Question 5.P.10: Find the angle between the angular momentum l = 4 and the z-...

Since m_{l} =0,\pm 1,\pm 2, … ,\pm l and the angle between the orbital angular momentum l and the z-axis is \cos \theta _{ml} =m_{l} /\sqrt{l(l+1)} we have

\theta _{ml} =\cos ^{-1} \left[\frac{m_{l}}{\sqrt{l(l+1)}} \right] =\cos ^{-1} \left[\frac{m_{l}}{2\sqrt{5} } \right];                (5.282)

hence

\theta _{0} =\cos ^{-1} (0)=90° ,                               (5.283)

\theta _{1} =\cos ^{-1} \left[\frac{1}{2\sqrt{5} } \right]=77.08° ,             \theta _{2} =\cos ^{-1} \left[\frac{2}{2\sqrt{5} } \right]=63.43° ,               (5.284)

\theta _{3} =\cos ^{-1} \left[\frac{3}{2\sqrt{5} } \right]=47.87° ,             \theta _{4} =\cos ^{-1} \left[\frac{4}{2\sqrt{5} } \right]=26.57° .               (5.285)

The angles for the remaining quantum numbers m_{4}= -1,-2,-3 ,-4 can be inferred at once from the relation

\theta _{-ml}=180° -\theta _{ml},                      (5.286)

hence

\theta _{-1} =180°-77.08°=102.92° ,        \theta _{-2} =180°-63.43°=116.57° ,         (5.287)

\theta _{-3} =180°-47.87°=132.13° ,        \theta _{-4} =180°-26.57°=153.43° .          (5.288)