)$ with degeneracy $2j+1$.