这是一份uwa西澳大学PHYS2001的成功案例

$$
c_{s}(t)=\frac{i}{\hbar}\left\langle s^{(0)}\left|\sum_{\alpha=1}^{N} \frac{q^{(\alpha)}}{m^{(\alpha)}} \Pi_{i}^{(\alpha)}\right| n^{(0)}\right\rangle \int_{0}^{t} A_{i}(0, t) e^{-i \omega_{n-s} t} d t
$$
The matrix element in is evaluated by using the commutator
$$
\left[H^{(0)}, p_{i}\right]=-i \hbar \sum_{\alpha=1}^{N} \frac{q^{(\alpha)}}{m^{(\alpha)}} \Pi_{i}^{(\alpha)}
$$
where $p_{i}$ is the electric dipole moment operator, and we have assumed a velocityindependent potential in $(2.3)$. Then
$$
\left\langle s^{(0)}\left|\sum_{\alpha=1}^{N} \frac{q^{(\alpha)}}{m^{(\alpha)}} \Pi_{i}^{(\alpha)}\right| n^{(0)}\right\rangle=i \omega_{s n}\left\langle p_{i}\right\rangle_{s n} .
$$
To proceed further we require an explicit form for the vector potential, which we take to be that for a harmonic plane wave
$$
\mathbf{A}(\mathbf{r}, t)=\mathbf{A}_{0} \cos (\mathbf{k} \cdot \mathbf{r}-\omega t) .
$$

MATH2001 COURSE NOTES ：

$$
H^{(1)}=-\sum_{\alpha=1}^{N} \frac{q^{(\alpha)}}{2 m(\alpha)}\left(\mathbf{B} \times \mathbf{r}^{(\alpha)}\right) \cdot \mathbf{\Pi}^{(\alpha)}=-\mathbf{m} \cdot \mathbf{B}
$$
and
$$
H^{(2)}=-\sum_{\alpha=1}^{N} \frac{\left(q^{(\alpha)}\right)^{2}}{8 m^{(\alpha)}}\left[r_{i}^{(\alpha)} r_{j}^{(\alpha)}-\left(r^{(\alpha)}\right)^{2} \delta_{i j}\right] B_{i} B_{j}
$$
Here
$$
\mathbf{m}=\sum_{\alpha=1}^{N} \frac{q^{(\alpha)}}{2 m(\alpha)} \mathbf{r}^{(\alpha)} \times \boldsymbol{\Pi}^{(\alpha)}
$$