Chemisty 101

Exercises

Preliminary provocations

1. Is the assumption that the atomic orbitals are orthogonal always a reasonable assumption?
2. What happens if the hopping $t$ is chosen to be negative?
3. How does the size of the Hamiltonian matrix change with the number of atoms?
4. How does the size of the Hamiltonian matrix change if each atom now has two orbitals?
5. Assuming that we have two atoms with a single orbital each, what is the size of the Hamiltonian matrix if we also consider the spin of the electron?

Exercise 1: Shell-filling model of atoms

1. Describe the shell-filling model of atoms.

2. Use Madelung’s rule to deduce the atomic shellfilling configuration of the element tungsten, which has atomic number 74.

   The atomic number of Tungsten is 74: $$1\textrm{s}^2 2\textrm{s}^2 2\textrm{p}^6 3\textrm{s}^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^{14}5d^4$$

3. Although Madelung’s rule for the filling of electronic shells holds extremely well, there are a number of exceptions to the rule. Here are a few of them: $$\textrm{Cu} = [\textrm{Ar}] 4\textrm{s}^1 3\textrm{d}^{10}$$ $$\textrm{Pd} = [\textrm{Kr}] 5\textrm{s}^0 4\textrm{d}^{10}$$ $$\textrm{Ag} = [\textrm{Kr}] 5\textrm{s}^1 4\textrm{d}^{10}$$ $$\textrm{Au} = [\textrm{Xe}] 6\textrm{s}^1 4\textrm{f}^{14} 5\textrm{d}^{10}$$ What should the electron configurations be if these elements followed Madelung’s rule and the Aufbau principle? What could be the reason for the deficiency of Madelung's rule?

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

   The wheels fall off as $V(\mathbf{r}) \ne 1/|r|$

Exercise 2: Application of the LCAO model to the delta-function potential

Consider an electron moving in 1D between two negative delta-function shaped potential wells. The complete Hamiltonian of this one-dimensional system is then: $$\hat{H} = \frac{\hat{p}^2}{2m}-V_0\delta(x_1-x)-V_0\delta(x_2-x),$$ where $V_0>0$ is the potential strength, $\hat{p}$ the momentum of the electron, and $x_1$, $x_2$ the positions of the potential wells.

Properties of a single $\delta$-function potential

A delta function $\delta(x_0 - x)$ centered at $x_0$ is defined to be zero everywhere except for $x_0$, where it tends to infinity. Further, a delta function has the property: $$\int_{-\infty}^{\infty} f(x)\delta(x_0-x)dx = f(x_0).$$

The procedure to find the energy and a wave function of a state bound in a $\delta$-function potential, $V=-V_0\delta(x-x_0)$, is similar to that of a quantum well:

1. Assume that we have a bound state with energy $E<0$.
2. Compute the wave function $\phi$ in different regions of space: namely $x < x_0$ and $x > x_0$.
3. Apply the boundary conditions at $x = x_0$. The wave function $\phi$ must be continuous, but $d\phi/dx$ is not. Instead due to the presence of the delta-function: $$\frac{d\phi}{dx}\Bigr|_{x_0+\epsilon} - \frac{d\phi}{dx}\Bigr|_{x_0-\epsilon}= -\frac{2mV_0}{\hbar^2}\phi(x_0).$$
4. Find at which energy the boundary conditions at $x = x_0$ are satisfied. This is the energy of the bound state.
5. Normalize the wave function.

Let us apply the LCAO model to solve this problem. Consider the trial wave function for the ground state to be a linear combination of two orbitals $|1\rangle$ and $|2\rangle$: $$|\psi\rangle = \phi_1|1\rangle + \phi_2|2\rangle.$$ The orbitals $|1\rangle$ and $|2\rangle$ correspond to the wave functions of the electron when there is only a single delta peak present:

$$H_1 |1\rangle = \epsilon_1 |1\rangle,$$ $$H_2 |2\rangle = \epsilon_2 |2\rangle.$$

We start of by calculating the wavefunction of an electron bound to a single delta-peak. To do so, you first need to set up the Schrödinger equation of a single electron bound to a single delta-peak. You do not have to solve the Schrödinger equation twice—you can use the symmetry of the system to calculate the wavefunction of the other electron bound to the second delta-peak.

1. Find the expressions for the wave functions of the states $|1\rangle$ and $|2\rangle$: $\psi_1(x)$ and $\psi_2(x)$. Also find an expression for their energies $\epsilon_1$ and $\epsilon_2$. Remember that you need to normalize the wave functions.

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

   Where $κ = \sqrt{\frac{-2mE}{ħ^2}} = \frac{mV_0}{ħ^2}$. The energy is given by $ϵ_1 = ϵ_2 = -\frac{mV_0^2}{2ħ^2}$. The wave function of a single delta peak is given by

   $$\psi_1(x) = \frac{\sqrt{mV_0}}{ħ}e^{-\frac{mV_0}{ħ^2}|x-x_1|}$$ $\psi_2(x)$ can be found by replacing $x_1$ by $x_2$

2. Construct the LCAO Hamiltonian. To simplify the calculations, assume that the orbitals are orthogonal.

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

3. Diagonalize the LCAO Hamiltonian and find an expression for the eigenenergies of the system. It was previously mentioned that $V_0>0$. Using this, determine which energy corresponds to the bonding energy.

   $$ϵ_{\pm} = \beta(1/2+\exp(-2\alpha) \pm \exp(-\alpha))$$ Where $\beta = -\frac{mV_0^2}{ħ^2}$ and $α = \frac{mV_0}{ħ^2}|x_2-x_1|$

Exercise 3: Polarisation of a hydrogen molecule

Consider a hydrogen molecule as a one-dimensional system with two identical nuclei at $x=-\frac{d}{2}$ and $x=+\frac{d}{2}$, so that the center of the molecule is at $x=0$. Each atom contains a single electron with charge $-e$. The LCAO Hamiltonian of the system is given by

$$H_{\textrm{eff}} = \begin{pmatrix} E_0&&-t \\ -t&&E_0 \end{pmatrix}.$$

The electric potential is given by

$$V_{\mathbf{E}}=-\int_{C} \mathbf{E} \cdot \mathrm{d} \boldsymbol{\ell}$$

1. Let us add an electric field $\mathcal{E} \hat{\bf{x}}$ to the system. Which term needs to be added to the Hamiltonian of each electron?

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

2. Compute the LCAO Hamiltonian of the system in presence of the electric field. What are the new onsite energies of the two orbitals?

   $$\hat{H} = \begin{pmatrix} E_0 & -t\\ -t & E_0 \end{pmatrix} +\begin{pmatrix} ⟨1|ex\mathcal{E}|1⟩ & ⟨1|ex\mathcal{E}|2⟩\\ ⟨2|ex\mathcal{E}|1⟩ & ⟨2|ex\mathcal{E}|2⟩ \end{pmatrix} = \begin{pmatrix} E_0 - \gamma & -t\\ -t & E_0 + \gamma \end{pmatrix},$$ where $\gamma = e d \mathcal{E}/2$ and have used $$⟨1|ex\mathcal{E}|1⟩ = -e d \mathcal{E}/2⟨1|1⟩ = -e d \mathcal{E}/2$$

3. Diagonalize the modified LCAO Hamiltonian. Find the ground state wavefunction $\psi$.

   The eigenstates of the Hamiltonian are given by: $$E_{\pm} = E_0\pm\sqrt{t^2+\gamma^2}$$ The ground state wave function is: $$\begin{split} |\psi⟩ &= \frac{t}{\sqrt{(\gamma+\sqrt{\gamma^2+t^2})^2+t^2}}\begin{pmatrix} \frac{\gamma+\sqrt{t^2+\gamma^2}}{t}\\ 1 \end{pmatrix}\\ |\psi⟩ &= \frac{\gamma+\sqrt{t^2+\gamma^2}}{\sqrt{(\gamma+\sqrt{\gamma^2+t^2})^2+t^2}}|1⟩+\frac{t}{\sqrt{(\gamma+\sqrt{\gamma^2+t^2})^2+t^2}}|2⟩ \end{split}$$

4. Find the polarisation $P$ of the molecule in the ground state.

   Reminder: polarisation

   The polarisation $P$ of a molecule with $n\leq 2$ electrons at its ground state $|\psi\rangle$ is: $$P = n e \langle\psi|\hat{x}|\psi\rangle.$$ Use that ground state you found in 3.2 is a linear superposition of two orthogonal orbitals centered at $-\frac{d}{2}$ and $+\frac{d}{2}$.

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

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Last update: October 30, 2023

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