Solutions to lecture 5: LCAO model

Solutions to the LCAO model

src/5_atoms_and_lcao_solutions.md

@@ -20,69 +20,63 @@
 1. $$
     \psi(x) =
     \begin{cases}
-        &\sqrt{\kappa}e^{\kappa(x-x_1)}, x<x_1\\
-        &\sqrt{\kappa}e^{-\kappa(x-x_1)}, x>x_1
+        &\sqrt{κ}e^{κ(x-x_1)}, x<x_1\\
+        &\sqrt{κ}e^{-κ(x-x_1)}, x>x_1
     \end{cases}
 $$
 
-  Where $\kappa = \sqrt{\frac{-2mE}{\hbar^2}} = \frac{mV_0}{\hbar^2}$.
+  Where $κ = \sqrt{\frac{-2mE}{ħ^2}} = \frac{mV_0}{ħ^2}$.
 
-  The energy is given by $\epsilon_1 = \epsilon_2 = -\frac{mV_0}{\hbar^2}$
+  The energy is given by $ϵ_1 = ϵ_2 = -\frac{mV_0}{ħ^2}$
 
   The wave function of a single delta peak is given by
 
   $$
-      \psi_1(x) = \frac{\sqrt{mV_0}}{\hbar}e^{-\frac{mV_0}{\hbar^2}|x-x_1|}
+      \psi_1(x) = \frac{\sqrt{mV_0}}{ħ}e^{-\frac{mV_0}{ħ^2}|x-x_1|}
   $$
 
   $\psi_2(x)$ can be found by replacing $x_1$ by $x_2$
 
 2. $$
-    H = -\frac{mV_0^2}{\hbar^2}\begin{pmatrix}
-        1/2+\exp(-\frac{mV_0}{\hbar^2}(x_2-x_1)) &
-        \exp(\frac{mV_0}{\hbar^2}(x_2-x_1))\\
-        \exp(-\frac{mV_0}{\hbar^2}(x_2-x_1)) &
-        1/2+\exp(+\frac{mV_0}{\hbar^2}(x_2-x_1))
+    H = -\frac{mV_0^2}{ħ^2}\begin{pmatrix}
+        1/2+\exp(-\frac{mV_0}{ħ^2}(x_2-x_1)) &
+        \exp(\frac{mV_0}{ħ^2}(x_2-x_1))\\
+        \exp(-\frac{mV_0}{ħ^2}(x_2-x_1)) &
+        1/2+\exp(+\frac{mV_0}{ħ^2}(x_2-x_1))
     \end{pmatrix}
   $$
 
 3. $$
-    \epsilon_{\pm} = \beta(3/2+\cosh{2\alpha}+2\cosh{\alpha}\pm \cosh{\alpha})
+    ϵ_{\pm} = \beta(3/2+\cosh{2α}+2\cosh{α}\pm \cosh{α})
    $$
 
-   Where $\beta = -\frac{mV_0^2}{\hbar^2}$ and $\alpha = \frac{mV_0}{\hbar^2}(x_2-x_1)$
+   Where $\beta = -\frac{mV_0^2}{ħ^2}$ and $α = \frac{mV_0}{ħ^2}(x_2-x_1)$
 
 ### Question 3
 
 1.
 
 $$
-    H_{\mathcal{E}} = eRE
+    H_{\mathcal{E}} = eR\mathcal{E},
 $$
-
-Where R is the distance between the negatively charged electrons and the positive charged nuclei.
+where R is the distance between the negatively charged electrons and the positive charged nuclei.
 
 2.
-
 $$
     H_{eff} = \begin{pmatrix}
     E_0 - \gamma & -t\\
     -t & E_0 + \gamma
     \end{pmatrix}
 $$
-
 Where $\gamma = e d \mathcal{E}/2$ and where we have used that $$⟨1|H_{eff}|1⟩ = -e d \mathcal{E}/2⟨1|1⟩ = e d \mathcal{E}/2$$
 
 3.
 
 The eigenstates of the Hamiltonian are given by:
-
 $$
     E_{\pm} = E_0\pm\sqrt{t^2+\gamma^2}
 $$
-
 The ground state wave function is:
-
 $$
     \begin{split}
         |\psi⟩ &= \frac{t}{\sqrt{(\gamma+\sqrt{\gamma^2+t^2})^2+t^2}}\begin{pmatrix}