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Energy expression

We begin with point nuclei with coordinates $ \mathbf{R}$ and momenta $ \mathbf{P}$, and electrons defined by spherical Gaussian wave packets with positions $ \mathbf{x}$, translational momenta $ \mathbf{p_{x}}$, sizes $ s$, and radial momenta $ p_{s}$:

$\displaystyle \Psi \propto \prod_{j} \,\exp\left[-\left(\frac{1}{s^{2}}-\frac{2...
...(\mathbf{r}-\mathbf{x})^{2}\right]\cdot\exp[i \mathbf{p_{x}} \cdot \mathbf{x}].$ (4.1)

Then the overall energy is a sum of the Hartree product kinetic energy, Hartree product electrostatic energy, and antisymmetrization (Pauli) energy:

$\displaystyle E = E_{ke} + E_{nuc \cdot nuc} + E_{nuc \cdot elec} + E_{elec \cdot elec} + E_{Pauli}
$

which can be broken down further as follows:
$\displaystyle E_{ke}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sum_{i}\int \left\vert\nabla \psi_{i}\right\vert^{2} \, dV = \sum_{i} \frac{3}{2} \ \frac{1}{s_{i}^{2}}$  
$\displaystyle E_{nuc \cdot nuc}$ $\displaystyle =$ $\displaystyle \sum_{i<j} \frac{Z_{i} Z_{j}}{R_{ij}}$  
$\displaystyle E_{nuc \cdot elec}$ $\displaystyle =$ $\displaystyle -\sum_{i, j} Z_{i} \int \frac{\vert\psi_{j}\vert^{2}}{R_{ij}} \, ...
...\frac{Z_{i}}{R_{ij}}\ \mathrm{Erf}\left(\frac{\sqrt{2} \, R_{ij}}{s_{i}}\right)$  
$\displaystyle E_{elec \cdot elec}$ $\displaystyle =$ $\displaystyle \sum_{i<j} \int \frac{\vert\psi_{i}\vert^{2}\vert\psi_{j}\vert^{2...
...athrm{Erf}\left( \frac{\sqrt{2} \, x_{ij}}{\sqrt{s_{i}^{2} + s_{j}^{2}}}\right)$  
$\displaystyle E_{Pauli}$ $\displaystyle =$ $\displaystyle \sum_{\sigma_{i} = \sigma_{j}} E(\uparrow \uparrow)_{ij} + \sum_{\sigma_{i} \neq \sigma_{j}}
E(\uparrow \downarrow)_{ij}$  

where $ E(\uparrow \uparrow)$ and $ E(\uparrow \downarrow)$ are the Pauli potential functions:
$\displaystyle E(\uparrow \uparrow)_{ij}$ $\displaystyle =$ $\displaystyle \left(\frac{S_{ij}^{2}}{1-S_{ij}^{2}} + (1 - \rho) \frac{S_{ij}^{2}}{1 + S_{ij}^{2}}\right) \Delta T_{ij}$  
$\displaystyle E(\uparrow \downarrow)_{ij}$ $\displaystyle =$ $\displaystyle \frac{\rho S_{ij}^{2}}{1 + S_{ij}^{2}} \Delta T_{ij}$  

where $ \Delta T$ is a measure of the kinetic energy change upon antisymmetrization, and $ S$ is the overlap between two wave packets:
    $\displaystyle \Delta T_{ij} = \frac{3}{2}\left(\frac{1}{\bar{s}_{1}^{2}}+\frac{...
...2}+\bar{s}_{2}^{2})-2 \bar{x}_{12}^{2})}{(\bar{s}_{1}^{2}+\bar{s}_{2}^{2})^{2}}$  
    $\displaystyle S_{ij} = \left(\frac{2}{\bar{s}_{i}/\bar{s}_{j}+\bar{s}_{j}/\bar{s}_{i}}\right)^{3/2} \exp(-\bar{x}_{ij}^{2}/(\bar{s}_{i}^{2}+\bar{s}_{j}^{2}))$  

where $ \rho = -0.2$, $ \bar{x}_{ij} = x_{ij} \cdot 1.125$, and $ \bar{s}_{i} = s_{i} \cdot 0.9$. We will explain the motivation for the Pauli expression, and the consequences of the combined energy terms in more detail below.


next up previous contents
Next: Bonding comes from balancing Up: General theory of the Previous: General theory of the   Contents
Julius 2008-04-29