Differential matrix element \[sec.2.7.2\]

Let us define the following differential components of the angular integral $I_{\rm dip}$ by solid angle $\Omega$ and by angle $\theta$: $$\begin{array}{lcl}
  \displaystyle\frac{d\, I_{\rm dip}\, (l_{f})}{d\Omega} & = &
  \displaystyle\frac{d\, I_{\rm dip}\, (l_{f})}{\sin{\theta}\,d\theta\, d\varphi} =
    \sqrt{\displaystyle\frac{1}{3}}\;
    \displaystyle\sum\limits_{\mu = \pm 1}
    \displaystyle\sum\limits_{m}
      h_{\mu} \cdot
        Y_{l_{f}m_{f}}^{*} ({\mathbf n}_{r})\;
        Y_{1,-\mu}(\mathbf{n}_{r}), \\

  \displaystyle\frac{d\, I_{\rm dip}\, (l_{f})}{\sin{\theta}\,d\theta} & = &
    \sqrt{\displaystyle\frac{1}{3}}\;
    \displaystyle\sum\limits_{\mu = \pm 1}
    \displaystyle\sum\limits_{m}
      h_{\mu} \cdot
    \displaystyle\int\limits_{0}^{2\pi}
        Y_{l_{f}m_{f}}^{*} ({\mathbf n}_{r})\;
        Y_{1,-\mu}(\mathbf{n}_{r})\; d\varphi.
\end{array}
\label{eq.2.7.2.1}$$ But, in their calculations we shall assume that the selection rules (\[eq.2.7.1.8\]) are fulfilled. We find: $$\begin{array}{lcl}
  \displaystyle\frac{d\, I_{\rm dip}\, (l_{f})}{d\Omega} & = &
    \sqrt{\displaystyle\frac{1}{3}}\;
      \displaystyle\sum\limits_{\mu = \pm 1} h_{\mu} \cdot
      \displaystyle\frac{3}{8\pi} \cdot
      P_{1}^{1} (\cos{\theta}) \cdot P_{1}^{1} (\cos{\theta}) =
    -i\; \displaystyle\frac{\sqrt{6}}{8\pi}\; \sin^{2}{\theta}, \\

  \displaystyle\frac{d\, I_{\rm dip}\, (l_{f})}{\sin{\theta}\,d\theta} & = &
  \displaystyle\int\limits_{0}^{2\pi}
    \displaystyle\frac{d\, I_{\rm dip}\, (l_{f})}{d\Omega}\; d\varphi =
    -i\; \displaystyle\frac{\sqrt{6}}{4}\; \sin^{2}{\theta}.
\end{array}
\label{eq.2.7.2.2}$$ On their basis we shall define the following differential components of the matrix element: $$\begin{array}{lcl}
  \displaystyle\frac{d\, p_{\rm dip}\, (l_{f}=1)}{d\Omega} & = &
  -\,\displaystyle\frac{J_{\rm dip}\,(1)}{w_{fi}} \cdot \displaystyle\frac{d\, I_{\rm dip}\, (l_{f})}{d\Omega} =
  i\;\displaystyle\frac{\sqrt{6}}{8\pi}\;
    \displaystyle\frac{J_{\rm dip}\, (1)}{w_{fi}} \cdot \sin^{2}{\theta}, \\
  \displaystyle\frac{d\, p_{\rm dip}\, (l_{f}=1)}{\sin{\theta}\,d\theta} & = &
  -\,\displaystyle\frac{J_{\rm dip}\, (1)}{w_{fi}} \cdot
    \displaystyle\frac{d\, I_{\rm dip}\, (l_{f})}{\sin{\theta}\,d\theta} =
  i\;\displaystyle\frac{\sqrt{6}}{4}\; \displaystyle\frac{J_{\rm dip}\, (1)}{w_{fi}} \cdot \sin^{2}{\theta}.
\end{array}
\label{eq.2.7.2.3}$$ One can 