Rank-one angular matrix elements

Apply the Wigner–Eckart contract with $k=1$. In the order $V_J\otimes V_1$ there is a scalar $R_0$, independent of $m,n,q$, such that the matrix element is $R_0{C}_{J,m;1,q}^{J^{\mathrm{\prime }},n}$, with the stated selection rules. The angular-basis notation gives these coefficients and matrix elements their displayed meaning.

When $J^{\mathrm{\prime }}$ is in the coupling list, the factor-swap map sends $\mathrm{\mid }J,m⟩\otimes \mathrm{\mid }1,q⟩$ to $\mathrm{\mid }1,q⟩\otimes \mathrm{\mid }J,m⟩$ and intertwines the total spin operators. It therefore sends the chosen highest-weight vector in the first order to a unit highest-weight vector in the unique spin-$J^{\mathrm{\prime }}$ summand in the second order. The multiplicity-one decomposition in F5 makes that highest-weight space one-dimensional, so the two chosen vectors differ by one unit scalar, independent of magnetic label. Their positively normalized lowering constructions preserve this same scalar for every $n$. Consequently $C_{J,m;1,q}^{J^{\mathrm{\prime }},n}$ is one fixed scalar multiple of $C_{1,q;J,m}^{J^{\mathrm{\prime }},n}$ for all $m,n,q$. Absorb that scalar into $R_0$ to obtain $R$. If $J^{\mathrm{\prime }}$ is outside the coupling list both the matrix element and the stipulated coefficient vanish, so $R=0$ works. The selection rules remain those of step 1.1; the qualification about Cartesian components follows from their definition. This proves the statement.