Template:!: Difference between revisions
(New page: Group work of Team 1 '''Questions''' (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...) |
No edit summary |
||
| Line 7: | Line 7: | ||
:<math> \hat{R}_{\Delta \phi} \equiv \exp \left( \frac{i \Delta \phi \hat{L}_z}{\hbar} \right)</math> | :<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(\pi) \!<\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 | when acting on the function <math> f(\pi) \!<\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> \hat{R}_{\Delta\phi} f(\phi) = f \left( \phi + \Delta \phi \right) </math>. | ||
(b) Show that the operator | (b) Show that the operator | ||
| Line 15: | Line 15: | ||
:<math> \hat{R}_{\Delta\phi} = \exp \left( \frac{i \Delta \mathbf{\phi} \cdot \mathbf{\hat{L}}}{\hbar} \right) | :<math> \hat{R}_{\Delta\phi} = \exp \left( \frac{i \Delta \mathbf{\phi} \cdot \mathbf{\hat{L}}}{\hbar} \right) | ||
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 \mathbf{\phi} \! </math>, we may write | 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 \mathbf{\phi} \!</math>, we may write | ||
:<math> \hat{R}_{\delta \mathbf{\phi}} f(\mathbf{r}) = f\left( \mathbf{r} + \delta \mathbf{r} \right) </math> | :<math> \hat{R}_{\delta \mathbf{\phi}} f(\mathbf{r}) = f\left( \mathbf{r} + \delta \mathbf{r} \right) </math> | ||
| Line 27: | Line 27: | ||
(a) | (a) | ||
:<math> \hat{R}_{\Delta \phi} f = \left[ \exp \left( \Delta \phi \frac{\partial}{\partial \phi} \right) \right] f </math> | :<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> = 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 \gt 1 \! </math>. For the infinitesimal rotation | (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 \gt 1 \! </math>. For the infinitesimal rotation | ||
:<math> \mathbf{r}' = \mathbf{r} + \delta \mathbf{r} = \mathbf{r} + \delta \mathbf{\phi} \times \mathbf{r} </math> | :<math> \mathbf{r}' = \mathbf{r} + \delta \mathbf{r} = \mathbf{r} + \delta \mathbf{\phi} \times \mathbf{r} </math> | ||
| Line 40: | Line 40: | ||
:<math> = f(\mathbf{r})+ \frac{i}{\hbar}\delta \mathbf{\phi} \cdot \mathbf{\hat{L}} f(\mathbf{r}) </math>. | :<math> = f(\mathbf{r})+ \frac{i}{\hbar}\delta \mathbf{\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> \cal{O} \left(\delta \phi \right) \! </math> . [ The expression <math> \delta \mathbf{r} = \delta \mathbf{\phi} \times \mathbf{r} \! </math> is valid only to terms of <math> \cal{O} \left(\delta \phi \right) \! </math>.] In this manner we obtain | In the Taylor series expansion of <math> f\left( \mathbf{r}+\delta\mathbf{r} \right) \!</math> above we have only kept terms of <math> \cal{O} \left(\delta \phi \right) \!</math> . [ The expression <math> \delta \mathbf{r} = \delta \mathbf{\phi} \times \mathbf{r} \!</math> is valid only to terms of <math> \cal{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 \mathbf{\phi} \cdot \mathbf{\hat{L}} \right) f(\mathbf{r}) = \hat{R}_{\delta\mathbf{\phi}} f(\mathbf{r}) </math> | :<math> f\left( \mathbf{r} + \delta \mathbf{r} \right) = \left( \hat{I} + \frac{i}{\hbar} \delta \mathbf{\phi} \cdot \mathbf{\hat{L}} \right) f(\mathbf{r}) = \hat{R}_{\delta\mathbf{\phi}} f(\mathbf{r}) </math> | ||
For a finite rotational displacement through the angle <math> \mathbf{\Delta} \mathbf{\phi} = n \delta \mathbf{\phi} \! </math>, we apply the operator <math> \hat{R}_{\delta \mathbf{\phi}} \! </math>, <math> n \! </math> times: | For a finite rotational displacement through the angle <math> \mathbf{\Delta} \mathbf{\phi} = n \delta \mathbf{\phi} \!</math>, we apply the operator <math> \hat{R}_{\delta \mathbf{\phi}} \!</math>, <math> n \!</math> times: | ||
:<math> \hat{R}_{n\delta\mathbf{\phi}} = \left( \hat{R}_{\delta\mathbf{\phi}} \right)^n = \left( \hat{I} + \frac{i}{\hbar} \delta \mathbf{\phi} \cdot \mathbf{\hat{L}} \right)^n </math> | :<math> \hat{R}_{n\delta\mathbf{\phi}} = \left( \hat{R}_{\delta\mathbf{\phi}} \right)^n = \left( \hat{I} + \frac{i}{\hbar} \delta \mathbf{\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>. | 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 \mathbf{\phi} \cdot \mathbf{\hat{L}} \right)^{\Delta \phi / \delta \phi} = e^{i \Delta \mathbf{\phi} \cdot \mathbf{\hat{L}} \hbar </math>. | :<math> \hat{R}_{\Delta\phi} = \lim_{\Delta \phi / \delta \phi \rightarrow \infty} \left( \hat{I} + \frac{i}{\hbar} \delta \mathbf{\phi} \cdot \mathbf{\hat{L}} \right)^{\Delta \phi / \delta \phi} = e^{i \Delta \mathbf{\phi} \cdot \mathbf{\hat{L}} \hbar </math>. | ||
The operator <math> \hat{R}_{\delta\mathbf{\phi}} \! </math> rotates <math> \mathbf{r} \! </math> to <math> \mathbf{r} + \delta\mathbf{\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 \mathbf{\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 \mathbf{\phi} \times \mathbf{r} \! </math>. Thus, rotation of coordinates through <math> \delta \mathbf{\phi} \! </math> is generated by the operator <math> \hat{R}_{-\delta \mathbf{\phi}}. | The operator <math> \hat{R}_{\delta\mathbf{\phi}} \!</math> rotates <math> \mathbf{r} \!</math> to <math> \mathbf{r} + \delta\mathbf{\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 \mathbf{\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 \mathbf{\phi} \times \mathbf{r} \!</math>. Thus, rotation of coordinates through <math> \delta \mathbf{\phi} \!</math> is generated by the operator <math> \hat{R}_{-\delta \mathbf{\phi}}. | ||
-------- | -------- | ||
(Note: This problem is excerpted from {\it "Introductory Quantum Mechanics", 2nd edition, p377-p379 which is written by Richard L. Liboff.) | (Note: This problem is excerpted from {\it "Introductory Quantum Mechanics", 2nd edition, p377-p379 which is written by Richard L. Liboff.) | ||
Revision as of 16:31, 29 November 2009
Group work of Team 1
Questions
(a) Show that the operator
- 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 \hat{R}_{\Delta \phi} \equiv \exp \left( \frac{i \Delta \phi \hat{L}_z}{\hbar} \right)}
when acting on the function 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 f(\pi) \!<\math> changes <math> f \!} by a rotation of coordinates about 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 so that the radius through 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 \phi \!} is rotated to the radius through . That is, show 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 \hat{R}_{\Delta\phi} f(\phi) = f \left( \phi + \Delta \phi \right) } .
(b) Show that the operator
- 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 \hat{R}_{\Delta\phi} = \exp \left( \frac{i \Delta \mathbf{\phi} \cdot \mathbf{\hat{L}}}{\hbar} \right) when action on <math> f(\mathbf{r}) \!} changes 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 f \!} by rotating 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{r} \!} to a new value on the surface of the sphere of radius 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 r \!} , but rotated away from 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{r} \!} through the azimuth 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 \Delta \phi \!} , 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 \mathbf{r} \left( \theta, \phi \right) \rightarrow \mathbf{r}' = \mathbf{r} \left( \theta, \phi + \Delta \phi \right) \!} . For infinitesimal displacement 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 \delta \mathbf{\phi} \!} , we may write
- 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 \hat{R}_{\delta \mathbf{\phi}} f(\mathbf{r}) = f\left( \mathbf{r} + \delta \mathbf{r} \right) }
- 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 \delta \mathbf{r} = \delta \mathbf{\phi} \times \mathbf{r}. }
Solutions
(a)
- 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 \hat{R}_{\Delta \phi} f = \left[ \exp \left( \Delta \phi \frac{\partial}{\partial \phi} \right) \right] f }
- 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 = 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).}
(b) Let 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 \delta \phi \!} be an infinitesimal angle 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 \Delta \phi = n \delta \phi \!} in the limit 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 n \gt 1 \! } . For the infinitesimal rotation
- 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{r}' = \mathbf{r} + \delta \mathbf{r} = \mathbf{r} + \delta \mathbf{\phi} \times \mathbf{r} }
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 f \left( \mathbf {r} + \delta \mathbf{r} \right) = f(\mathbf{r})+ \delta \mathbf{\phi} \times \mathbf{r} \cdot \mathbf{\nabla} f(\mathbf{r}) }
- 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 = f(\mathbf{r})+ \delta \mathbf{\phi} \cdot \mathbf{r} \times \mathbf{\nabla} f(\mathbf{r}) }
- 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 = f(\mathbf{r})+ \frac{i}{\hbar} \delta \mathbf{\phi} \cdot \mathbf{r} \cdot \mathbf{\hat{p}} f(\mathbf{r}) }
- 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 = f(\mathbf{r})+ \frac{i}{\hbar}\delta \mathbf{\phi} \cdot \mathbf{\hat{L}} f(\mathbf{r}) } .
In the Taylor series expansion of 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 f\left( \mathbf{r}+\delta\mathbf{r} \right) \!} above we have only kept terms of 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 \cal{O} \left(\delta \phi \right) \!} . [ The expression 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 \delta \mathbf{r} = \delta \mathbf{\phi} \times \mathbf{r} \!} is valid only to terms of 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 \cal{O} \left(\delta \phi \right) \!} .] In this manner 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 f\left( \mathbf{r} + \delta \mathbf{r} \right) = \left( \hat{I} + \frac{i}{\hbar} \delta \mathbf{\phi} \cdot \mathbf{\hat{L}} \right) f(\mathbf{r}) = \hat{R}_{\delta\mathbf{\phi}} f(\mathbf{r}) }
For a finite rotational displacement through 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 \mathbf{\Delta} \mathbf{\phi} = n \delta \mathbf{\phi} \!} , we apply the operator 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 \hat{R}_{\delta \mathbf{\phi}} \!} , 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 n \!} times:
- 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 \hat{R}_{n\delta\mathbf{\phi}} = \left( \hat{R}_{\delta\mathbf{\phi}} \right)^n = \left( \hat{I} + \frac{i}{\hbar} \delta \mathbf{\phi} \cdot \mathbf{\hat{L}} \right)^n }
and pss to the limit 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 n \rightarrow \infty \!} or, equivalently, 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 \Delta \phi / \delta \phi \rightarrow \infty \!} .
- 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 \hat{R}_{\Delta\phi} = \lim_{\Delta \phi / \delta \phi \rightarrow \infty} \left( \hat{I} + \frac{i}{\hbar} \delta \mathbf{\phi} \cdot \mathbf{\hat{L}} \right)^{\Delta \phi / \delta \phi} = e^{i \Delta \mathbf{\phi} \cdot \mathbf{\hat{L}} \hbar } .
The operator 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 \hat{R}_{\delta\mathbf{\phi}} \!} rotates 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{r} \!} to 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{r} + \delta\mathbf{\phi}\times\mathbf{r} \!} with respect to a fixed coordinate frame. If, on the other hand, the coordinate frame is rotated through 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 \delta \mathbf{\phi} \!} 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 \mathbf{r} \!} fixed in space, then in the new coordinate frame this vector has the 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 \mathbf{r} - \delta \mathbf{\phi} \times \mathbf{r} \!} . Thus, rotation of coordinates through 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 \delta \mathbf{\phi} \!} is generated by the operator <math> \hat{R}_{-\delta \mathbf{\phi}}.
(Note: This problem is excerpted from {\it "Introductory Quantum Mechanics", 2nd edition, p377-p379 which is written by Richard L. Liboff.)