# @surface

## Contents

### Keep Out

ab_initio:交換相関ポテンシャル

# 交換相関ポテンシャル

$$\Psi(q_1, q_2)=\frac{1}{\sqrt{2}}\left(\phi_a(q_1)\phi_b(q_2)-\phi_a(q_2)\phi_b(q_1)\right)$$

$$q_1 = (\vec{r_1}, \sigma_1)=(x_1, y_1, z_1, \sigma_1)$$

$$V(\vec{r_1}, \vec{r_2})=\displaystyle\frac{e^2}{|\vec{r_1}- \vec{r_2}|}$$

$$\langle\Psi|V|\Psi\rangle$$ $$=\frac{1}{2}$$

$$\phi_a(q_1)\phi_b(q_2)-\phi_a(q_2)\phi_b(q_1) |V| \phi_a(q_1)\phi_b(q_2)-\phi_a(q_2)\phi_b(q_1) \rangle$$

$$=\frac{1}{2}\langle \phi_a(q_1)\phi_b(q_2) |V| \phi_a(q_1)\phi_b(q_2) \rangle$$

$$-\frac{1}{2}\langle \phi_a(q_1)\phi_b(q_2) |V| \phi_a(q_2)\phi_b(q_1) \rangle$$ $$-\frac{1}{2}\langle \phi_a(q_2)\phi_b(q_1) |V| \phi_a(q_1)\phi_b(q_2) \rangle$$ $$+\frac{1}{2}\langle \phi_a(q_2)\phi_b(q_1) |V| \phi_a(q_2)\phi_b(q_1) \rangle$$

$$=\frac{1}{2}\langle \phi_a(q)\phi_b(q') |V| \phi_a(q)\phi_b(q') \rangle$$ $$-\frac{1}{2}\langle \phi_a(q)\phi_b(q') |V| \phi_a(q')\phi_b(q) \rangle$$ $$-\frac{1}{2}\langle \phi_a(q)\phi_b(q') |V| \phi_a(q')\phi_b(q) \rangle$$ $$+\frac{1}{2}\langle \phi_a(q)\phi_b(q') |V| \phi_a(q)\phi_b(q') \rangle$$

ab_initio/交換相関ポテンシャル.txt · 最終更新: 2020/08/17 15:43 by kimi