fix issue #5

chap1/chap1.tex

@@ -666,13 +666,13 @@ \subsubsection{The Linear Variational Problem}
 \begin{equation}\label{key}
 \begin{aligned}
 H_{11} &= \Braket{1s | \mathscr{H} | 1s} = -\dfrac{1}{2} + F\Braket{1s| r\cos\theta | 1s} = -\dfrac{1}{2}\\
-H_{12} &= H_{21} = \Braket{1s | \mathscr{H} | 2p_z} = 0 + F\Braket{1s| r\cos\theta | 2p_z} = \dfrac{512 \sqrt{2} \pi }{243}F\\
+H_{12} &= H_{21} = \Braket{1s | \mathscr{H} | 2p_z} = 0 + F\Braket{1s| r\cos\theta | 2p_z} = \dfrac{128 \sqrt{2} \pi }{243}F\\
 H_{22} &= \Braket{2p_z | \mathscr{H} | 2p_z} = -\dfrac{1}{8} + F\Braket{2p_z| r\cos\theta | 2p_z} = -\dfrac{1}{8}
 \end{aligned}
 \end{equation}
-Suppose $ \vb{c} = \mqty(\cos p\\ \sin p) $,
+Suppose $ \vb{c} = \mqty(\cos p\\ \sin p) $, with the result of Ex 1.20, we have
 \begin{equation}\label{key}
-p = \dfrac{1}{2}\arctan\dfrac{2H_{12}}{H_{11} - H_{22}} = -\dfrac{1}{2} \tan ^{-1}\qty(\dfrac{2048 \sqrt{2} F}{729})
+p = \dfrac{1}{2}\arctan\dfrac{2H_{12}}{H_{11} - H_{22}} = -\dfrac{1}{2} \arctan \qty(\dfrac{2048 \sqrt{2} F}{729})
 \end{equation}
 thus
 \begin{equation}\label{key}