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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.

In an irreducible spin representation, the angular momentum lowering operator has squared norm factor $(I+m)(I-m+1)$. Multiplying these factors from the highest-weight vector gives the factorial normalization displayed above. All factorial arguments are integers even when $I$ is half-integral. This produces a consistent positive-coefficient phase convention for all weight states and fixes signs in subsequent isospin rotations.

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.

An irreducible spin representation is a nonzero Hilbert space with no proper nonzero closed invariant subspace under all three angular momentum operators. Equivalently, the associated unitary representation of SU(2) cannot be decomposed into two nonzero orthogonal invariant pieces. The irreducible possibilities are labelled by a spin $j\in \{0,\frac{1}{2},1,\frac{3}{2},\dots \}$ and have dimension $2j+1$.

Choose an orthonormal basis $|j,m_s⟩$, with $m_s=-j,-j+1,\dots ,j$. The angular momentum commutation relations imply $[{\mathbf{S}}^2,S_i]=0$, $[S_3,S_{\pm }]=\pm \hslash S_{\pm }$, $[S_{+},S_{-}]=2\hslash S_3$, and $S_{+}^{†}=S_{-}$. The complete action isIn particular, the spin raising operator annihilates $|j,j⟩$ and the spin lowering operator annihilates $|j,-j⟩$. The coefficient follows by taking the squared norm and using $S_{\mp }{S}_{\pm }={\mathbf{S}}^2-S_3^2\mp \hslash S_3$. Thus the ladder operators connect every member of the basis, while terminating at its two endpoints.

On the tensor product of spins $j_1$ and $j_2$, the combined angular momentum operator is $\mathbf{J}={\mathbf{S}}^{(1)}\otimes I+I\otimes {\mathbf{S}}^{(2)}$. The Clebsch-Gordan decomposition iswhere $J$ increases by one and each irreducible spin representation occurs once. The dimensions agree: ${\sum }_J(2J+1)=(2j_1+1)(2j_2+1)$.

The uncoupled orthonormal basis diagonalizes $S_3^{(1)},S_3^{(2)}$; the coupled orthonormal basis diagonalizes ${\mathbf{J}}^2,J_3$. The Clebsch-Gordan coefficients are their change-of-basis entries, with conventionThey vanish unless $m_1+m_2=M$. Since both bases are orthonormal, this change-of-basis matrix obeys unitarity. Overall phases of the coupled basis vectors are fixed by a convention and do not change the energy eigenvalues.

Set $\hslash =1$ and let $N=2I\in {\mathbb{Z}}_{\ge 0}$. A concrete irreducible spin representation is the symmetric powerStart with $N$ copies of the defining SU(2) doublet and restrict their tensor product representation to the completely symmetric subspace. Its orthonormal basis consists of symmetric states with $I+m$ up components and $I-m$ down components, where $m=I,I-1,\dots ,-I$. There are $N+1=2I+1$ such states. The total generators are $J_i={\sum }_{a=1}^N{\sigma }_i^{(a)}/2$, acting on this subspace, where ${\sigma }_i$ are the Pauli matrices. This also constructs the trivial representation when $I=0$.

The spin ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfyThe Casimir operator $J^2={\sum }_i{J}_i^2$ commutes with every generator. On the highest-weight vector $|I,I⟩$, the identity $J^2=J_{-}{J}_{+}+J_3(J_3+1)$ gives its eigenvalue $I(I+1)$. With phases chosen to make the lowering coefficients positive,The squared norm of the lowered state follows from $J_{+}{J}_{-}=J^2-J_3(J_3-1)$. The ladder stops exactly at $m=-I$. Every weight is connected to every other by these operators, and $J_3$ has distinct eigenvalues. Hence any invariant subspace contains a weight vector and then the whole ladder: the representation is irreducible.

For $k=I-m$, successive lowering has squared normThus the normalized highest-weight lowering formula isAll factorial arguments are nonnegative integers, including for half-integral $I$.

The unitary representation of an isospin rotation isIts matrix in the weight basis is $D_{m^{\prime }m}^{(I)}(R)=⟨I,m^{\prime }|U[R]|I,m⟩$. Insert the completeness relation between two operators to obtainThese verify the group representation and unitarity properties, and the ladder argument gives irreducibility. Here rotations carry their SU(2) lifts: $U(2\pi )=(-1{)}^{2I}I$. Integer $I$ descends to the ordinary SO(3) group; half-integral $I$ is a representation of its double cover, not a single-valued representation of $SO(3)$.

For $I=1$, order the basis as $(|1,1⟩,|1,0⟩,|1,-1⟩)$. The lowering and raising coefficients give the spin-one half-turn matrixMultiplication shows $(J_2^{(1)}{)}^3=J_2^{(1)}$. Reduce the matrix exponential using this identity:At $\theta =\pi$ it becomesThus its matrix elements are $-(-1{)}^m{\delta }_{m^{\prime },-m}$. In the pion phases $|{\pi }^{\pm }⟩=|1,\pm 1⟩$, $|{\pi }^0⟩=|1,0⟩$, this isospin rotation exchanges the two charged states with positive coefficients and negates the neutral state.

Charge conjugation is linear and unitary here. Similarity preserves commutators, soConsequently the conjugated generators obey the same Lie algebra. The specified signs also give $\mathcal{C}{J}_{\pm }{\mathcal{C}}^{-1}=-J_{\mp }$. Since $|{\pi }^{\pm }⟩=J_{\pm }|{\pi }^0⟩/\sqrt{2}$ and the neutral state has positive charge conjugation eigenvalue,The signs depend on the chosen charged-state phases, which have been fixed by the ladder convention.

Conjugation by $R_{\pi }=e^{-i\pi J_2}$ changes $(J_1,J_2,J_3)$ to $(-J_1,J_2,-J_3)$, exactly as charge conjugation does. Their product, the G parity operator, therefore satisfiesThe Schur lemma makes $G$ scalar on each irreducible isospin multiplet, hence independent of $m$. On the neutral pion, $G|{\pi }^0⟩=-\mathcal{C}|{\pi }^0⟩=-|{\pi }^0⟩$. Therefore the pion G-parity is

The angular momentum commutation relations arewhere ${\epsilon }_{ijk}$ is the Levi-Civita symbol. The angular momentum eigenstates havewith angular momentum quantum number $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and magnetic quantum number $\mu =-j,-j+1,\dots ,j$. There are $2j+1$ such states in one irreducible spin representation, so the $J^2$ eigenspace has dimension $2j+1$ within that multiplet. The use of $\mu$ avoids confusing the magnetic quantum number with the particle mass in the earlier questions.

For completeness, the ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfy $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$ andThe upper and lower ends are $\mu =\pm j$; integer steps between them require $2j$ to be a nonnegative integer. This explains both the spectrum and the multiplet dimension. Pure orbital angular momentum permits only integer $j$, whereas general spin allows half-integers. If a larger Hilbert space contains several copies of the same irreducible spin representation, the full $J^2$ eigenspace includes all those copies: its dimension is $(2j+1)$ times their multiplicity, and can be infinite. The paper's dimension question is understood per irreducible multiplet.

Choose the ordered angular momentum eigenstates $(|1,1⟩,|1,0⟩,|1,-1⟩)$, so $J_3=\hslash \mathrm{diag}(1,0,-1)$. Since $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$, a nonzero matrix element of $J_{+}$ must increase the magnetic quantum number by one, and one of $J_{-}$ must decrease it by one. Thus only entries $(1,2),(2,3)$ of $J_{+}$ and $(2,1),(3,2)$ of $J_{-}$ can be nonzero. The adjoint operator identity $J_{-}=J_{+}^{†}$ givesFor the $j=1$ irreducible spin representation, both adjacent transitions are present; the following calculation determines their sizes.

A one-dimensional quantum spin model with transverse XY exchange interactions; $J>0$ is the ferromagnetic sign in the displayed convention. Unlike the classical XY model, its site variables are noncommuting spin operators in a fixed irreducible spin representation. With periodic boundary conditions and $L\ge 2$, $H=-\frac{J}{2}{\sum }_n(S_n^{+}{S}_{n+1}^{-}+S_n^{-}{S}_{n+1}^{+})$. It preserves total $S^z$ and the associated global U(1) symmetry.

For active rotations with $U=e^{-i\varphi J_3}{e}^{-i\theta J_2}{e}^{-i\psi J_3}$, the irreducible spin representation matrix is $D_{m{m}^{\prime }}^{(j)}=e^{-im\varphi }{d}_{m{m}^{\prime }}^{(j)}(\theta )e^{-i{m}^{\prime }\psi }$, where $d_{m{m}^{\prime }}^{(j)}=⟨j,m|e^{-i\theta J_2}|j,m^{\prime }⟩$. Its spin-one-half middle matrix is $\left(\begin{array}{cc}\cos (\theta /2)& -\sin (\theta /2)\\ \sin (\theta /2)& \cos (\theta /2)\end{array}\right)$. A $2\pi$ rotation acts by $(-1{)}^{2j}$; only integer-spin representations descend to the special orthogonal group in three dimensions.