Intertwining operator

An intertwining operator between representations $\rho$ and $\sigma$ is a linear equivariant map $T$ satisfying $T\rho (g)=\sigma (g)T$ for every group element $g$.

A tensor operator transforming in an irreducible representation $R$ between states in $V$ defines an intertwining operator $R\otimes V\to V$. In a basis of these intertwining maps, its scalar coefficients $g_{\nu }$ are reduced matrix elements, independent of the component indices. Schur lemma makes the number of independent coefficients the multiplicity of $V$ in $R\otimes V$. The component dependence is fixed by the associated Clebsch-Gordan coefficients.

The Wigner-Eckart theorem separates the dependence of a spherical tensor operator on angular momentum projections from its dynamical content. In one standard convention,The reduced matrix element on the right is independent of $m,m^{\prime },q$. The Clebsch-Gordan coefficients impose $m^{\prime }=m+q$ and $|j-k|\le j^{\prime }\le j+k$; additional selection rules, including parity, can make the reduced matrix element vanish.

To see why the factorization holds, combine the rank-$k$ transformation law of the operator with the spin-$j$ transformation law of the initial state. This gives an intertwining operator from the tensor product representation $j\otimes k$ to the final representation $j^{\prime }$. The Clebsch-Gordan decomposition contains each allowed $j^{\prime }$ once. Therefore the corresponding intertwining operator is unique up to an overall scalar, which is the reduced matrix element after the chosen normalization. Ratios within the same pair of multiplets depend only on Clebsch-Gordan coefficients, whenever the common reduced matrix element is nonzero.

An octet operator between baryon octet states has two reduced matrix elements, because the tensor square of the SU(3) adjoint representation contains two octets. The symmetric invariant $d_{ris}$ and antisymmetric invariant $f_{ris}$ give independent component tensors. Equivalently, represent baryons by traceless matrices $B$ and let the operator act as $F[t_i,B]+D(\{t_i,B\}-\frac{2}{3}\mathrm{tr}(t_{i}B)I)$. Normalization choices are absorbed into $F,D$; Hermitian operators permit real coefficients in a Hermitian octet basis.