Phy5645/AngularMomentumExercise: Difference between revisions

Latest revision as of 13:40, 18 January 2014

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

@@ Line 1: / Line 1: @@
-If the earth (classically) revolves around the sun counter-clockwise in the x-y plane with the sun at the origin, what is the minimum angle between the angular momentum vector of the earth and the z axis? Ignore the intrinsic spin of the earth. The angular momentum of the earth around the sun is <math>\ 4.83*10^{31} </math> <math> \frac{kg*m^2}{s}</math>.
+In quantum mechanics,
-Compare the minimum angle to that of a quantum particle with <math>\ l=4 </math>.
+<math>L=\sqrt{\langle\hat{\mathbf{L}}^2\rangle}=\hbar\sqrt{l(l+1)}</math>
+and
-Solution:
+<math>L_z=\langle\hat{L}_z\rangle=m\hbar.</math>
+The angle <math>\theta</math> between <math>\mathbf{L}</math> and the <math>z\!</math> axis is given by
-Recall that in QM: <math>\ L = \hbar\sqrt{l(l+1)}</math>;  <math>\ L_z = m\hbar</math>.
+<math>\cos{\theta}= \frac{L_z}{L} = \frac{m}{\sqrt{l(l+1)}}.</math>
-The angle <math>\ \theta </math> between <math> \mathbf{L} </math> and the z-axis fulfills: <math>\ \cos \theta = \frac{L_z}{L} = \frac{m}{\sqrt{l(l+1)}}</math>.
+To make <math>\theta </math> as small as possible, <math>m\!</math> must be at its maximum value, <math>m=l.\!</math>
-To make <math>\ \theta </math> as small as possible, <math>\ m </math> must be maximum. This is when <math>\ m=l </math>.
+Therefore, the minimum angle <math>\theta=\alpha\!</math> is given by
-Therefore, the minimum angle <math>\ \alpha </math> obeys: <math>\ \cos \alpha = \frac{l}{\sqrt{l(l+1)}} </math>
-We solve <math>\ L = \hbar\sqrt{l(l+1)}</math> to find <math>\ l </math>.
+<math>\cos{\alpha}=\frac{l}{\sqrt{l(l+1)}},</math>
-Since <math>\ l </math> will be very large we invoke the approximation: <math>\ \sqrt{l(l+1)} \approx l \sqrt{(1+\frac{1}{l})} = l+\frac{1}{2}</math>. We resist the urge to discard the <math> \frac{1}{2} </math> because without it our result will be trivial.
-<math>\ L= \hbar(l+\frac{1}{2})</math>, therefore <math>\ l = \frac{L}{\hbar}-\frac{1}{2}.
+or by
+<math>\sin{\alpha}=\frac{1}{\sqrt{l+1}}.</math>
+We now solve <math>\ L = \hbar\sqrt{l(l+1)}</math> to find <math>l.\!</math>
+Since <math>\ l</math> will be very large, we make the approximation, <math>\ \sqrt{l(l+1)} \approx l \sqrt{\left(1+\frac{1}{l}\right)} = l,</math> so that <math>L\approx l\hbar.</math>  Because <math>l\!</math> is large, we see that <math>\sin{\alpha}</math> is small, and thus we may make the approximation,
+<math>\alpha\approx\sqrt{\frac{\hbar}{L}}.</math>
+If we now substitute in <math>\ L = 4.83\times 10^{31}\,\text{J}\cdot\text{s}</math> and <math>\ \hbar = 1.055 \times 10^{-34}\,\text{J} \cdot \text{s},</math> we obtain
+<math>\ \alpha \approx \sqrt{\frac{1.055\times 10^{-34}}{4.83\times 10^{31}}}\approx 1.48 \times 10^{-33}\,\text{rad}.</math>
+This is the smallest angle that <math>\mathbf{L}</math> can make with the <math>z\!</math> axis in the case of the Earth going around the sun.
+In the case of a quantum particle with <math>\ l = 4</math>, we must use the exact expression for the angle.
+<math>\ \sin \alpha = \frac{1}{\sqrt{5}},</math>
+which gives us
+<math>\ \alpha \approx 0.464\,\text{rad} = 26.6 \deg. </math>
+This is the smallest angle that the angular momentum vector of a particle with <math>\ l=4 </math> can make with the <math>z\!</math> axis.  This angle is much larger than that for the Earth orbiting the sun, as we would expect.
+Back to [[Orbital Angular Momentum Eigenfunctions#Problems|Orbital Angular Momentum Eigenfunctions]]

Phy5645/AngularMomentumExercise: Difference between revisions

Latest revision as of 13:40, 18 January 2014

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