Phy5645/AngularMomentumExercise

Recall that in QM: $\ L=\hbar {\sqrt {l(l+1)}}$; $\ L_{z}=m\hbar$.

The angle $\ \theta$ between $\mathbf {L}$ and the z-axis fulfills: $\ \cos \theta ={\frac {L_{z}}{L}}={\frac {m}{\sqrt {l(l+1)}}}$.

To make $\ \theta$ as small as possible, $\ m$ must be maximum ($\ l$ is fixed in this problem). This is when $\ m=l$. Therefore, the minimum angle $\ \alpha$ obeys: $\ \cos \alpha ={\frac {l}{\sqrt {l(l+1)}}}$

We solve $\ L=\hbar {\sqrt {l(l+1)}}$ to find $\ l$. Since $\ l$ will be very large we invoke the approximation: $\ {\sqrt {l(l+1)}}\approx l{\sqrt {\left(1+{\frac {1}{l}}\right)}}=l+{\frac {1}{2}}$. We resist the urge to discard the ${\frac {1}{2}}$ because without it our result will be trivial. $\ L=\hbar \left(l+{\frac {1}{2}}\right)$. Therefore $\ l={\frac {L}{\hbar }}-{\frac {1}{2}}$. Plugging this expression into the equation for $\ \alpha$ and using the previous approximation again, we have: $\ \cos \alpha \approx {\frac {l}{l+{\frac {1}{2}}}}={\frac {{\frac {L}{\hbar }}-{\frac {1}{2}}}{\frac {L}{\hbar }}}={\frac {2L-\hbar }{2L}}$.

$\ \alpha \approx cos^{-1}{\frac {2L-\hbar }{2L}}$.

Plugging in $\ L=4.83\cdot 10^{31}J\cdot s$ and $\ \hbar =1.055\cdot 10^{-34}J\cdot s$ we obtain:

$\ \alpha \approx cos^{-1}(0.999999999999999999999999999999999999999999999999999999999999999998908)\approx 1.48\cdot 10^{-33}rad$.

This is the smallest angle that $\ \mathbf {L}$ can make with the z-axis in the case of the earth going around the sun.

In the case of a quantum particle with $\ l=4$, we must use the exact expression: $\ \cos \alpha ={\frac {l}{\sqrt {l(l+1)}}}$.

$\ \cos \alpha ={\frac {4}{\sqrt {4(4+1)}}}={\frac {4}{\sqrt {4(4+1)}}}={\frac {4}{\sqrt {20}}}={\frac {2}{\sqrt {5}}}$.

$\ \alpha \approx 0.464rad\approx 26.6\deg$. This is the smallest angle that the angular momentum vector of a particle with $\ l=4$ can make with the z-axis.