Solution
= Solution
Because $H$ is independent of $\phi$, Hamilton's equation gives $\dot p_\phi=0$. Put
$$
L^2=p_\theta^2+\frac{p_\phi^2}{\sin^2\theta}.
$$
Using
$$
\dot\theta=\frac{p_\theta}{mr^2},\qquad
\dot p_\theta=\frac{p_\phi^2\cos\theta}{mr^2\sin^3\theta},
$$
one obtains $dL^2/dt=0$. Physically, $p_\phi=L_z$ is the component of <angular momentum> along the polar axis, while $L^2$ is the squared magnitude of the <angular momentum>.