|
|
| (9 intermediate revisions by 2 users not shown) |
| Line 1: |
Line 1: |
| Submitted by Team 1
| | :<math> \hat{R}(\Delta\phi)f(\phi) = \left[ \exp \left( \Delta \phi \frac{\partial}{\partial \phi} \right) \right] f(\phi)</math> |
| | :<math> = f(\phi) + \frac{df(\phi)}{d\phi}\Delta\phi + \tfrac{1}{2}\frac{d^2 f(\psi)}{d\phi^2}\left(\Delta\phi\right)^2 + \cdots = f \left( \phi + \Delta \phi \right).</math> |
|
| |
|
| '''Questions'''
| | Back to [[Spherical Coordinates#Problem|Spherical Coordinates]] |
| | |
| (a) Show that the operator
| |
| | |
| :<math> \hat{R}_{\Delta \phi} \equiv \exp \left( \frac{i \Delta \phi \hat{L}_z}{\hbar} \right)</math>
| |
| | |
| when acting on the function <math> f(\phi) \!</math> changes <math> f \!</math> by a rotation of coordinates about the <math> z \!</math> axis so that the radius through <math> \phi \!</math> is rotated to the radius through <math> \phi + \Delta \phi \!</math>. That is, show that
| |
| | |
| :<math> \hat{R}_{\Delta\phi} f(\phi) = f \left( \phi + \Delta \phi \right) </math>.
| |
| | |
| (b) Show that the operator
| |
| | |
| :<math> \hat{R}_{\Delta\vec{\phi}} = \exp \left( \frac{i \Delta \vec{\phi} \cdot \mathbf{\hat{L}}}{\hbar} \right) </math>
| |
| | |
| when action on <math> f(\mathbf{r}) \!</math> changes <math> f \!</math> by rotating <math> \mathbf{r} \!</math> to a new value on the surface of the sphere of radius <math> r \!</math>, but rotated away from <math> \mathbf{r} \!</math> through the azimuth <math> \Delta \phi \!</math>, so that <math> \mathbf{r} \left( \theta, \phi \right) \rightarrow \mathbf{r}' = \mathbf{r} \left( \theta, \phi + \Delta \phi \right) \!</math>. For infinitesimal displacement <math> \delta \vec{\phi} \!</math>, we may write
| |
| | |
| :<math> \hat{R}_{\delta \vec{\phi}} f(\mathbf{r}) = f\left( \mathbf{r} + \delta \mathbf{r} \right) </math>
| |
| | |
| :<math> \delta \mathbf{r} = \delta \vec{\phi} \times \mathbf{r}. </math>
| |
| | |
| ---------
| |
| | |
| '''Solutions'''
| |
| | |
| (a)
| |
| :<math> \hat{R}_{\Delta \phi} f = \left[ \exp \left( \Delta \phi \frac{\partial}{\partial \phi} \right) \right] f </math>
| |
| :<math> = f(\phi) + \Delta \phi \frac{\partial f}{\partial \phi} + \frac{\left(\Delta\phi\right)^2}{2} \frac{\partial^2 f}{\partial \phi^2} + \cdots = f \left( \phi + \Delta \phi \right).</math>
| |
| | |
| (b) Let <math> \delta \phi \!</math> be an infinitesimal angle so that <math> \Delta \phi = n \delta \phi \!</math> in the limit that <math> n >> 1 \!</math>. For the infinitesimal rotation
| |
| | |
| :<math> \mathbf{r}' = \mathbf{r} + \delta \mathbf{r} = \mathbf{r} + \delta \vec{\phi} \times \mathbf{r} </math>
| |
| | |
| so that
| |
| | |
| :<math> f \left( \mathbf {r} + \delta \mathbf{r} \right) = f(\mathbf{r})+ \delta \vec{\phi} \times \mathbf{r} \cdot \mathbf{\nabla} f(\mathbf{r}) </math>
| |
| :<math> = f(\mathbf{r})+ \delta \vec{\phi} \cdot \mathbf{r} \times \mathbf{\nabla} f(\mathbf{r}) </math>
| |
| :<math> = f(\mathbf{r})+ \frac{i}{\hbar} \delta \vec{\phi} \cdot \mathbf{r} \cdot \mathbf{\hat{p}} f(\mathbf{r}) </math>
| |
| :<math> = f(\mathbf{r})+ \frac{i}{\hbar}\delta \vec{\phi} \cdot \mathbf{\hat{L}} f(\mathbf{r}) </math>.
| |
| | |
| In the Taylor series expansion of <math> f\left( \mathbf{r}+\delta\mathbf{r} \right) \!</math> above we have only kept terms of <math> O \left(\delta \phi \right) \!</math>. [The expression <math> \delta \mathbf{r} = \delta \vec{\phi} \times \mathbf{r} \!</math> is valid only to terms of <math> O \left(\delta \phi \right) \!</math>.] In this manner we obtain
| |
| | |
| :<math> f\left( \mathbf{r} + \delta \mathbf{r} \right) = \left( \hat{I} + \frac{i}{\hbar} \delta \vec{\phi} \cdot \mathbf{\hat{L}} \right) f(\mathbf{r}) = \hat{R}_{\delta\vec{\phi}} f(\mathbf{r}) </math>
| |
| | |
| For a finite rotational displacement through the angle <math> \mathbf{\Delta} \vec{\phi} = n \delta \vec{\phi} \!</math>, we apply the operator <math> \hat{R}_{\delta \vec{\phi}} \!</math>, <math> n \!</math> times:
| |
| | |
| :<math> \hat{R}_{n\delta\vec{\phi}} = \left( \hat{R}_{\delta\vec{\phi}} \right)^n = \left( \hat{I} + \frac{i}{\hbar} \delta \vec{\phi} \cdot \mathbf{\hat{L}} \right)^n </math>
| |
| | |
| and pss to the limit <math> n \rightarrow \infty \!</math> or, equivalently, <math> \Delta \phi / \delta \phi \rightarrow \infty \!</math>.
| |
| | |
| :<math> \hat{R}_{\Delta \phi} = \lim_{\Delta \phi / \delta \phi \rightarrow \infty} \left( \hat{I} + \frac{i}{\hbar} \delta \vec{\phi} \cdot \mathbf{\hat{\mathbf{L}}} \right)^{\Delta \phi / \delta \phi} = e^{i \Delta \vec{\phi} \cdot \mathbf{\hat{\mathbf{L}}} \hbar} </math>.
| |
| | |
| The operator <math> \hat{R}_{\delta\vec{\phi}} \!</math> rotates <math> \mathbf{r} \!</math> to <math> \mathbf{r} + \delta\vec{\phi}\times\mathbf{r} \!</math> with respect to a fixed coordinate frame. If, on the other hand, the coordinate frame is rotated through <math> \delta \vec{\phi} \!</math> with <math> \mathbf{r} \!</math> fixed in space, then in the new coordinate frame this vector has the value <math> \mathbf{r} - \delta \vec{\phi} \times \mathbf{r} \!</math>. Thus, rotation of coordinates through <math> \delta \vec{\phi} \!</math> is generated by the operator <math> \hat{R}_{-\delta \vec{\phi}}.</math>
| |
| | |
| --------
| |
| | |
| (Note: This problem is excerpted from {\it{"Introductory Quantum Mechanics"}}, 2nd edition, p377-p379, which is written by Richard L. Liboff.)
| |
![{\displaystyle {\hat {R}}(\Delta \phi )f(\phi )=\left[\exp \left(\Delta \phi {\frac {\partial }{\partial \phi }}\right)\right]f(\phi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4fcdc54995e0ddd72a6eebbc4b84d0827208000d)
