Solutions to the LCAO model

src/5_LCAO_model_solutions.md

+# Solutions for LCAO model exercises
+
+### Question 1
+
+1. See lecture notes\newline
+
+2. The atomic number of Tungsten is 74:
+
+\begin{equation*}
+	1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^4	
+\end{equation*}
+
+3.
+
+\begin{equation*}
+    \begin{split}
+        Cu &= [Ar]4s^23d^9\\
+		Pd &= [Kr]5s^24d^8\\
+		Ag &= [Kr]5s^24d^9\\
+		Au &= [Xe]6s^24f^145d^9
+    \end{split}
+\end{equation*}
+
+## Question 2
+1. 
+
+\begin{equation*}
+    \psi(x) = 
+    \begin{cases}
+        &\sqrt{\kappa}e^{\kappa(x-x_1)}, x<x_1\\
+        &\sqrt{\kappa}e^{-\kappa(x-x_1)}, x>x_1
+    \end{cases}
+\end{equation*}
+
+Where $\kappa = \sqrt{\frac{-2mE}{\hbar^2}} = \frac{mV_0}{\hbar^2}$.
+
+The energy is given by $\epsilon1 = \epsilon2 = -\frac{mV_0}{\hbar^2}$
+
+The wavefunction of a single delta peak is given by
+\begin{equation*}
+    \psi_1(x) = \frac{\sqrt{mV_0}}{\hbar}e^{-\frac{mV_0}{\hbar^2}|x-x_1|}
+\end{equation*}
+$\psi_2(x)$ can be found by replacing $x_1$ by $x_2$
+
+2. 
+
+\begin{equation*}
+    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))
+    \end{pmatrix}
+\end{equation*}
+
+3. 
+
+\begin{equation*}
+    \epsilon_{\pm} = \beta(3/2+\cosh{2\alpha}+2\cosh{\alpha}\pm \cosh{\alpha})
+\end{equation*}
+
+Where $\beta = -\frac{mV_0^2}{\hbar^2}$ and $\alpha = \frac{mV_0}{\hbar^2}(x_2-x_1)$
+
+### Question 3
+1.
+
+\begin{equation*}
+    H_{\mathcal{E}} = eRE
+\end{equation*}
+
+Where R is the distance between the negatively charged electrons and the positive charged nuclei.
+
+2.
+
+\begin{equation*}
+    H_{eff} = \begin{pmatrix}
+    E_0 - \gamma & -t\\
+    -t & E_0 + \gamma
+    \end{pmatrix}
+\end{equation*}
+
+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:
+
+\begin{equation*}
+    E_{\pm} = E_0\pm\sqrt{t^2+\gamma^2}
+\end{equation*}
+
+The wavefunction corresponding to the ground state energie is:
+
+\begin{equation*}
+    \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}
+\end{equation*}
+
+4.
+\begin{equation*}
+    P = -\frac{2\gamma^2}{\mathcal{E}}(\frac{1}{\sqrt{\gamma^2+t^2}})
+\end{equation*}
+