5_atoms_and_lcao_solutions.md

Solutions for LCAO model exercises

Exercise 1

Question 1.

See lecture notes.

Question 2.

The atomic number of Tungsten is 74:

1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^4

Question 3.

\begin{align} \textrm{Cu} &= [\textrm{Ar}]4s^23d^9\ \textrm{Pd} &= [\textrm{Kr}]5s^24d^8\ \textrm{Ag} &= [\textrm{Kr}]5s^24d^9\ \textrm{Au} &= [\textrm{Xe}]6s^24f^{14}5d^9 \end{align}

Exercise 2

Question 1.

\psi(x) = \begin{cases} &\sqrt{κ}e^{κ(x-x_1)}, x<x_1\\ &\sqrt{κ}e^{-κ(x-x_1)}, x>x_1 \end{cases}

Where κ = \sqrt{\frac{-2mE}{\hbar^2}} = \frac{mV_0}{\hbar^2}. The energy is given by ϵ_1 = ϵ_2 = -\frac{mV_0^2}{2\hbar^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_2(x) can be found by replacing x_1 by x_2

Question 2.

H = -\frac{mV_0^2}{\hbar^2}\begin{pmatrix} 1/2+\exp(-\frac{2mV_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{2mV_0}{\hbar^2}|x_2-x_1|) \end{pmatrix}

Question 3.

ϵ_{\pm} = \beta(1/2+\exp(-2\alpha) \pm \exp(-\alpha))

Where \beta = -\frac{mV_0^2}{\hbar^2} and α = \frac{mV_0}{\hbar^2}|x_2-x_1|

Exercise 3

Question 1.

H_{\mathcal{E}} = ex\mathcal{E},

Question 2.

\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} = \begin{pmatrix} E_0 - \gamma & -t\\ -t & E_0 + \gamma \end{pmatrix}, 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.

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} + \gamma }{t}\alpha Then, using the normalization condition \alpha^2+\beta^2=1, we find the normalized eigenfunction.

%The ground state wave function is: % % |\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} %

Question 4.

P = -\frac{2\gamma^2}{\mathcal{E}}(\frac{1}{\sqrt{\gamma^2+t^2}})