Update 5_atoms_and_lcao_solutions.md - more fixes

docs/5_atoms_and_lcao_solutions.md

@@ -72,19 +72,19 @@ $$
 
 $$
     \hat{H} = \begin{pmatrix}
-    E_0  & -t\\
-    -t & E_0 
+    E_0  & -t \\
+    -t   & E_0 
     \end{pmatrix} +e\mathcal{E}\begin{pmatrix}
-    \langle 1|\hat{x}|1\rangle & \langle 1|\hat{x}|2\rangle\\
+    \langle 1|\hat{x}|1\rangle & \langle 1|\hat{x}|2\rangle \\
     \langle 2|\hat{x}|1\rangle & \langle 2|\hat{x}|2\rangle
     \end{pmatrix} = \begin{pmatrix}
-    E_0 - \gamma & -t\\
-    -t & E_0 + \gamma
+    E_0-\gamma & -t \\
+    -t & E_0+\gamma
     \end{pmatrix},
 $$
 where we defined $\gamma = e d \mathcal{E}/2$ and used 
 $$
-    ⟨1|\hat{x}|1⟩ = -\frac{d}{2}.
+    \langle 1|\hat{x}|1\rangle = -\frac{d}{2}.
 $$
 
 #### Question 3.
@@ -114,5 +114,5 @@ $$
 #### Question 4.
 We find the polarization using 
 $$
-P=2e\langle\psi|\hat{x}|\psi\rangle = 2e\phi_1^2\langle 1|\hat{x}| 1 \rangle + \phi_1^2\langle 2|\hat{x}| 2 \rangle = ed(\phi_2^2-\phi_1^2) =ed\frac{\eta}{1-\eta}
+P=2e\langle\psi|\hat{x}|\psi\rangle = 2e\left(\phi_1^2\langle 1|\hat{x}| 1 \rangle + \phi_1^2\langle 2|\hat{x}| 2 \rangle \right) = ed(\phi_2^2-\phi_1^2) =ed\frac{\eta}{1-\eta}
 $$