Update 5_atoms_and_lcao_solutions.md - more fixes

docs/5_atoms_and_lcao_solutions.md

@@ -65,7 +65,7 @@ Where $\beta = -\frac{mV_0^2}{\hbar^2}$ and $α = \frac{mV_0}{\hbar^2}|x_2-x_1|$
 #### Question 1.
 
 $$
-    H_{\mathcal{E}} = ex\mathcal{E},
+    \hat{H}_{\mathcal{E}} = e\mathcal{E}\hat{x},
 $$
 
 #### Question 2.
@@ -74,17 +74,17 @@ $$
     \hat{H} = \begin{pmatrix}
     E_0  & -t\\
     -t & E_0 
-    \end{pmatrix} +\begin{pmatrix}
-    ⟨1|ex\mathcal{E}|1⟩ & ⟨1|ex\mathcal{E}|2⟩\\
-    ⟨2|ex\mathcal{E}|1⟩ & ⟨2|ex\mathcal{E}|2⟩
+    \end{pmatrix} +e\mathcal{E}\begin{pmatrix}
+    \langle 1|\hat{x}|1\rangle & \langle 1|\hat{x}|2\rangle\\
+    \langle 2|\hat{x}|1\rangle & \langle 2|\hat{x}|2\rangle
     \end{pmatrix} = \begin{pmatrix}
     E_0 - \gamma & -t\\
     -t & E_0 + \gamma
     \end{pmatrix},
 $$
-where $\gamma = e d \mathcal{E}/2$ and we have used 
+where we defined $\gamma = e d \mathcal{E}/2$ and used 
 $$
-    ⟨1|ex\mathcal{E}|1⟩ = -e d \mathcal{E}/2⟨1|1⟩ = -e d \mathcal{E}/2
+    ⟨1|\hat{x}|1⟩ = -\frac{d}{2}.
 $$
 
 #### Question 3.