# 量子力学的基础 Foundations of Quantum Mechanics PHYS96018

To begin our analysis it is helpful to consider a simpler problem, i.e. how to solve the linear equation
$$\frac{\mathrm{d} \varphi}{\mathrm{d} t}=A \varphi$$
on a finite-dimensional vector space $V$. If the matrix $A$ is diagonalizable and all eigenvalues are simple (i.e. without degeneracy), the space $V$ has a basis of right eigenvectors $\left{v_{j}\right}$ :
$$A v_{j}=\varepsilon_{j} v_{j}$$

$$\varphi(0)=\sum_{k=1}^{N} B_{k} v_{k},$$
where the coefficients of linear combination are obtained by using the left eigenvectors $v^{j}$ satisfying
$$v^{j} A=\varepsilon_{j} v^{j}$$
in the form
$$B_{j}=\left(v^{j}, \varphi(0)\right)$$

## PHYS96018 COURSE NOTES ：

The operator of the electric dipole moment reads $d=-e x$, or, upon transformation to a spherical basis, defined by $d_{\pm 1}=\mp\left(d_{1} \pm i d_{2}\right) / \sqrt{2}$, $d_{0}=d_{3}$,
$d_{\mu}=-e \sqrt{\frac{4 \pi}{3}} r Y_{1 \mu} .$
Converting the field $\boldsymbol{E}$ to the same basis,
$$E_{+1}=-\frac{1}{\sqrt{2}}\left(E_{1}+\mathrm{i} E_{2}\right), \quad E_{-1}=\frac{1}{\sqrt{2}}\left(E_{1}-\mathrm{i} E_{2}\right), \quad E_{0}=E_{3},$$
the scalar product becomes
$$\boldsymbol{d} \cdot \boldsymbol{E}=\sum_{\mu} d_{\mu}^{} E_{\mu}=\sum_{v} d_{v} E_{v}^{} .$$