Phy5645/AngularMomentumExercise: Difference between revisions
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<math>\alpha\approx\sqrt{\frac{\hbar}{L}}.</math> | <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 | 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} rad.</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> | ||
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. | 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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which gives us | which gives us | ||
<math>\ \alpha \approx 0.464 rad = 26.6 \deg. </math> | <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. | 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]] | Back to [[Orbital Angular Momentum Eigenfunctions]] | ||
Revision as of 23:12, 29 August 2013
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, 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 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.