Wigner–Eckart with explicit magnetic-label contract

Write $K={⨁}_{\beta \in B}{V}_{{\mathrm{\ell }}_{\beta }}$ and let ${\beta }_0$ index the chosen output copy, so ${\mathrm{\ell }}_{{\beta }_0}=J^{\mathrm{\prime }}$. Define $P:K\to K$ by $P(g_{\beta }{)}_{\beta \in B}=(h_{\beta }{)}_{\beta \in B}$, where $h_{{\beta }_0}=g_{{\beta }_0}$ and $h_{\beta }=0$ for $\beta \mathrm{\ne }{\beta }_0$. This linear map satisfies $P^2=P$ and has the chosen output copy as its range. For $a=z,+,-$, both $P{J}_a^{K}g$ and $J_a^{K}Pg$ have chosen coordinate $J_a^{(J^{\mathrm{\prime }})}{g}_{{\beta }_0}$ and every other coordinate zero, by the blockwise definition in F1. Hence $P{J}_a^K=J_a^{K}P$. Identify the range of $P$ with $V_{J^{\mathrm{\prime }}}$ and embed each $e_{J,m}$ in the chosen input coordinate, with zeros elsewhere. Define the linear map $F:V_J\otimes V_k\to V_{J^{\mathrm{\prime }}}$ by $F(e_{J,m}\otimes e_{k,q})=P{T}_q{e}_{J,m}$ under this identification. Applying the tensor-family equations and the ladder formulas to each product-basis vector gives $J_a^{(J^{\mathrm{\prime }})}F=F{J}_a^{\mathrm{t}\mathrm{o}\mathrm{t}}$ for $a=z,+,-$.

Suppose $J^{\mathrm{\prime }}$ is in the coupling list. The product-basis matrices of $J_z^{\mathrm{t}\mathrm{o}\mathrm{t}}$ and $J_{+}^{\mathrm{t}\mathrm{o}\mathrm{t}}$ are real. The highest-weight line of the $J^{\mathrm{\prime }}$ summand is therefore spanned by a real vector: the real and imaginary parts of any vector in that line satisfy the same real equations, and at least one is nonzero. Choose a normalized real highest-weight vector temporarily. Its coupled ladder vectors have real product-basis coefficients. The recorded highest-weight vector differs from this temporary choice by one unit scalar $\lambda$, so all recorded coefficients $C_{J,m;k,q}^{J^{\mathrm{\prime }},M}$ equal $\lambda$ times real coefficients and their complex conjugates equal ${\lambda }^{-2}{C}_{J,m;k,q}^{J^{\mathrm{\prime }},M}$.

For a coupled highest-weight vector $u_j$, the intertwining equations show that $F{u}_j$ has $J_z$ weight $j$ and is killed by $J_{+}$. In the output spin block the ladder formulas show that a nonzero vector killed by $J_{+}$ has weight $J^{\mathrm{\prime }}$; hence $F{u}_j=0$ if $j\mathrm{\ne }{J}^{\mathrm{\prime }}$, while for $j=J^{\mathrm{\prime }}$ it is a scalar multiple $a{e}_{J^{\mathrm{\prime }},J^{\mathrm{\prime }}}$. Applying $J_{-}$ and the identical normalized ladder formulas gives $F{v}_{J^{\mathrm{\prime }},M}=a{e}_{J^{\mathrm{\prime }},M}$ for every $M$, and $F$ vanishes on all other coupled summands. If $J^{\mathrm{\prime }}$ is absent from the coupling list, $F=0$.

Expanding a product-basis vector in the orthonormal coupled basis and applying step 2.1 gives $A_{m^{\mathrm{\prime }},m,q}=a\stackrel{‾}{C_{J,m;k,q}^{J^{\mathrm{\prime }},m^{\mathrm{\prime }}}}=a{\lambda }^{-2}{C}_{J,m;k,q}^{J^{\mathrm{\prime }},m^{\mathrm{\prime }}}$. Set $R=a{\lambda }^{-2}$; if $J^{\mathrm{\prime }}$ is absent set $R=0$. The total $J_z$ eigenvalue of $v_{J^{\mathrm{\prime }},m^{\mathrm{\prime }}}$ is $m^{\mathrm{\prime }}$, whereas that of $e_{J,m}\otimes e_{k,q}$ is $m+q$, so its coefficient is zero unless these weights agree. The coupling list gives the other selection rule, and $R$ can be zero, for example for the zero tensor family. This proves the contract.