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| We now write down the Cartesian components of the angular momentum operator in spherical coordinates. We will make use of this result later in determining the eigenfunctions of the angular momentum squared and of one of its components. | | We now write down the Cartesian components of the angular momentum operator in spherical coordinates. We will make use of this result later in determining the eigenfunctions of the angular momentum squared and of one of its components. |
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| The Cartesian coordinates <math>x,\!</math <math>y,\!</math> and <math>z\!</math> can be written in terms of the spherical coordinates <math>r,\!</math> <math>\theta,\!</math> and <math>\phi\!</math> as follows: | | The Cartesian coordinates <math>x,\!</math> <math>y,\!</math> and <math>z\!</math> can be written in terms of the spherical coordinates <math>r,\!</math> <math>\theta,\!</math> and <math>\phi\!</math> as follows: |
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| <math>x=r\sin\theta\cos\phi,\!</math> <math>y=r\sin\theta\sin\phi,\!</math> <math>z=r\cos\theta\!</math> | | <math>x=r\sin\theta\cos\phi,\!</math> <math>y=r\sin\theta\sin\phi,\!</math> <math>z=r\cos\theta\!</math> |
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| Denote the state <math> \langle \mathbf{r} \!\, | = \langle \mathbf{r}\! \, \theta \phi | </math>
| | Let us start with the <math>x\!</math> component of the angular momentum, <math>\hat{L}_x.</math> In Cartesian coordinates, this is |
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| If you choose <math> \alpha \! </math> along the z-axis then the only coordinate that will change is phi such that <math> \phi \rightarrow \phi + \alpha </math>. Now the state is written as:
| | <math>\hat{L}_x=-i\hbar\left (y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right ).</math> |
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| <math> \langle \mathbf{r}\! \, \theta \phi | \left(1 + \frac{i}{\hbar} \alpha L_{z}\right) = \langle \mathbf{r} \! \, \theta \phi + \alpha | </math>
| | If we make use of the chain rule, then we obtain |
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| Working to first order in alpha the right hand side becomes:
| | <math>\hat{L}_{x} = -i\hbar\left( -\sin\phi \frac{\partial}{\partial \theta} \, -\cot\theta\cos\phi \frac{\partial}{\partial\phi} \! \right).</math> |
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| <math> \langle \mathbf{r}\! \, \theta \phi | + \alpha \frac{\partial}{\partial \phi} \langle \mathbf{r}\! \, \theta \phi | </math> | | Similarly, the <math>y\!</math> and <math>z\!</math> components may be found to be |
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| Therefore <math> L_{z} = \frac{\hbar}{i} \frac{\partial}{\partial \phi} </math>
| | <math>\hat{L}_{y} = -i\hbar\left(\cos\phi \frac{\partial}{\partial\theta} - \cot\theta\sin\phi \frac{\partial}{\partial\phi} \! \right) </math> |
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| | and |
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| Now choose <math> \alpha \! </math> along the x-axis then the cartesian coordinates are changed such that:
| | <math>\hat{L}_{z} = -i\hbar\frac{\partial}{\partial \phi}.</math> |
| <math> x \rightarrow x \! </math> ,
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| <math> y \rightarrow y - \alpha z \!</math>, and
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| <math> z \rightarrow z + \alpha y \! </math>,
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| from these transformations it can be determined that <math> \delta \theta = -\alpha\sin\phi \! </math> since <math> \delta z = \alpha y \! </math>
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| and since x does not change it can be determined that <math> \delta \phi = -\alpha\cot\theta \cos\phi \! </math>.
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| This means that the original state is now written as:
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| <math> \langle \mathbf{r}\! \, \theta \phi | \left(1 + \frac{i}{\hbar} \alpha L_{x}\right) = \langle \mathbf{r} \! \, \,\theta - \alpha\sin\phi \, \phi - \alpha\cot\theta\cos\phi | </math>
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| Expanding the right hand side of the above equation as before to the first order of alpha the whole equation becomes:
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| <math> \langle \mathbf{r}\! \, \theta \phi | L_{x} = \frac{\hbar}{i} \left( -\sin\phi \frac{\partial}{\partial \theta} \, -\cot\theta\cos\phi \frac{\partial}{\partial\phi} \! \right) \langle \mathbf{r}\! \, \theta \phi | </math>
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| Therfore <math> L_{x} = \frac{\hbar}{i}\left( -\sin\phi \frac{\partial}{\partial \theta} \, -\cot\theta\cos\phi \frac{\partial}{\partial\phi} \! \right) </math>
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| Using the same techinque, choose <math> \alpha \! </math> along the y-axis and the coordinates will change in a similar fashion such that it can be shown that <math> L_{y} = \frac{\hbar}{i} \left(\cos\phi \frac{\partial}{\partial\theta} - \cot\theta\sin\phi \frac{\partial}{\partial\phi} \! \right) </math>
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| ==Problem== | | ==Problem== |
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| Show, using the above results, that the operator, | | Show, using the above results, that the operator, |
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| :<math> \hat{R}_{\Delta\phi}=\exp \left (\frac{i}{\hbar}\Delta\phi\hat{L}_z\right ),</math> | | :<math> \hat{R}(\Delta\phi)=\exp \left (\frac{i}{\hbar}\Delta\phi\hat{L}_z\right ),</math> |
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| when applied to a function <math> f(\phi)\!</math> of the azimuthal angle <math>\phi,\!</math> rotates the angle <math>\phi\!</math> to <math>\phi+\Delta\phi.</math> That is, show that | | when applied to a function <math> f(\phi)\!</math> of the azimuthal angle <math>\phi,\!</math> rotates the angle <math>\phi\!</math> to <math>\phi+\Delta\phi.\!</math> That is, show that |
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| <math> \hat{R}_{\Delta\phi}f(\phi)=f(\phi+\Delta\phi).</math> | | <math>\hat{R}(\Delta\phi)f(\phi)=f(\phi+\Delta\phi).</math> |
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| [[Phy5645/Angular Momentum Problem 1|Solution]] | | [[Phy5645/Angular Momentum Problem 1|Solution]] |
We now write down the Cartesian components of the angular momentum operator in spherical coordinates. We will make use of this result later in determining the eigenfunctions of the angular momentum squared and of one of its components.
The Cartesian coordinates 
and 
can be written in terms of the spherical coordinates 
and 
as follows:
Let us start with the 
component of the angular momentum, 
In Cartesian coordinates, this is
If we make use of the chain rule, then we obtain
Similarly, the 
and components may be found to be
and
Problem
(Richard L. Liboff, Introductory Quantum Mechanics, 2nd Edition, pp. 377-379)
Show, using the above results, that the operator,
when applied to a function 
of the azimuthal angle 
rotates the angle to 
That is, show that
Solution
