1.5. 附录#
s1 - 势能面的形式#
全分子波函数为:
\[
\hat{H}^\mathrm{tot} = \sum_{i} -\frac{1}{2}\nabla^2_i + \sum_A -\frac{1}{2M_A}\nabla^2_A - \sum_{iA}\frac{Z_{iA}}{r_{iA}} + \sum_{i>j} \frac{1}{r_{ij}} + \sum_{A>B} \frac{Z_AZ_B}{R_{AB}}
\]
我们希望的核波函数为:
\[
\left[
E^\mathrm{el}(R) - \sum_A \frac{1}{2M_A}\nabla^2_A
\right] \ket{\chi(R)} = E \ket{\chi(R)}
\]
这种假设带来的全波函数为:
\[
\ket{\Psi(R, r)} = \ket{\phi(R;r)}\ket{\chi(R)}
\]
假如有:
\[
E^\mathrm{el}(R) = \braket{
\phi(R;r) | \sum_{i} -\frac{1}{2}\nabla^2_i - \sum_{iA}\frac{Z_{iA}}{r_{iA}} + \sum_{i>j} \frac{1}{r_{ij}} + \sum_{A>B} \frac{Z_AZ_B}{R_{AB}} | \phi(R;r)
}
\]
则:
\[ \begin{align}\begin{aligned}\begin{split}
\begin{split}
& \bigg[
\braket{
\phi(R;r) | \sum_{i} -\frac{1}{2}\nabla^2_i - \sum_{iA}\frac{Z_{iA}}{r_{iA}} + \sum_{i>j} \frac{1}{r_{ij}} + \sum_{A>B} \frac{Z_AZ_B}{R_{AB}} | \phi(R;r)
} \\ &- \sum_A \frac{1}{2M_A}\nabla^2_A
\bigg] \ket{\chi(R)} = E \ket{\chi(R)} \\
& \\
& \bigg[
\braket{
\phi(R;r) | \sum_{i} -\frac{1}{2}\nabla^2_i - \sum_{iA}\frac{Z_{iA}}{r_{iA}} + \sum_{i>j} \frac{1}{r_{ij}} + \sum_{A>B} \frac{Z_AZ_B}{R_{AB}} | \phi(R;r)
} \\ &- \bra{\phi(R;r)}\sum_A \frac{1}{2M_A}\nabla^2_A
\ket{\phi(R;r)} \bigg] \ket{\chi(R)} = \bra{\phi(R;r)} E \ket{\phi(R;r)}\ket{\chi(R)} \\
& \\
& \bigg[
\ket{ \sum_{i} -\frac{1}{2}\nabla^2_i - \sum_{iA}\frac{Z_{iA}}{r_{iA}} + \sum_{i>j} \frac{1}{r_{ij}} + \sum_{A>B} \frac{Z_AZ_B}{R_{AB}} | \phi(R;r)
} \\ &- \sum_A \frac{1}{2M_A}\nabla^2_A
\ket{\phi(R;r)} \bigg] \ket{\chi(R)} = E \ket{\phi(R;r)}\ket{\chi(R)} \\
& \\
& \bigg[
\sum_{i} -\frac{1}{2}\nabla^2_i - \sum_{iA}\frac{Z_{iA}}{r_{iA}} + \sum_{i>j} \frac{1}{r_{ij}} + \sum_{A>B} \frac{Z_AZ_B}{R_{AB}} \\ &- \sum_A \frac{1}{2M_A}\nabla^2_A
\bigg] \ket{\phi(R;r)}\ket{\chi(R)} = E \ket{\phi(R;r)}\ket{\chi(R)} \\
& \\
& \hat{H}^\mathrm{tot} \ket{\phi(R;r)}\ket{\chi(R)} = E \ket{\phi(R;r)}\ket{\chi(R)} \\\end{split}\\\end{split}
\end{aligned}\end{align} \]
s2 - 多电子概率密度#
考察\(r_i=r_1\)这一项的情况:
\[\begin{split}
\begin{split}
n(r_1) =& \braket{\Psi|r=r_1} \braket{r|\Psi} \\
=& \braket{\Psi|\int dr_1 |r1}\braket{r_1|r} \braket{r|\int dr_1 |r_1}\braket{r_1|\Psi} \\
=& \left[
\int dr_1 \braket{\Psi|r1}\delta(r_1-r)
\right]\left[
\int dr_1 \delta(r_1-r)\braket{r_1|\Psi}
\right] \\
=& \left[
\int dr_1 \braket{\Psi|r1}\delta(r_1-r)
\right]
\int dr_2\cdots dr_n \ket{r_2}\bra{r_2}\cdots\ket{r_n}\bra{r_n} \\
& \cdot \left[
\int dr_1 \delta(r_1-r)\braket{r_1|\Psi}
\right] \\
=& \int dr_2\cdots dr_n \left[
\int dr_1 \Psi^*(r_1,\cdots,r_n) \delta(r_1-r)
\right] \\
& \cdot \left[
\int dr_1 \delta(r_1-r)\Psi(r_1,\cdots,r_n)
\right] \\
=& \int dr_2\cdots dr_n \Psi^*(r,\cdots,r_n)\Psi(r,\cdots,r_n)
\end{split}
\end{split}\]
将上式结果逐项代入\(r_i\)，加和即可得到结果。