# 高能量天体物理学 High Energy Astrophysics PHYS201501

$$H=\frac{p^{2}}{2 m_{e}}-V(r)$$
Making the usual substitution: $\boldsymbol{p}=-i \hbar \nabla$ we get the relevant form of (114):
$$\left(-\frac{\hbar^{2}}{2 m_{e}} \nabla^{2}-V(r)\right) \psi(\boldsymbol{r})=E \psi(\boldsymbol{r})$$
It is convenient to use atomic units where the natural unit of length is the Bohr radius: $a_{0} \equiv \hbar^{2} / m e^{2}=0.52910^{-8} \mathrm{~cm}$, and the natural unit of energy is twice the Rydberg constant: $e^{2} / a_{0} \equiv 2 R y=27.2 \mathrm{eV}=4.3610^{-11}$ erg. In these units, $e=\hbar=m=1$.
Equation (119) then takes the form:
$$\left(\frac{1}{2} \nabla^{2}+E+V(r)\right) \psi(\boldsymbol{r})=0$$

## PHYS201501 COURSE NOTES ：

$$\frac{1}{2} m v^{2}=\frac{Z e^{2}}{2 r}$$
For the ground-state:
$$r \simeq \frac{a_{0}}{Z},$$
and thus:
$$v \simeq\left(\frac{Z^{2} e^{2}}{m a_{0}}\right)^{1 / 2}=(Z \alpha) c$$