这是一份liverpool利物浦大学PHYS306的成功案例

$$
\cos \phi \sin \phi=-\frac{2 . J}{2 \Delta}
$$
i.e.
$$
\sin ^{2} \phi \cos ^{2} \phi=\frac{J^{2}}{\Delta^{2}} .
$$
Using similar manipulations we obtain
$$
\sin ^{4} \phi=\frac{(\delta-\Delta)^{2}}{4 \Delta^{2}}
$$
and
$$
\cos ^{4} \phi=\frac{(\delta+\Delta)^{2}}{4 \Delta^{2}}
$$

PHYS387 COURSE NOTES ：

which is well-known to us from theoretical mechanics since it describes the precession of the vector $S$ around the fixed vector $\boldsymbol{\Omega}$. In fact, the vector $\boldsymbol{S}$ has a fixed length. To prove this, we define
$$
\begin{aligned}
\bar{\rho}{r l} &=e^{i \delta t / \hbar} \rho{r l} \
\bar{J} &=e^{i \delta t / \hbar} J
\end{aligned}
$$
and obtain
$$
\begin{aligned}
&\hbar \frac{d}{d t} \bar{\rho}{r l}=-i . \bar{J}\left(1-2 \rho{r r}\right) \
&\hbar \frac{d}{d t} \rho_{r r}=-i \bar{J}\left(\bar{\rho}{r l}\right)^{}+i \bar{J}^{} \bar{\rho}{r l}
\end{aligned}
$$