Solutions for LCAO model exercises

Question 1

1. See lecture notes.
2. The atomic number of Tungsten is 74: $$1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^{6}5s^{2}4d^{10}5p^{6}6s^{2}4f^{14}5d^4$$
3. $$\begin{array}{rl}\text{Cu}& =[\text{Ar}]4s^{2}3d^9\\ \text{Pd}& =[\text{Kr}]5s^{2}4d^8\\ \text{Ag}& =[\text{Kr}]5s^{2}4d^9\\ \text{Au}& =[\text{Xe}]6s^{2}4f^{14}5d^9\end{array}$$

Question 2

1. $$ \psi(x) = \begin{cases} &\sqrt{κ}e^{κ(x-x_1)}, xx_1 \end{cases} $$ Where $\kappa =\sqrt{\frac{-2mE}{{\hslash }^2}}=\frac{m{V}_0}{{\hslash }^2}$. The energy is given by ${ϵ}_1={ϵ}_2=-\frac{m{V}_0^2}{2{\hslash }^2}$ The wave function of a single delta peak is given by $${\psi }_1(x)=\frac{\sqrt{m{V}_0}}{\hslash }{e}^{-\frac{m{V}_0}{{\hslash }^2}|x-x_1|}$$ ${\psi }_2(x)$ can be found by replacing $x_1$ by $x_2$

2. $$H=-\frac{m{V}_0^2}{{\hslash }^2}\left(\begin{array}{cc}1/2+\exp (-\frac{2m{V}_0}{{\hslash }^2}|x_2-x_1|)& \exp (-\frac{m{V}_0}{{\hslash }^2}|x_2-x_1|)\\ \exp (-\frac{m{V}_0}{{\hslash }^2}|x_2-x_1|)& 1/2+\exp (-\frac{2m{V}_0}{{\hslash }^2}|x_2-x_1|)\end{array}\right)$$

3. $${ϵ}_{\pm }=\beta (1/2+\exp (-2\alpha )\pm \exp (-\alpha ))$$ Where $\beta =-\frac{m{V}_0^2}{{\hslash }^2}$ and $\alpha =\frac{m{V}_0}{{\hslash }^2}|x_2-x_1|$

Question 3

1.

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

2. $$\hat{H}=\left(\begin{array}{ccc}{E}_0& -t-t& E_0\end{array}\right)+\left(\begin{array}{ccc}⟨1|ex\mathcal{E}|1⟩& ⟨1|ex\mathcal{E}|2⟩⟨2|ex\mathcal{E}|1⟩& ⟨2|ex\mathcal{E}|2⟩\end{array}\right)=\left(\begin{array}{ccc}{E}_0-\gamma & -t-t& E_0+\gamma \end{array}\right),$$ where $\gamma =ed\mathcal{E}/2$ and have used \($⟨1|ex\mathcal{E}|1⟩=-ed\mathcal{E}/2⟨1|1⟩=-ed\mathcal{E}/2$\)

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: $$\begin{array}{rlr}|\psi ⟩& =\frac{t}{\sqrt{(\gamma +\sqrt{{\gamma }^2+t^2}{)}^2+t^2}}\left(\begin{array}{c}\frac{\gamma +\sqrt{t^2+{\gamma }^2}}{t}1\end{array}\right)|\psi ⟩& =\frac{\gamma +\sqrt{t^2+{\gamma }^2}}{\sqrt{(\gamma +\sqrt{{\gamma }^2+t^2}{)}^2+t^2}}|1⟩+\frac{t}{\sqrt{(\gamma +\sqrt{{\gamma }^2+t^2}{)}^2+t^2}}|2⟩\end{array}$$

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