这是一份uwa西澳大学PHYS3001的成功案例

The direct comparison of angle resolved Auger spectra yields further results. Hemmers determined the energy resolved Auger spectrum for two different polar angles: $\theta=0^{\circ}$ and $\theta=54.7^{\circ}$. It follows from (3.116) for the Auger angular distribution
$$
\begin{gathered}
I\left(0^{\circ}\right)=\frac{I_{0}}{4 \pi}(1+\beta) \
I\left(54.7^{\circ}\right)=\frac{I_{0}}{4 \pi}
\end{gathered}
$$

PHYS3001 COURSE NOTES ：

The first term in corresponds to the electric dipole $E 1$ approximation with the transition operator (in the length form of the long-wave length approximation)
$$
T_{\lambda}^{E 1}=D_{\lambda}=\sum_{n}\left(r_{n}\right){\lambda}=\sqrt{\frac{4 \pi}{3}} \sum{n} r_{n} Y_{1 \lambda}\left(\theta_{n}, \phi_{n}\right) .
$$
The ‘first-order’ corrections to the dipole approximation are given by the second, magnetic dipole $M 1$ term with the transition operator
$$
T_{\lambda}^{M 1}=-\mathrm{i} M_{\lambda}=-\mathrm{i} \frac{\alpha}{2} \sum_{n}\left[\left(\ell_{n}\right){\lambda}+2\left(s{n}\right){\lambda}\right], $$ and by the third, electric quadrupole $E 2$ term with the transition operator $$ T{\lambda}^{E 2}=\frac{\mathrm{i} \alpha \omega}{2 \sqrt{3}} Q_{\lambda}=\frac{\mathrm{i} \alpha \omega}{2 \sqrt{3}} \sqrt{\frac{4 \pi}{5}} \sum_{n} r_{n}^{2} Y_{2 \lambda}\left(\theta_{n}, \phi_{n}\right) .
$$