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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 neutral pion has positive charge conjugation eigenvalue and its spin-one isospin half-turn has eigenvalue minus one. Hence the G parity eigenvalue is minus one. Since G parity commutes with isospin, both charged members have the same value. Charged-state phase changes alter individual charge-conjugation and rotation formulas together, leaving this eigenvalue unchanged.

In the ordered spin-one basis $(m=1,0,-1)$, the second angular momentum operator satisfies $J_2^3=J_2$. Therefore its matrix exponential reduces to $I-i\sin \theta J_2+(\cos \theta -1)J_2^2$. A half-turn exchanges the two outer weights and negates the middle weight. Applied to the pion isospin triplet, this fixes the rotation factor in G parity.