    from matplotlib import pyplot

import numpy as np

from common import draw_classic_axes, configure_plotting

configure_plotting()

pi = np.pi

Atoms and bonds¶

(based on chapters 5 and 6.2 of the book)

Expected prior knowledge

Before the start of this lecture, you should be able to:

Write down the Schrödinger equation
Compute eigenvectors and eigenvalues of a matrix
Solve the Schrödinger equation of a bound state with a $δ$-function potential in 1D (for the exercises)
Write down the quantum numbers of the hydrogen atom
Describe the orbitals of the hydrogen atom using the quantum numbers

Learning goals

After this lecture you will be able to:

Describe the shell-filling model of atoms
Derive the LCAO model
Obtain the energy spectrum of the LCAO model of several orbitals

Looking back¶

So far we have:

Introduced the $k$-space (reciprocal space)
Postulated the dispersion relation of free electrons and phonons
Calculated the heat capacity of free electrons and phonons

As a result we:

Understood how phonons store heat (Debye model)
Understood how free electrons conduct (Drude model) and store heat/energy (Sommerfeld model)

We made several approximations and postulations through these models. However, there are still several mysteries:

Why is there a phonon cutoff frequency? Why are there no more phonon modes beyond this cutoff frequency?
Why don't electrons scatter off from every single atom in the Drude model? Atoms are charged and should provide a lot of scattering.
Why are some materials not metals? (Think if you know a crystal that isn't a metal)

To answer these questions we will need to study atoms in more detail.

A quick review of atoms¶

Why chemistry is not physics¶

Everything is described by the Schrödinger equation:

\[ Hψ = Eψ, \]

with $H$ the sum of kinetic energy and the potential energy. In the hydrogen atom, the potential energy is due to the Coulomb interaction between the electron and the nucleus:

\[ H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial {\mathbf r^2}} - \frac{e^2}{4\pi\varepsilon_0|r|}. \]

In the case of helium, the Hamiltonian becomes more complex. The Hamiltonian contains not just the Coulomb attraction between the electrons and the nuclei, but also Coulomb repulsion between the two electrons:

\[ H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial {\mathbf r_1^2}} -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial {\mathbf r_2^2}}- \frac{2e^2}{4\pi\varepsilon_0|r_1|} - \frac{2e^2}{4\pi\varepsilon_0|r_2|} + \frac{e^2}{4\pi\varepsilon_0|r_1 - r_2|}, \]

which means we need to find eigenvalues and eigenvectors of a 6-dimensional partial differential equation, instead of a 3-dimensional one descirbing hydrogen.

"Mundane" copper has 29 electrons, so to find the electron spectrum of copper we would need to solve an 87-dimensional Schrödinger equation. There is no way in the world we can do so even with the most powerful modern computers!

The growth in complexity with the number of interacting quantum particles is why many-body quantum physics is very much an open area in solid state physics and quantum chemistry.

However we need to focus on what is possible to do, and apply heuristic rules based on the accumulated knowledge of how atoms work (hence we will need a bit of chemistry).

The complexity of solving the problem is the reason why we need to accept empirical observations as chemical laws even though they work with limited precision, and are instead consequences of the Schrödinger equation.

Quantum numbers and shell filling¶

The electrons in a hydrogen atom can be described by the following four quantum numbers: $|n, l, l_z, m_s⟩$

Quantum numbers:

$n=1,2,\ldots$ is the azimuthal (principal) quantum number
$l=0, 1, \ldots, n-1$ is the angular momentum (also known as $s, p, d, f$ orbitals)
$l_z=-l, -l+1\ldots, l$ is the $z$-component of angular momentum
$m_s$ is the $z$-compoment of the spin

Below is an illustration of several lowest energy orbitals in hydrogen:

(image source: Wikipedia © Geek3 CC-BY-SA)

It turns out that electrons in other atoms occupy orbitals similar to those of hydrogen, however the electron energies are very different due to the Coulomb interaction. Therefore, we can use our knowledge of the hydrogen orbitals to describe other atoms.

When we consider an atom with multiple electrons, we need to determine which orbitals are filled and which are not. The order of orbital filling is set by several rules:

Aufbau principle: electrons first fill a complete shell (all electrons with the same $n$) before going to the next one
Madelung's rule: electrons first occupy the shells with the lowest $n+l$. If there are several orbitals with equal $n+l$, electrons occupy those with smaller $n$

Combining the two rules, we obtain the shell filing order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc.

While these rules accurately predict the electronic structure of most elements, they are only approximate, and fail to describe some of the heavier elements.

Shell filing is important to us because the valence electrons (those in the outermost shell) are the only ones participating in chemical reactions and electric conduction. From the valence electrons' point of view, the inner shell electrons act like a negatively charged cloud. The electrostatic repulsion between them reduces the effective charge of the atomic nucleus, but does not play any further role.

Covalent bonds and linear combination of atomic orbitals¶

Two atoms¶

Consider two atoms next to each other which form a diatomic molecule. The total Hamiltonian of the system is

\[ H = V_1 + V_2 + K, \]

with $V_1$ the potential due to the first nucleus, $V_2$ due to the second nucleus, and $K$ is the kinetic energy of the electron.

Since different orbitals of an atom are separated in energy, we consider one orbital per atom (even though this is often a bad starting point and it should only work for $s$-orbitals).

Let's additionally consider the atoms being sufficiently far apart, such that the shape of the orbitals barely changes due to the presence of the other atom.

Let's denote the wave function of an electron bound to the first and second atom $|1⟩$ and $|2⟩$ respectively:

\[\begin{align} (V_1 + K)|1⟩ = ɛ_0|1⟩,\\ (V_2 + K)|2⟩ = ɛ_0|2⟩. \end{align}\]

Our main idea is to search for a solution in the form:

\[ |ψ⟩ = φ_{1}|1⟩ + φ_{2}|2⟩. \]

where $φ_{1}$ and $φ_{2}$ are the probability amplitudes of the respective orbitals. The orbital $|ψ⟩$ is called a molecular orbital because it describes the entire orbital of the diatomic molecule. The molecular orbital is created as a Linear Combination of Atomic Orbitals (LCAO).

For simplicity, we assume that the atomic orbitals are orthogonal¹, i.e. $⟨1|2⟩=0$. Orthogonality ensures that $|ψ⟩$ is normalized whenever $|φ_1|^2 + |φ_2|^2 = 1$.

We apply the Hamiltonian to the molecular orbital $|ψ⟩$:

\[ H|ψ⟩ = E|ψ⟩ = φ_{1}H|1⟩ + φ_{2}H|2⟩. \]

Taking the left inner product with $⟨1|$, we obtain

\[ ⟨1|E|ψ⟩ = φ_{1}⟨1|H|1⟩ + φ_{2}⟨1|H|2⟩ = E φ_{1}. \]

Similarly, taking the inner product with $⟨2|$ yields:

\[ E φ_2 = φ_{1}⟨2|H|1⟩ + φ_{2}⟨2|H|2⟩. \]

We combine these two equations into an eigenvalue problem:

\[ E \begin{pmatrix} φ_1 \\ φ_2 \end{pmatrix} = \begin{pmatrix} ⟨1|H|1⟩ & ⟨1|H|2⟩ \\ ⟨2|H|1⟩ & ⟨2|H|2⟩ \end{pmatrix} \begin{pmatrix} φ_1 \\ φ_2\end{pmatrix}. \]

The eigenvalue problem depends only on two parameters: the onsite energy $⟨1|H|1⟩ = ⟨2|H|2⟩ \equiv E_0$ that gives the energy of an electron occupying either of the orbitals, and the hopping integral (or just hopping) $⟨1|H|2⟩ \equiv -t$ that characterizes the energy associated with the electron moving between the two orbitals.

Let us examine what contitutes the onsite energy and the hopping:

\[ E_0 = ⟨1|H|1⟩ = ⟨1|V_1 + V_2 + K|1⟩ = ɛ_0 + ⟨1|V_2|1⟩, \]

where we used that $V_1 + K|1⟩ = ɛ_0|1⟩$. In other words the onsite energy is the combination of the energy of the original orbital plus the energy shift $⟨1|V_2|1⟩$ of the electron due to the potential of the neighboring atom. Turning to the hopping, we obtain

\[ t = -⟨1|H|2⟩ = -⟨1|V_1 + V_2 + K|2⟩ = -⟨1|V_1|2⟩. \]

All orbitals $|n⟩$ are purely real because we consider bound state solutions of the Schrödinger equation. Hence $t$ is real as well.

The eigenvalue problem we obtained describes a particle with a discrete $2×2$ Hamiltonian:

\[ H = \begin{pmatrix} E_0 & -t \\ -t & E_0 \end{pmatrix}. \]

Diagonalizing this LCAO Hamiltonian yields the following two eigenvalues:

\[ E_{±} = E_0 ∓ t. \]

The eigenvector corresponding to the eigenvalue $E_+ = E_0 - t$ is even and symmetric:

\[ |ψ_{+}⟩ = \tfrac{1}{\sqrt{2}}(|1⟩ + |2⟩), \]

while the eigenvector with energy $E_- = E_0 + t$

\[ |ψ_{-}⟩ = \tfrac{1}{\sqrt{2}}(|1⟩ - |2⟩) \]

is odd/antisymmetric.

The molecular orbitals are shown in the figure below. According to the node theorem of quantum mechanics, wave functions with lower energies have fewer points where $ψ=0$. Because $ψ_- = 0$ between the two atoms, and $ψ_+$ is not, we conclude that $E_+ < E_-$, and therefore $t > 0$.

Bonding vs antibonding¶

If we decrease the interatomic distance, the two atoms get closer and their atomic orbitals have more overlap. The increase in orbital overlap increases the hopping $t$. We plot the symmetric and anti-symmetric energies as a function of the inter-atomic distance:

When an electron (or two, because there are two states with opposite spin) occupies $|ψ_+⟩$, the atoms attract (or bond) because the total energy is lowered. Therefore, if $t$ is positive, $|ψ_+⟩$ is called the bonding orbital.

If an electron occupies the $|ψ_{-}⟩$ orbital, the molecular energy increases with decreasing interatomic distance. This means that the atoms repel each other. Hence, if $t$ is positive, $|ψ_{-}⟩$ is called the antibonding orbital.

Therefore if each atom has a single electron in the outermost shell, these atoms attract because the bonding orbital hosts two electrons with opposite spins. On the other hand, if each atom has 0 or 2 electrons in the outermost shell, the net force from the bonding and antibonding orbitals cancels out, but Coulomb repulsion remains.

Summary¶

Electrons in atoms occupy shells, with only electrons in the outermost shell (valence electrons) contributing to interatomic interactions.
The molecular orbital can be written as a Linear Combination of Atomic Orbitals (LCAO)
The LCAO method reduces the full Hamiltonian to a finite size problem written in the basis of individual orbitals.
If two atoms have one orbital and one electron each, they occupy the bonding orbital.

Exercises¶

Warm-up questions¶

Is the assumption that the atomic orbitals are orthogonal always a reasonable assumption?
What happens if the hopping $t$ is chosen to be negative?
How does the size of the Hamiltonian matrix change with the number of atoms?
How does the size of the Hamiltonian matrix change if each atom now has two orbitals?
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¶

Describe the shell-filling model of atoms.
Use Madelung’s rule to deduce the atomic shellfilling configuration of the element tungsten, which has atomic number 74.
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: $\textrm{Cu} = [\textrm{Ar}] 4s^1 3d^{10}$, $\textrm{Pd} = [\textrm{Kr}] 5s^0 4d^{10}$, $\textrm{Ag} = [\textrm{Kr}] 5s^1 4d^{10}$, $\textrm{Au} = [\textrm{Xe}] 6s^1 4f^{14} 5d^{10}$. 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?

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:

Assume that we have a bound state with energy $E<0$.
Compute the wave function $\phi$ in different regions of space: namely $x < x_0$ and $x > x_0$.
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:

$$ \frac{d\phi}{dx}\Bigr|{x_0+\epsilon} - \frac{d\phi}{dx}\Bigr|= -\frac{2mV_0}{\hbar^2}\phi(x_0). $$

Find at which energy the boundary conditions at $x = x_0$ are satisfied. This is the energy of the bound state.
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 $|1⟩$ and $|2⟩$:

\[|ψ⟩ = φ_1|1⟩ + φ_2|2⟩.\]

The orbitals $|1⟩$ and $|2⟩$ correspond to the wave functions of the electron when there is only a single delta peak present:

\[\begin{align} H_1 |1⟩ &= \epsilon_1 |1⟩,\\ H_2 |2⟩ &= \epsilon_2 |2⟩. \end{align}\]

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.

Find the expressions for the wave functions of the states $|1⟩$ and $|2⟩$: $ψ_1(x)$ and $ψ_2(x)$. Also find an expression for their energies $\epsilon_1$ and $\epsilon_2$. Remember that you need to normalize the wave functions.
Construct the LCAO Hamiltonian. To simplify the calculations, assume that the orbitals are orthogonal.
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.

Exercise 3: Polarization of a hydrogen molecule¶

Consider a hydrogen molecule as a one-dimensional system with two identical nuclei at $x=-d/2$ and $x=+d/2$, 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

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

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?

The electric potential is given by

\[ V_{\mathbf{E}}=-\int_{C} \mathbf{E} \cdot \mathrm{d} \boldsymbol{\ell} \]
Compute the LCAO Hamiltonian of the system in presence of the electric field. What are the new onsite energies of the two orbitals?
Diagonalize the modified LCAO Hamiltonian. Find the ground state wavefunction $ψ$.
Find the polarization $P$ of the molecule in the ground state.

Reminder: polarization

The polarization $P$ of a molecule with $n\leq 2$ electrons at its ground state $|ψ⟩$ is:

\[ P = n e ⟨ψ|x|ψ⟩. \]

Use that ground state you found in 3.2 is a linear superposition of two orthogonal orbitals centered at $-d/2$ and $+d/2$.

See the book exercise 6.5 for relaxing the orthogonality assumption. ↩