Wigner 𝔇 Derivation
This is just the long-form derivation of the formula for Wigner's \(\mathfrak{D}\) matrices, discussed more on this page.
\[\begin{align}
\mathbf{e}_{(m')}(\mathbf{R}\, \mathbf{Q})
&=
\frac{(\mathbf{R}\, \mathbf{Q})_{a}^{\ell+m'}\, (\mathbf{R}\, \mathbf{Q})_{b}^{\ell-m'}}
{\sqrt{ (\ell+m')!\, (\ell-m')! }} \\\\
&=
\frac{(\mathbf{R}_a\, \mathbf{Q}_a - \bar{\mathbf{R}}_b\, \mathbf{Q}_b)^{\ell+m'}\,
(\mathbf{R}\, \mathbf{Q})_{b}^{\ell-m'}}
{\sqrt{ (\ell+m')!\, (\ell-m')! }} \\\\
&=
\sum_{\rho} \binom{\ell+m'} {\rho}
\frac{(\mathbf{R}_a\, \mathbf{Q}_a)^{\ell+m'-\rho} (- \bar{\mathbf{R}}_b\, \mathbf{Q}_b)^{\rho}\,
(\mathbf{R}_b\, \mathbf{Q}_a + \bar{\mathbf{R}}_a\, \mathbf{Q}_b)^{\ell-m'}} {\sqrt{ (\ell+m')!\, (\ell-m')! }} \\\\
&=
\sum_{\rho,\rho'} \binom{\ell+m'} {\rho} \binom{\ell-m'} {\rho'}
\frac{(\mathbf{R}_a\, \mathbf{Q}_a)^{\ell+m'-\rho} (- \bar{\mathbf{R}}_b\, \mathbf{Q}_b)^{\rho}\,
(\mathbf{R}_b\, \mathbf{Q}_a)^{\ell-m'-\rho'} (\bar{\mathbf{R}}_a\, \mathbf{Q}_b)^{\rho'}}
{\sqrt{ (\ell+m')!\, (\ell-m')! }} \\\\
&=
\sum_{\rho,m} \binom{\ell+m'} {\rho} \binom{\ell-m'} {\ell-m-\rho}
\frac{(\mathbf{R}_a\, \mathbf{Q}_a)^{\ell+m'-\rho} (- \bar{\mathbf{R}}_b\, \mathbf{Q}_b)^{\rho}\,
(\mathbf{R}_b\, \mathbf{Q}_a)^{m-m'+\rho} (\bar{\mathbf{R}}_a\, \mathbf{Q}_b)^{\ell-m-\rho}}
{\sqrt{ (\ell+m')!\, (\ell-m')! }} \\\\
&=
\sum_{\rho,m} \binom{\ell+m'} {\rho} \binom{\ell-m'} {\ell-m-\rho}
\mathbf{R}_a^{\ell+m'-\rho} (- \bar{\mathbf{R}}_b)^{\rho}\, \mathbf{R}_b^{m-m'+\rho} \bar{\mathbf{R}}_a^{\ell-m-\rho}
\frac{\mathbf{Q}_a^{\ell+m} \mathbf{Q}_b^{\ell-m}} {\sqrt{ (\ell+m')!\, (\ell-m')! }} \\\\
&=
\sum_{m} \mathbf{e}_{(m)}(\mathbf{Q}) \sum_{\rho} \binom{\ell+m'} {\rho} \binom{\ell-m'} {\ell-m-\rho}
\mathbf{R}_a^{\ell+m'-\rho} (- \bar{\mathbf{R}}_b)^{\rho}\, \mathbf{R}_b^{m-m'+\rho} \bar{\mathbf{R}}_a^{\ell-m-\rho}
\frac{\sqrt{ (\ell+m)!\, (\ell-m)! }} {\sqrt{ (\ell+m')!\, (\ell-m')! }} \\\\
\end{align}\]
We have introduced a new summation variable \(m\) and used the substitution \(\rho' \mapsto \ell-m-\rho\) to bring this into the form we need to express Wigner's \(\mathfrak{D}\) matrix. Alternatively, we could have made an equivalent substitution for \(\rho\), so that \(\mathfrak{D}\) would be given as a sum over \(\rho'\). This would have the effect of reversing the roles of \(a\) and \(b\), which is what we do on this page when \(\lvert \mathbf{R}_a \rvert < \lvert \mathbf{R}_b \rvert\).