Difference between revisions of "Theory Notes"
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| − | < | + | ==Helium Calculations== |
| + | <math> [-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} - | ||
| + | \frac{Ze^2}{r_2}+\frac{e^2}{r^2_{12}} ]\psi = E\psi\nonumber | ||
| + | </math> | ||
| + | |||
| + | Define <math>\rho = \frac{Zr}{a_0}</math> where <math>a_0 = \frac{\hbar^2}{me^2}</math> (Bohr | ||
| + | radius). Then | ||
| + | |||
| + | <math>[-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) | ||
| + | - Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 + | ||
| + | \frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber</math> | ||
| + | |||
| + | But <math>\frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0}</math> is in | ||
| + | atomic units (au) of energy. Therefore | ||
| + | |||
| + | <math>[-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) - | ||
| + | \frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}]\psi = | ||
| + | \varepsilon\psi\nonumber</math> | ||
| + | where <math>\varepsilon = \frac{Ea_0}{Z^2e^2}</math> | ||
| + | |||
| + | The problem to be solved is thus | ||
| + | <math>[\frac{1}{2}(\nabla^2_1+\nabla^2_2) - | ||
| + | \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi = | ||
| + | \varepsilon\psi\nonumber</math> | ||
| + | |||
| + | [figure to be inserted] | ||
Revision as of 04:57, 29 November 2010
Helium Calculations
\( [-\frac{\hbar^2}{2m}(\nabla^2_1 +\nabla^2_2) - \frac{Ze^2}{r_1} - \frac{Ze^2}{r_2}+\frac{e^2}{r^2_{12}} ]\psi = E\psi\nonumber \)
Define \(\rho = \frac{Zr}{a_0}\) where \(a_0 = \frac{\hbar^2}{me^2}\) (Bohr radius). Then
\([-\frac{\hbar^2}{2m}Z^2(\frac{me^2}{\hbar^2})^2(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) - Z^2\frac{e^2}{a_0}\rho^{-1}_1 - Z^2\frac{e^2}{a_0}\rho^{-1}_2 + \frac{e^2}{a_0}Z\rho^{-1}_{12}]\psi= E\psi\nonumber\)
But \(\frac{\hbar^2}{m}(\frac{me^2}{\hbar^2})^2 = \frac{e^2}{a_0}\) is in atomic units (au) of energy. Therefore
\([-\frac{1}{2}(\nabla^2_{\rho_1}+\nabla^2_{\rho_2}) - \frac{1}{\rho_1} - \frac{1}{\rho_2} + \frac{Z^{-1}}{\rho_{12}}]\psi = \varepsilon\psi\nonumber\) where \(\varepsilon = \frac{Ea_0}{Z^2e^2}\)
The problem to be solved is thus \([\frac{1}{2}(\nabla^2_1+\nabla^2_2) - \frac{1}{r_1}-\frac{1}{r_2} + \frac{Z^{-1}}{r_{12}}]\psi = \varepsilon\psi\nonumber\)
[figure to be inserted]
