Phy5645/AngularMomentumExercise: Difference between revisions

From PhyWiki
Jump to navigation Jump to search
No edit summary
No edit summary
 
(6 intermediate revisions by 2 users not shown)
Line 1: Line 1:
Team 2
In quantum mechanics,


Suppose the earth revolves around the sun counter-clockwise in the x-y plane with the sun at the origin. Quantum mechanically, what is the minimum angle the angular momentum vector of the earth makes with the z axis? Ignore the intrinsic spin of the earth. The angular momentum of the earth around the sun is <math>\ 4.83 \cdot 10^{31} J \cdot s</math>.
<math>L=\sqrt{\langle\hat{\mathbf{L}}^2\rangle}=\hbar\sqrt{l(l+1)}</math>
Compare the minimum angle with that of a quantum particle with <math>\ l=4 </math>.


and


<math>L_z=\langle\hat{L}_z\rangle=m\hbar.</math>


Solution:
The angle <math>\theta</math> between <math>\mathbf{L}</math> and the <math>z\!</math> axis is given by


<math>\cos{\theta}= \frac{L_z}{L} = \frac{m}{\sqrt{l(l+1)}}.</math>


Recall that in QM: <math>\ L = \hbar\sqrt{l(l+1)}</math>;  <math>\ L_z = m\hbar</math>.
To make <math>\theta </math> as small as possible, <math>m\!</math> must be at its maximum value, <math>m=l.\!</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>.
Therefore, the minimum angle <math>\theta=\alpha\!</math> is given by


To make <math>\ \theta </math> as small as possible, <math>\ m </math> must be maximum (<math>\ l </math> is fixed in this problem). This is when <math>\ m=l </math>.
<math>\cos{\alpha}=\frac{l}{\sqrt{l(l+1)}},</math>
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>.
or by
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}</math>. Plugging this expression into the equation for <math>\ \alpha </math> and using the previous approximation again, we have: <math>\ \cos \alpha \approx \frac{l}{l+\frac{1}{2}} = \frac{\frac{L}{\hbar}-\frac{1}{2}}{\frac{L}{\hbar}} = \frac{2L-\hbar}{2L}</math>.


<math>\ \alpha \approx cos^{-1}\frac{2L-\hbar}{2L}</math>.
<math>\sin{\alpha}=\frac{1}{\sqrt{l+1}}.</math>


Plugging in <math>\ L = 4.83 \cdot 10^{31} J \cdot s </math> and <math>\ \hbar = 1.055 \cdot 10^{-34} J \cdot s</math> we obtain:
We now solve <math>\ L = \hbar\sqrt{l(l+1)}</math> to find <math>l.\!</math>


<math>\ \alpha \approx cos^{-1}(0.999999999999999999999999999999999999999999999999999999999999999998908) \approx 1.48 \cdot 10^{-33} rad</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,


This is the smallest angle that <math>\ \mathbf{L} </math> makes with the z-axis in the case of the earth going around the sun.
<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


In the case of a quantum particle with <math>\ l = 4</math>, we must use the exact expression <math>\ \cos \alpha = \frac{l}{\sqrt{l(l+1)}} </math>.
<math>\ \alpha \approx \sqrt{\frac{1.055\times 10^{-34}}{4.83\times 10^{31}}}\approx 1.48 \times 10^{-33}\,\text{rad}.</math>


<math>\ \cos \alpha = \frac{4}{\sqrt{4(4+1)}} = \frac{4}{\sqrt{4(4+1)}} = \frac{4}{\sqrt{20}} = \frac{2}{\sqrt{5}}</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.


<math>\ \alpha \approx 0.464 rad \approx 26.6 \deg </math>.
In the case of a quantum particle with <math>\ l = 4</math>, we must use the exact expression for the angle.
This is the smallest angle that the angular momentum vector of a particle with <math>\ l=4 </math> makes with the z-axis.
 
<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]]

Latest revision as of 13:40, 18 January 2014

In quantum mechanics,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\sqrt{\langle\hat{\mathbf{L}}^2\rangle}=\hbar\sqrt{l(l+1)}}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_z=\langle\hat{L}_z\rangle=m\hbar.}

The angle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta} between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{L}} and the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z\!} axis is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{\theta}= \frac{L_z}{L} = \frac{m}{\sqrt{l(l+1)}}.}

To make Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta } as small as possible, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\!} must be at its maximum value, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m=l.\!}

Therefore, the minimum angle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta=\alpha\!} is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos{\alpha}=\frac{l}{\sqrt{l(l+1)}},}

or by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin{\alpha}=\frac{1}{\sqrt{l+1}}.}

We now solve Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ L = \hbar\sqrt{l(l+1)}} to find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l.\!}

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ l} will be very large, we make the approximation, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \sqrt{l(l+1)} \approx l \sqrt{\left(1+\frac{1}{l}\right)} = l,} so that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L\approx l\hbar.} Because Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l\!} is large, we see that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin{\alpha}} is small, and thus we may make the approximation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha\approx\sqrt{\frac{\hbar}{L}}.}

If we now substitute in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ L = 4.83\times 10^{31}\,\text{J}\cdot\text{s}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \hbar = 1.055 \times 10^{-34}\,\text{J} \cdot \text{s},} we obtain

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{L}} can make with the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z\!} axis in the case of the Earth going around the sun.

In the case of a quantum particle with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ l = 4} , we must use the exact expression for the angle.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \sin \alpha = \frac{1}{\sqrt{5}},}

which gives us

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \alpha \approx 0.464\,\text{rad} = 26.6 \deg. }

This is the smallest angle that the angular momentum vector of a particle with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ l=4 } can make with the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Retrieved from "http://wiki.physics.fsu.edu/index.php?title=Phy5645/AngularMomentumExercise&oldid=15815"