# 核与粒子物理学 Nuclear and Particle Physics PHYS204

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$$V(r)=-\frac{Z e^{2}}{\rho}\left(\frac{3}{2}-\frac{r^{2}}{2 \rho^{2}}\right)=-\frac{3 Z e^{2}}{2 \rho}+\frac{1}{2}\left(\frac{Z e^{2}}{\rho^{3}}\right) r^{2}$$
The dependence of the potential on $r$ suggests that the $\mu^{-}$may be treated as an isotropic harmonic oscillator of eigenfrequency $\omega=\sqrt{\frac{Z_{e^{2}}}{m_{\mu} \rho^{3}}}$. The energy levels are therefore
$$E_{n}=\hbar \omega\left(n+\frac{3}{2}\right)-\frac{3 Z e^{2}}{2 \rho}$$
where $n=0,1,2, \ldots, \rho \approx 1.2 \times 10^{-13} A^{1 / 3} \mathrm{~cm}$.

## PHYS204COURSE NOTES ：

$$\frac{d E}{d t}=-P$$
i.e.,
$$\frac{e^{2}}{2 r^{2}} \frac{d r}{d t}=-\frac{2 e^{2}}{3 c^{3}} \cdot \frac{e^{4}}{r^{4} M^{2}}$$
Integration gives
$$R^{3}-r^{3}=\frac{4}{c^{3}} \cdot \frac{e^{4}}{M^{2}} t$$
where $R$ is the radius of the initial orbit of the $\mu$ mesion, being
$$R \approx \frac{\hbar^{2}}{m e^{2}}$$
At the $\mu$ ground state the radius of its orbit is the Bohr radius of the mesic atom
$$r_{0}=\frac{\hbar^{2}}{M e^{2}},$$