Update 5_atoms_and_lcao_solutions.md - typo fix

docs/5_atoms_and_lcao_solutions.md

@@ -82,7 +82,10 @@ $$
     -t & E_0 + \gamma
     \end{pmatrix},
 $$
-where $\gamma = e d \mathcal{E}/2$ and have used $$⟨1|ex\mathcal{E}|1⟩ = -e d \mathcal{E}/2⟨1|1⟩ = -e d \mathcal{E}/2$$
+where $\gamma = e d \mathcal{E}/2$ and we have used 
+$$
+    ⟨1|ex\mathcal{E}|1⟩ = -e d \mathcal{E}/2⟨1|1⟩ = -e d \mathcal{E}/2
+$$
 
 #### Question 3.
 
@@ -90,13 +93,18 @@ The eigenstates of the Hamiltonian are given by:
 $$
     E_{\pm} = E_0\pm\sqrt{t^2+\gamma^2}
 $$
+Calling the elements of the eigenvector $\alpha$ and $\beta$, we find
+$$
+\alpha(E_0-\gamma)-\beta t = \alpha E_\pm
+$$
+From this we find 
+$$
+beta = -\frac{E_\pm- E_0 + \gamma}{t}\alpha = -\frac{\pm\sqrt{t^2+\gamma^2}}{t}
+$$
+Then, using the normalization condition $\alpha^2+\beta^2$=1, we find the normalized eigenfunction.
+
 The ground state wave function is:
 $$
-    \begin{split}
-        |\psi⟩ &= \frac{t}{\sqrt{(\gamma+\sqrt{\gamma^2+t^2})^2+t^2}}\begin{pmatrix}
-        \frac{\gamma+\sqrt{t^2+\gamma^2}}{t}\\
-        1
-        \end{pmatrix}\\
     |\psi⟩ &= \frac{\gamma+\sqrt{t^2+\gamma^2}}{\sqrt{(\gamma+\sqrt{\gamma^2+t^2})^2+t^2}}|1⟩+\frac{t}{\sqrt{(\gamma+\sqrt{\gamma^2+t^2})^2+t^2}}|2⟩
     \end{split}
 $$