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| | In quantum mechanics, |
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| Suppose that classically, 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 can make with the z axis? Ignore the intrinsic spin of the earth. The angular momentum of the earth 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>.
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| | <math>L_z=\langle\hat{L}_z\rangle=m\hbar.</math> |
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| Solution:
| | The angle <math>\theta</math> between <math>\mathbf{L}</math> and the <math>z\!</math> axis is given by |
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| | <math>\cos{\theta}= \frac{L_z}{L} = \frac{m}{\sqrt{l(l+1)}}.</math> |
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| 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> |
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| 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 |
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| 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>
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| We solve <math>\ L = \hbar\sqrt{l(l+1)}</math> to find <math>\ l </math>.
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| Since <math>\ l </math> will be very large we invoke the approximation: <math>\ \sqrt{l(l+1)} \approx l \sqrt{\left(1+\frac{1}{l}\right)} = 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.
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| <math>\ L= \hbar\left(l+\frac{1}{2}\right)</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>.
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| <math>\ \alpha \approx cos^{-1}\frac{2L-\hbar}{2L}</math>. | | <math>\sin{\alpha}=\frac{1}{\sqrt{l+1}}.</math> |
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| 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> |
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| | 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, |
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| | <math>\alpha\approx\sqrt{\frac{\hbar}{L}}.</math> |
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| <math>\ \alpha \approx cos^{-1}(0.999999999999999999999999999999999999999999999999999999999999999998908) \approx 1.48 \cdot 10^{-33} rad</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 |
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| This is the smallest angle that <math>\ \mathbf{L} </math> can make with the z-axis in the case of the earth going around the sun.
| | <math>\ \alpha \approx \sqrt{\frac{1.055\times 10^{-34}}{4.83\times 10^{31}}}\approx 1.48 \times 10^{-33}\,\text{rad}.</math> |
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| | 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. |
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| 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>. | | In the case of a quantum particle with <math>\ l = 4</math>, we must use the exact expression for the angle. |
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| <math>\ \cos \alpha = \frac{4}{\sqrt{4(4+1)}} = \frac{4}{\sqrt{4(4+1)}} = \frac{4}{\sqrt{20}} = \frac{2}{\sqrt{5}}</math>. | | <math>\ \sin \alpha = \frac{1}{\sqrt{5}},</math> |
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| <math>\ \alpha \approx 0.464 rad \approx 26.6 \deg </math>. | | which gives us |
| This is the smallest angle that the angular momentum vector of a particle with <math>\ l=4 </math> can make with the z-axis. | | |
| | <math>\ \alpha \approx 0.464\,\text{rad} = 26.6 \deg. </math> |
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| | 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. |
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| | Back to [[Orbital Angular Momentum Eigenfunctions#Problems|Orbital Angular Momentum Eigenfunctions]] |
In quantum mechanics,
and
The angle 
between 
and the 
axis is given by
To make as small as possible, 
must be at its maximum value, 
Therefore, the minimum angle 
is given by
or by
We now solve 
to find 
Since 
will be very large, we make the approximation, 
so that 
Because 
is large, we see that 
is small, and thus we may make the approximation,
If we now substitute in 
and 
we obtain
This is the smallest angle that can make with the axis in the case of the Earth going around the sun.
In the case of a quantum particle with 
, we must use the exact expression for the angle.
which gives us
This is the smallest angle that the angular momentum vector of a particle with can make with the axis. This angle is much larger than that for the Earth orbiting the sun, as we would expect.
Back to Orbital Angular Momentum Eigenfunctions