5_atoms_and_lcao_solutions.md



Solutions for LCAO model exercises

Question 1

See lecture notes.
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
  

  \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}
  

Question 2


\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


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}
  

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|


Question 3


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


\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 have used ⟨1|ex\mathcal{E}|1⟩ = -e d \mathcal{E}/2⟨1|1⟩ = -e d \mathcal{E}/2


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}
\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}


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