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@@ -52,13 +52,15 @@ However, there are still several mysteries:
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:
$$H\psi = E\psi,$$
$$Hψ = Eψ,$$
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:
@@ -70,20 +72,20 @@ The Hamiltonian contains not just the Coulomb attraction between the electrons a
$$H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial {\mathbf r_1^2}} -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial {\mathbf r_2^2}}- \frac{2e^2}{4\pi\varepsilon_0|r_1|} - \frac{2e^2}{4\pi\varepsilon_0|r_2|} + \frac{e^2}{4\pi\varepsilon_0|r_1 - r_2|},$$
which means we need to find eigenvalues and eigenvectors of a 6-dimensional partial differential equation!
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 electronic spectrum of Copper we would need to solve an 87-dimensional Schrödinger equation.
"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 "merely" consequences of the Schrödinger equation.
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: $|n, l, l_z, m_s\rangle$
The electrons in a hydrogen atom can be described by the following four quantum numbers: $|n, l, l_z, m_s⟩$
Quantum numbers:
@@ -92,13 +94,14 @@ Quantum numbers:
* $l_z=-l, -l+1\ldots, l$ is the $z$-component of angular momentum
* $m_s$ is the $z$-compoment of the spin
In the figure below, the orbitals for the first three azimuthal numbers are shown.
![](figures/single_atom.svg)
<!--
Instead of the image above, we plan to show the actual orbitals of hydrogen via this image https://en.wikipedia.org/wiki/Quantum_number#/media/File:Atomic_orbitals_n123_m-eigenstates.png .
-->
Below is an illustration of several lowest energy orbitals in hydrogen:
[![Hydrogen atomic orbitals](https://upload.wikimedia.org/wikipedia/commons/5/5c/Atomic_orbitals_n123_m-eigenstates.png)](https://commons.wikimedia.org/wiki/File:Atomic_orbitals_n123_m-eigenstates.png#/media/File:Atomic_orbitals_n123_m-eigenstates.png)
(image source: Wikipedia © Geek3 CC-BY-SA)
It turns out that electrons in other atoms occupy very similar orbitals, only the energies are very different due to the Coulomb interaction.
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.
@@ -107,13 +110,13 @@ 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 $n$
If we take the Aufbau principle and Madelung's rule into account, the shell filling order is 1s, 2s, 2p, 3s, 3p, 4s, 3d, ...
Note that there are exceptions to these rules.
The exceptions are especially relevant for very large atoms.
Combining the two rules, we obtain the shell filing order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc.
The valence electrons (those in the outermost shell) are the only ones participating in chemical reactions and electric conduction.
While these rules accurately predict the electronic structure of most elements, they are only approximate, and fail to describe some of the [heavier elements](https://en.wikipedia.org/wiki/Aufbau_principle#Exceptions_in_the_d-block).
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.
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
@@ -128,45 +131,40 @@ Let's imagine that the atoms are sufficiently far apart, such that the shape of
Let's denote the wave function of an electron bound on the first and second atom $|1⟩$ and $|2⟩$ respectively.
We search for a solution in the form:
$$
|\psi⟩ = \varphi_{1}|1⟩ + \varphi_{2}|2⟩.
|ψ⟩ = φ_{1}|1⟩ + φ_{2}|2⟩.
$$
where $\varphi_{1}$ and $\varphi_{2}$ are the probability amplitudes of the respective orbitals.
The orbital $|\psi⟩$ is called a _molecular orbital_ because it describes the entire orbital of the diatomic molecule.
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 orthogonal (i.e. $⟨1|2⟩=0$).
Orthogonality ensures that $|\psi⟩$ is normalized whenever $|\varphi_1|^2 + |\varphi_2|^2 = 1$.
(See the book exercise 6.5 for relaxing this assumption)
For simplicity, we assume that the atomic orbitals are orthogonal[^1], 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 $|\psi⟩$:
We apply the Hamiltonian to the molecular orbital $|ψ⟩$:
$$
\begin{align}
H|\psi> &= E |\psi>\\
&= \varphi_{1}H|1⟩ + \varphi_{2}H|2⟩.\\
\end{align}
H|ψ⟩ = E|ψ⟩ = φ_{1}H|1⟩ + φ_{2}H|2⟩.
$$
Taking the left inner product with $\langle 1|$, we obtain
Taking the left inner product with $⟨1|$, we obtain
$$
\begin{align}
\langle 1|E|\psi> &= \varphi_{1}\langle 1|H|1⟩ + \varphi_{2}\langle 1|H|2⟩\\
&= E \varphi_{1}.
\end{align}
⟨1|E|ψ⟩ = φ_{1}⟨1|H|1⟩ + φ_{2}⟨1|H|2⟩ = E φ_{1}.
$$
We do the same with $\langle 2|$:
Similarly, taking the inner product with $⟨2|$ yields:
$$
E \varphi_2 = \varphi_{1}\langle 2|H|1⟩ + \varphi_{2}\langle 2|H|2⟩.
E φ_2 = φ_{1}⟨2|H|1⟩ + φ_{2}⟨2|H|2⟩.
$$
We combine these two equations into an eigenvalue problem:
$$
E \begin{pmatrix} \varphi_1 \\ \varphi_2 \end{pmatrix}
E \begin{pmatrix} φ_1 \\ φ_2 \end{pmatrix}
= \begin{pmatrix}
⟨1|H|1⟩ & ⟨1|H|2⟩ \\ ⟨2|H|1⟩ & ⟨2|H|2⟩
\end{pmatrix}
\begin{pmatrix} \varphi_1 \\ \varphi_2\end{pmatrix}.
\begin{pmatrix} φ_1 \\ φ_2\end{pmatrix}.
$$
The eigenvalue problem depends only on two parameters:
@@ -180,50 +178,42 @@ $$ H = \begin{pmatrix} E_0 & -t \\ -t & E_0 \end{pmatrix}$$
Diagonalizing the Hamiltonian yields the following two eigenvalues:
$$
E = E_0 \pm t.
E_{±} = E_0 ∓ t.
$$
The corresponding eigenstates are:
The eigenvector corresponding to the eigenvalue $E_+ = E_0 - t$ is even and symmetric:
$|\psi_{\rm{symmetric}}⟩ = \tfrac{1}{\sqrt{2}}(|1⟩ + |2⟩)$ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (even/symmetric superposition with $E_{\rm{symmetric}} = E_0 - t$);
$|\psi_{\rm{antisymmetric}}⟩ = \tfrac{1}{\sqrt{2}}(|1⟩ - |2⟩)$ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (odd/antisymmetric superposition with $E_{\rm{antisymmetric}} = E_0 + t$).
$$
|ψ_{+}⟩ = \tfrac{1}{\sqrt{2}}(|1⟩ + |2⟩),
$$
while the eigenvector with energy $E_- = E_0 + t$
$$
|ψ_{-}⟩ = \tfrac{1}{\sqrt{2}}(|1⟩ - |2⟩)
$$
is odd/antisymmetric.
Here $E_{\rm{symmetric}}$ is the energy of the _symmetric_ molecular orbital and $E_{\rm{symmetric}}$ the energy of the _antisymmetric_ orbital.
The molecular orbitals are shown in the figure below.
The energy ordering of the molecular orbitals depends on the sign of $t$.
If $t>0$, then $E_{\rm{symmetric}}$ corresponds to the lower energy and when $t<0$, $E_{\rm{antisymmetric}}$ has the lowest energy.
From now on we assume that $t>0$, hence $E_{\rm{symmetric}}$ corresponds to the lowest energy.
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 $t > 0$.
![](figures/two_atoms.svg)
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Remake this figure in python
-->
### 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 absolute value of the hopping $|t|$.
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:
![](figures/bonding.svg)
<!--
Remake this figure in python
-->
??? Question "What is wrong with this image?"
If we bring the atoms together, the bonding energy diverges towards $-\infty$.
The lower binding energy implies that the most energetically favorable position is when atoms are on top of each other.
This is rather unphysical.
When an electron (or two, because there are two states with opposite spin) occupies $|\psi_{\rm{symmetric}}⟩$, the atoms attract (or *bond*) because the total energy is lowered.
Therefore, if $t$ is positive, $|\psi_{\rm{symmetric}}⟩$ is called the **bonding orbital**.
If an electron occupies the $|\psi_{\rm{antisymmetric}}⟩$ orbital, the molecular energy increases with decreasing interatomic 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 electrons repel each other.
Hence, if $t$ is positive, $|\psi_{\rm{antisymmetric}}⟩$ is called the **antibonding orbital**.
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.
On the other hand, if each atom has 0 or 2 electrons in the outermost shell, the net electron force cancels (but Coulomb repulsion remains).
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
@@ -232,7 +222,6 @@ On the other hand, if each atom has 0 or 2 electrons in the outermost shell, the
* 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
1. Is the assumption that the atomic orbitals are orthogonal always a reasonable assumption?
@@ -241,7 +230,6 @@ On the other hand, if each atom has 0 or 2 electrons in the outermost shell, the
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.
@@ -253,7 +241,6 @@ $$Au = [Xe] 6s^1 4f^{14} 5d^{10}$$
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.
@@ -282,18 +269,18 @@ where $V_0>0$ is the potential strength, $\hat{p}$ the momentum of the electron,
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 $|2\rangle$:
$$|\psi⟩ = \varphi_1|1⟩ + \varphi_2|2\rangle.$$
The orbitals $|1\rangle$ and $|2\rangle$ correspond to the wave functions of the electron when there is only a single delta peak present:
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:
$$H_1 |1\rangle = \epsilon_1 |1\rangle,$$
$$H_2 |2\rangle = \epsilon_2 |2\rangle.$$
$$H_1 |1⟩ = \epsilon_1 |1⟩,$$
$$H_2 |2⟩ = \epsilon_2 |2⟩.$$
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 $|1⟩$ and $|2⟩$: $\psi_1(x)$ and $\psi_2(x)$.
1. 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.
2. Construct the LCAO Hamiltonian. To simplify the calculations, assume that the orbitals are orthogonal.
@@ -316,21 +303,25 @@ H_{\textrm{eff}} = \begin{pmatrix}
\end{pmatrix}.
$$
1. Let us add an electric field $\mathcal{E} \hat{\bf{x}}$ to the system.
??? hint "The electric potential is given by"
$$
V_{\mathbf{E}}=-\int_{C} \mathbf{E} \cdot \mathrm{d} \boldsymbol{\ell}
$$
Which term needs to be added to the Hamiltonian of each electron?
1. Let us add an electric field $\mathcal{E} \hat{\bf{x}}$ to the system.
Which term needs to be added to the Hamiltonian of each electron?
??? hint "The electric potential is given by"
$$
V_{\mathbf{E}}=-\int_{C} \mathbf{E} \cdot \mathrm{d} \boldsymbol{\ell}
$$
2. Compute the LCAO Hamiltonian of the system in presence of the electric field.
What are the new onsite energies of the two orbitals?
3. Diagonalize the modified LCAO Hamiltonian. Find the ground state wavefunction $\psi$.
3. Diagonalize the modified LCAO Hamiltonian. Find the ground state wavefunction $ψ$.
4. Find the polarization $P$ of the molecule in the ground state.
??? hint "Reminder: polarization"
??? hint "Reminder: polarization"
The polarization $P$ of a molecule with $n\leq 2$ electrons at its ground state $|\psi\rangle$ is:
$$
P = n e \langle\psi|x|\psi\rangle.
$$
Use that ground state you found in 3.2 is a linear superposition of two orthogonal orbitals centered at $-\frac{d}{2}$ and $+\frac{d}{2}$.
The polarization $P$ of a molecule with $n\leq 2$ electrons at its ground state $|ψ⟩$ is:
$$
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 $-\frac{d}{2}$ and $+\frac{d}{2}$.
[^1]: See the book exercise 6.5 for relaxing the orthogonality assumption.
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