Phy5645/AngularMomentumExercise

In quantum mechanics,

$L={\sqrt {\langle {\hat {\mathbf {L} }}^{2}\rangle }}=\hbar {\sqrt {l(l+1)}}$

and

$L_{z}=\langle {\hat {L}}_{z}\rangle =m\hbar .$

The angle $\theta$ between $\mathbf {L}$ and the $z\!$ axis is given by

$\cos {\theta }={\frac {L_{z}}{L}}={\frac {m}{\sqrt {l(l+1)}}}.$

To make $\theta$ as small as possible, $m\!$ must be at its maximum value, $m=l.\!$

Therefore, the minimum angle $\theta =\alpha \!$ is given by

$\cos {\alpha }={\frac {l}{\sqrt {l(l+1)}}},$

or by

$\sin {\alpha }={\frac {1}{\sqrt {l+1}}}.$

We now solve $\ L=\hbar {\sqrt {l(l+1)}}$ to find $l.\!$

Since $\ l$ will be very large, we make the approximation, $\ {\sqrt {l(l+1)}}\approx l{\sqrt {\left(1+{\frac {1}{l}}\right)}}=l,$ so that $L\approx l\hbar .$ Because $l\!$ is large, we see that $\sin {\alpha }$ is small, and thus we may make the approximation,

$\alpha \approx {\sqrt {\frac {\hbar }{L}}}.$

If we now substitute in $\ L=4.83\times 10^{31}\,{\text{J}}\cdot {\text{s}}$ and $\ \hbar =1.055\times 10^{-34}\,{\text{J}}\cdot {\text{s}},$ we obtain

$\ \alpha \approx {\sqrt {\frac {1.055\times 10^{-34}}{4.83\times 10^{31}}}}\approx 1.48\times 10^{-33}\,{\text{rad}}.$

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 for the angle.

$\ \sin \alpha ={\frac {1}{\sqrt {5}}},$

which gives us

$\ \alpha \approx 0.464\,{\text{rad}}=26.6\deg .$

This is the smallest angle that the angular momentum vector of a particle with $\ l=4$ can make with the $z\!$ axis. This angle is much larger than that for the Earth orbiting the sun, as we would expect.

Back to Orbital Angular Momentum Eigenfunctions