# 原子轨道线性组合

（重定向自原子軌域線性組合

LCAO的数学形式为：

${\displaystyle \ \Psi _{i}=\sum _{j}^{n}c_{ji}\varphi _{j}}$

## 基本计算过程

${\displaystyle |\Psi \rangle =\sum \limits _{k}{{c_{k}}\left|{\varphi _{k}}\right\rangle }}$

${\displaystyle \sum \limits _{k}{{c_{k}}{\hat {H}}\left|{\varphi _{k}}\right\rangle }=E\sum \limits _{k}{{c_{k}}\left|{\varphi _{k}}\right\rangle }}$
${\displaystyle \sum \limits _{k}{{c_{k}}\underbrace {\left\langle {\varphi _{i}}\right|{\hat {H}}\left|\varphi _{k}\right\rangle } _{H_{ik}}}=E\sum \limits _{k}{{c_{k}}\underbrace {\left\langle {\varphi _{i}}|{\varphi _{k}}\right\rangle } _{S_{ik}}}}$
${\displaystyle \sum \limits _{k}{{c_{k}}\left({{H_{ik}}-E{S_{ik}}}\right)=0}}$

## 实例：双原子分子

${\displaystyle \Psi ={c_{A}}{\varphi _{A}}+{c_{B}}{\varphi _{B}}}$

${\displaystyle {\hat {H}}\left({{c_{A}}{\varphi _{A}}+{c_{B}}{\varphi _{B}}}\right)=E\left({{c_{A}}{\varphi _{A}}+{c_{B}}{\varphi _{B}}}\right)}$

${\displaystyle \int {d\tau \;\varphi _{A}^{*}{\hat {H}}\left({{c_{A}}{\varphi _{A}}+{c_{B}}{\varphi _{B}}}\right)}=E\int {d\tau \;\left({{c_{A}}\varphi _{A}^{*}{\varphi _{A}}+{c_{B}}\varphi _{A}^{*}{\varphi _{B}}}\right)}}$
${\displaystyle \int {d\tau \;\varphi _{B}^{*}{\hat {H}}\left({{c_{A}}{\varphi _{A}}+{c_{B}}{\varphi _{B}}}\right)}=E\int {d\tau \;\left({{c_{A}}\varphi _{B}^{*}{\varphi _{A}}+{c_{B}}\varphi _{B}^{*}{\varphi _{B}}}\right)}}$

${\displaystyle {c_{A}}\underbrace {\int {d\tau \;\varphi _{A}^{*}{\hat {H}}{\varphi _{A}}}} _{H_{AA}}+{c_{B}}\underbrace {\int {d\tau \;\varphi _{A}^{*}{\hat {H}}{\varphi _{B}}}} _{H_{AB}}=E{c_{A}}\underbrace {\int {d\tau \;\varphi _{A}^{*}{\varphi _{A}}}} _{1}+E{c_{B}}\underbrace {\int {d\tau \;\varphi _{A}^{*}{\varphi _{B}}}} _{S_{AB}}}$
${\displaystyle {c_{A}}\underbrace {\int {d\tau \;\varphi _{B}^{*}{\hat {H}}{\varphi _{A}}}} _{H_{BA}}+{c_{B}}\underbrace {\int {d\tau \;\varphi _{B}^{*}{\hat {H}}{\varphi _{B}}}} _{H_{BB}}=E{c_{A}}\underbrace {\int {d\tau \;\varphi _{B}^{*}{\varphi _{A}}}} _{S_{BA}}+E{c_{B}}\underbrace {\int {d\tau \;\varphi _{B}^{*}{\varphi _{B}}}} _{1}}$

${\displaystyle {c_{A}}\left({{H_{AA}}-E}\right)+{c_{B}}\left({{H_{AB}}-E{S_{AB}}}\right)=0}$
${\displaystyle {c_{A}}\left({{H_{BA}}-E{S_{BA}}}\right)+{c_{B}}\left({{H_{BB}}-E}\right)=0}$

${\displaystyle {\begin{bmatrix}{{H_{AA}}-E}&{{H_{AB}}-E{S_{AB}}}\\{{H_{BA}}-E{S_{BA}}}&{{H_{BB}}-E}\\\end{bmatrix}}{\begin{bmatrix}{c_{A}}\\{c_{B}}\\\end{bmatrix}}=0}$

${\displaystyle {\frac {{H_{AA}}-E}{{H_{BA}}-E{S_{BA}}}}={\frac {{H_{AB}}-E{S_{AB}}}{{H_{BB}}-E}}}$

### 最简单的分子： H${\displaystyle _{2}^{+}}$

H${\displaystyle _{2}^{+}}$ 是由两个质子与一个电子组成的同核双原子分子，是最简单的分子形式。设想H${\displaystyle _{2}^{+}}$ 的分子轨道可以由两个氢原子的基态波函数1s线性叠加而成。此时满足${\displaystyle {H_{AA}}={H_{BB}}=\alpha ,{H_{AB}}={H_{BA}}=\beta ,{S_{AB}}={S_{BA}}=S}$ ，其中α为库仑积分，β为交换积分，S为重叠积分。于是，代入用于求能量的比值式：

${\displaystyle {\frac {\alpha -E}{\beta -ES}}={\frac {\beta -ES}{\alpha -E}}}$

${\displaystyle {E_{+}}={\frac {\alpha +\beta }{1+S}}}$ ，此时有${\displaystyle c_{A}=c_{B}}$
${\displaystyle {E_{-}}={\frac {\alpha -\beta }{1-S}}}$ ，此时有${\displaystyle c_{A}=-c_{B}}$

${\displaystyle {\Psi _{+}}=c\left({{\varphi _{A}}+{\varphi _{B}}}\right)}$
${\displaystyle {\Psi _{-}}=c\left({{\varphi _{A}}-{\varphi _{B}}}\right)}$

c可由归一化条件最终确定。

${\displaystyle {\Psi _{+}}(x,y)=c\left(e^{-{\frac {\sqrt {x^{2}+y^{2}}}{a_{0}}}}+e^{-{\frac {\sqrt {(x-x_{0})^{2}+y^{2}}}{a_{0}}}}\right)}$
${\displaystyle {\Psi _{-}}(x,y)=c\left(e^{-{\frac {\sqrt {x^{2}+y^{2}}}{a_{0}}}}-e^{-{\frac {\sqrt {(x-x_{0})^{2}+y^{2}}}{a_{0}}}}\right)}$

H2+分子的成键轨道${\displaystyle \Psi _{+}(x,y)}$ 的几率分布示意图

H2+分子的反键轨道${\displaystyle \Psi _{-}(x,y)}$ 的几率分布示意图