small fix in Chap 3.4

Chaps/Chap3.tex

@@ -1161,7 +1161,7 @@ \subsection{闭壳层H-F：限制性自旋轨道}\label{sec3.4.1}
 那么
 \begin{alignat}{3}
 	f(\mathbf{r}_1)\psi_j(\mathbf{r}_1) &= h(\mathbf{r}_1)\psi_j(\mathbf{r}_1) &+ \sum_c\int\dd\omega_1\dd{x}_2\,\alpha^*(\omega_1)\chi_c^*(\mathbf{x}_1)\twoe\chi_c(\mathbf{x}_2)\alpha(\omega_1)\psi_j(\mathbf{r}_1)\notag\\
-	&&-\sum_c\int\dd\omega_1\dd{x}_2\,\alpha^*(\omega_1)\chi_c^*(\mathbf{x}_1)\twoe\chi_c(\mathbf{x}_1)\alpha(\omega_2)\psi_j(\mathbf{r}_2)\notag\\
+	&&-\sum_c\int\dd\omega_1\dd{x}_2\,\alpha^*(\omega_1)\chi_c^*(\mathbf{x}_2)\twoe\chi_c(\mathbf{x}_1)\alpha(\omega_2)\psi_j(\mathbf{r}_2)\notag\\
 	&=\epsilon_j\psi_j(\mathbf{r}_1)&
 \end{alignat}
 式中含$h(\mathbf{r}_1)$的那项中$\dd\omega_1$的积分已经预先积完, 
@@ -1178,13 +1178,13 @@ \subsection{闭壳层H-F：限制性自旋轨道}\label{sec3.4.1}
 	& {}\quad + \sum_c^{N/2}\int\dd\omega_1\dd\omega_2\dd{r}_2\,\alpha^*(\omega_1)\psi_c^*(\mathbf{r}_2)\alpha^*(\omega_2)\twoe\psi_c(\mathbf{r}_2)\alpha(\omega_2)\alpha(\omega_1)\psi_j(\mathbf{r}_1)\notag\\
 	& {}\quad + \sum_c^{N/2}\int\dd\omega_1\dd\omega_2\dd{r}_2\,\,\alpha^*(\omega_1)\psi_c^*(\mathbf{r}_2)\beta^*(\omega_2)\twoe\psi_c(\mathbf{r}_2)\beta(\omega_2)\alpha(\omega_1)\psi_j(\mathbf{r}_1)\notag\\
 	& {}\quad - \sum_c^{N/2}\int\dd\omega_1\dd\omega_2\dd{r}_2\,\,\alpha^*(\omega_1)\psi_c^*(\mathbf{r}_2)\alpha^*(\omega_2)\twoe\psi_c(\mathbf{r}_1)\alpha(\omega_1)\alpha(\omega_2)\psi_j(\mathbf{r}_2)\notag\\
-	& {}\quad - \sum_c^{N/2}\int\dd\omega_1\dd\omega_2\dd{r}_2\,\alpha^*(\omega_1)\psi_c^*(\mathbf{r}_2)\beta^*(\omega_2)\twoe\psi_c(\mathbf{r}_2)\beta(\omega_1)\alpha(\omega_2)\psi_j(\mathbf{r}_1)\notag\\
+	& {}\quad - \sum_c^{N/2}\int\dd\omega_1\dd\omega_2\dd{r}_2\,\alpha^*(\omega_1)\psi_c^*(\mathbf{r}_2)\beta^*(\omega_2)\twoe\psi_c(\mathbf{r}_1)\beta(\omega_1)\alpha(\omega_2)\psi_j(\mathbf{r}_1)\notag\\
 	& = \epsilon_j\psi_j(\mathbf{r}_1)
 	\label{3.120}
 \end{align}
 现在可以进行对$\dd\omega_1$和$\dd\omega_2$的积分. 
 \autoref{3.120}中最后一项根据自旋正交性为零. 
-这反映自旋平行电子之间没有交换作用的事实. 
+这反映自旋反平行电子之间没有交换作用的事实. 
 两个库伦项是相等的, 
 那么可得
 \begin{alignat}{3}