In the case of helium, the Hamiltonian becomes more complex: not only is there a Coulomb attraction between the electrons and the nuclei, but there is also a Coulomb repulsion between the two electrons.
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:
which means we need to find eigenvalues and eigenvectors of a 6-dimensional partial differential equation!
"Mundane" copper has 29 electrons, so to find the electronic spectrum of Copper we would need to solve an 87-dimensional Schrödinger equation, and there is no way in the world we can do so currently.
"Mundane" copper has 29 electrons, so to find the electronic 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!
This 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.
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).
...
...
@@ -80,7 +83,7 @@ The complexity of solving the problem is the reason why we need to accept empiri
### Quantum numbers and shell filling
The electrons in a hydrogen orbital can be described by the folowing 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: $|n, l, l_z, m_s\rangle$
Quantum numbers:
...
...
@@ -95,21 +98,22 @@ In the figure below, the orbitals for the first three azimuthal numbers are show
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 .
-->
It turns out that electrons in other atoms approximately occupy very similar orbitals, only the energies are very different due to the Coulomb interaction.
It turns out that electrons in other atoms occupy very similar orbitals, only the 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 $n+l$, and of those with equal $n+l$ those with smaller $n$
* _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.
Especially for very large atoms.
The exceptions are especially relevant for very large atoms.
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 bound electrons provide a negatively charged cloud that reduces the effective charge of the atomic nucleus.
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.
## Covalent bonds and linear combination of atomic orbitals
...
...
@@ -121,68 +125,72 @@ Since different orbitals of an atom are separated in energy, we consider one orb
Let's imagine that the atoms are sufficiently far apart, such that the shape of the orbitals barely changes due to the presence of the other atom.
If $|1⟩$ is the wave function of an electron bound to the first atom, and $|2⟩$ is the wave function of the electron bound to the second atom, we will search for a solution in the form 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:
$$
\|\psi⟩ = \varphi_{1}|1⟩ + \varphi_{2}|2⟩
|\psi⟩ = \varphi_{1}|1⟩ + \varphi_{2}|2⟩.
$$
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.
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. $⟨1|2⟩=0$), such that $|\psi>$ is normalized whenever $|\varphi_1|^2 + |\varphi_2|^2 = 1$.
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)
If we apply the Hamiltonian to the molecular orbital $|\psi>$, we obtain
We apply the Hamiltonian to the molecular orbital $|\psi⟩$:
$$
\begin{align}
H|\psi> &= E |\psi>\\
&= \varphi_{1}H|1⟩ + \varphi_{2}H|2⟩\\
&= \varphi_{1}H|1⟩ + \varphi_{2}H|2⟩.\\
\end{align}
$$
Taking the left inner product with either $\langle 1|$, we obtain
Taking the left inner product with $\langle 1|$, we obtain
Here $E_{\rm{symmetric}}$ is the energy of the _symmetric_ molecular orbital and $E_{\rm{symmetric}}$ the energy of the _antisymmetric_ orbital, which can be seen in the figure below.
Which of these energies is higher depends on the sign of $t$.
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.
...
...
@@ -193,9 +201,9 @@ Remake this figure in python
-->
### Bonding vs antibonding
If we bring descrease the interatmic distance, the two atoms get closer and their atomic orbitals have more overlap.
This causes the absolute value of the hopping $|t|$ to become larger.
We can plot the symmetric and anti-symmetric energies as a function of the inter-atomic distance:
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 $|t|$.
We plot the symmetric and anti-symmetric energies as a function of the inter-atomic distance:
<!--
...
...
@@ -204,22 +212,25 @@ Remake this figure in python
??? Question "What is wrong with this image?"
If we bring the atoms together, the bonding energy diverges towards $-\infty$.
This would mean that the most energetically favourable position would be if the atoms are at the same positions.
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 $|\psi_{\rm{symmetric}}⟩$ is occupied by an electron (or two, because there are two states with opposite spin), it makes the atoms attract (or *bond*) because the total energy is lowered. Therefore, if $t$ is positive, $|\psi_{\rm{symmetric}}⟩$ is called the **bonding orbital**.
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**.
Occupation of the $|\psi_{\rm{antisymmetric}}⟩$ orbital increases the molecular energy as the interatomic distance gets smaller. This means that the electrons repel each other. Hence, if $t$ is positive, $|\psi_{\rm{antisymmetric}}⟩$ is called the **antibonding orbital**.
If an electron occupies the $|\psi_{\rm{antisymmetric}}⟩$ 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**.
Therefore if each atom has a single electron in the outermost shell, these atoms attract.
If each atom has 0 or 2 electrons in the outermost shell, the net electron force cancels (but Coulomb repulsion remains).
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).
### 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 bonding orbital.
* If two atoms have one orbital and one electron each, they occupy the bonding orbital.
## Exercises
...
...
@@ -228,7 +239,7 @@ If each atom has 0 or 2 electrons in the outermost shell, the net electron force
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 each a single orbital, what is the size of the Hamiltonian matrix if we also consider the spin of the electron?
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
...
...
@@ -268,7 +279,7 @@ where $V_0>0$ is the potential strength, $\hat{p}$ the momentum of the electron,
4. Find for which energy the boundary conditions at $x = x_0$ are satisfied. This is the energy of the bound state.
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 $|2\rangle$:
The orbitals $|1\rangle$ and $|2\rangle$ correspond to the wave functions of the electron when there is only a single delta peak present:
$$H_1 |1\rangle = \epsilon_1 |1\rangle,$$
$$H_2 |2\rangle = \epsilon_2 |2\rangle,$$
$$H_2 |2\rangle = \epsilon_2 |2\rangle.$$
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 for 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.
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 $\psi_2(x)$.
Also find an expression for their energies $\epsilon_1$ and $\epsilon_2$.
...
...
@@ -295,7 +306,7 @@ Using this, determine which energy corresponds to the bonding energy.
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 $-e$.
