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@@ -19,7 +19,9 @@ _(based on chapters 5 and 6.2 of the book)_
- Write down the Schrödinger equation
- Compute eigenvectors and eigenvalues of a matrix
- Find a state bound at a $δ$-function potential in 1D (for the exercises)
- 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
!!! summary "Learning goals"
@@ -27,24 +29,27 @@ _(based on chapters 5 and 6.2 of the book)_
- Describe the shell-filling model of atoms
- Derive the LCAO model
- Obtain the spectrum of the LCAO model of several orbitals
- Obtain the energy spectrum of the LCAO model of several orbitals
## Looking back
So far we have:
* Introduced the $k$-space (reciprocal space)
* Postulated electron and phonon dispersion relations
* 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 electrons conduct (Drude model) and store heat/energy (Sommerfeld model)
* Understood how free electrons conduct (Drude model) and store heat/energy (Sommerfeld model)
We used our best guess as a starting point, and there are several mysteries:
We made several approximations and postulations through these models.
However, there are still several mysteries:
* Why is there a cutoff frequency? Why are there no more phonon modes?
* Why do electrons not scatter off from every single atom in the Drude model? Atoms are charged and should provide a lot of scattering.
* 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)
## A quick review of atoms
@@ -55,114 +60,187 @@ We used our best guess as a starting point, and there are several mysteries:
$$H\psi = E\psi,$$
with $H$ the sum of kinetic energy and Coulomb interaction, so for hydrogen we have:
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:
$$H=-\hbar^2\frac{\partial^2}{2m\partial {\mathbf r^2}} - \frac{e^2}{4\pi\varepsilon_0|r|}$$
$$H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial {\mathbf r^2}} - \frac{e^2}{4\pi\varepsilon_0|r|}.$$
for helium it becomes more complex: $\psi({\mathbf r})\rightarrow \psi({\mathbf {r_1, r_2}})$, so
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:
$$H=-\hbar^2\frac{\partial^2}{2m\partial {\mathbf r_1^2}} -\hbar^2\frac{\partial^2}{2m\partial {\mathbf r_2^2}}- \frac{2e^2}{4\pi\varepsilon_0|r_1|} - \frac{2e^2}{4\pi\varepsilon_0|r_2|} + \frac{e^2}{4\pi\varepsilon_0|r_1 - r_2|},$$
$$H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial {\mathbf r_1^2}} -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial {\mathbf r_2^2}}- \frac{2e^2}{4\pi\varepsilon_0|r_1|} - \frac{2e^2}{4\pi\varepsilon_0|r_2|} + \frac{e^2}{4\pi\varepsilon_0|r_1 - r_2|},$$
which means we need to find eigenvalues and eigenvectors of a 6-dimensional differential equation!
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.
"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 exponential 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.
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).
However for us 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 "merely" consequences of the Schrödinger equation.
### Quantum numbers and shell filling
Single electron states have 4 quantum numbers: $|n, l, l_z, s\rangle$
![](figures/single_atom.svg)
The electrons in a hydrogen atom can be described by the following four quantum numbers: $|n, l, l_z, m_s\rangle$
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
* $s$ is the spin
* $m_s$ 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 .
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It turns out that electrons in all 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: *first fill a complete shell (all electrons with the same $n, l$) before going to the next one*
* Madelung's rule: *first occupy the shells with the lowest $l+n$, and of those with equal $l+n$ those with smaller $n$*
* _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 $n$
Therefore shell-filling order is 1s, 2s, 2p, 3s, 3p, 4s, 3d, ...
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.
The electrons in the outermost shell are the only ones participating in chemical reactions and electric conduction.
The rest provides a negatively charged cloud that reduces the attraction to the atomic nucleus.
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.
## Covalent bonds and linear combination of atomic orbitals
*Here we present the material in the order different from the book because the covalent bonds are the most important for the exercises*
### Two atoms
Consider two atoms next to each other.
Consider two atoms next to each other which form a diatomic molecule.
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 imagine that the atoms are sufficiently far apart, so that the shape of the orbitals or the energy of the orbitals doesn't change much.
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 near the second atom, we will search for a solution in form: $|ψ⟩ = ϕ_{1}|1⟩ + ϕ_{2}|2⟩$, or in other words as a *Linear Combination of Atomic Orbitals (LCAO)*.
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:
$$
|\psi⟩ = \varphi_{1}|1⟩ + \varphi_{2}|2⟩.
$$
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 let's assume now that $⟨1|2⟩=0$, so that $ψ$ is normalized whenever $|ϕ_1|^2 + |ϕ_2|^2 = 1$.
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)
Acting with the Hamiltonian on the LCAO wave function we get an eigenvalue problem:
We apply the Hamiltonian to the molecular orbital $|\psi⟩$:
$$
\begin{align}
H|\psi> &= E |\psi>\\
&= \varphi_{1}H|1⟩ + \varphi_{2}H|2⟩.\\
\end{align}
$$
Taking the left inner product with $\langle 1|$, we obtain
$$
E \begin{pmatrix} ϕ_1 \\ ϕ_2 \end{pmatrix}
\begin{align}
\langle 1|E|\psi> &= \varphi_{1}\langle 1|H|1⟩ + \varphi_{2}\langle 1|H|2⟩\\
&= E \varphi_{1}.
\end{align}
$$
We do the same with $\langle 2|$:
$$
E \varphi_2 = \varphi_{1}\langle 2|H|1⟩ + \varphi_{2}\langle 2|H|2⟩.
$$
We combine these two equations into an eigenvalue problem:
$$
E \begin{pmatrix} \varphi_1 \\ \varphi_2 \end{pmatrix}
= \begin{pmatrix}
⟨1|H|1⟩ & ⟨1|H|2⟩ \\ ⟨2|H|1⟩ & ⟨2|H|2⟩
\end{pmatrix}
\begin{pmatrix} ϕ_1 \\ ϕ_2\end{pmatrix}
\begin{pmatrix} \varphi_1 \\ \varphi_2\end{pmatrix}.
$$
It only depends on two parameters remaining of the original problem:
$⟨1|H|1⟩ = ⟨2|H|2⟩ ≡ E_0$ is the **onsite energy**, $⟨1|H|2⟩ ≡ -t$ is the **hopping integral** (or just **hopping**).
The eigenvalue problem depends only on two parameters:
$⟨1|H|1⟩ = ⟨2|H|2⟩ \equiv E_0$ the **onsite energy** and $⟨1|H|2⟩ \equiv -t$ the **hopping integral** (or just **hopping**).
Since we are considering bound states, all orbitals $|n⟩$ are purely real ⇒ $t$ is real as well.
All orbitals $|n⟩$ are purely real because we consider bound states.
Hence $t$ is real as well.
We use the previously defined eigenvalue problem to construct the Hamiltonian in matrix form:
$$ H = \begin{pmatrix} E_0 & -t \\ -t & E_0 \end{pmatrix}$$
Eigenstates & eigenvalues:
$|+⟩ = \tfrac{1}{\sqrt{2}}(|1⟩ + |2⟩)$ [even/symmetric superposition with $E_+ = E_0 - t$];
$|-⟩ = \tfrac{1}{\sqrt{2}}(|1⟩ - |2⟩)$ [odd/antisymmetric superposition with $E_- = E_0 + t$].
Diagonalizing the Hamiltonian yields the following two eigenvalues:
$$
E = E_0 \pm t.
$$
The corresponding eigenstates are:
Two atoms are a molecule, and $\psi_+$ and $\psi_-$ are **molecular orbitals**.
$|\psi_{\rm{symmetric}}⟩ = \tfrac{1}{\sqrt{2}}(|1⟩ + |2⟩)$ &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 $E_{\rm{antisymmetric}} = E_0 + t$).
An even superposition has a lower kinetic energy (derivative is smaller) ⇒ $t$ is positive.
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.
![](figures/two_atoms.svg)
### Bonding vs antibonding
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Remake this figure in python
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Hopping $t$ grows when the atoms are moved together because the overlap. Let's plot energy of molecular orbitals vs inter-atomic distance.
### 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 $|t|$.
We plot the symmetric and anti-symmetric energies as a function of the inter-atomic distance:
![](figures/bonding.svg)
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Remake this figure in python
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??? 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 $|+⟩$ 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. It is therefore called a **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**.
An occupied $|-⟩$ state increases the molecule energy as the atoms move closer ⇒ it makes atoms repel, and it is an **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 there are 0 or 2 electrons, the net electron force cancels (but electrostatic repulsion remains).
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).
### Summary
* Electrons in atoms occupy shells, with only the outermost shell contributing to interatomic interaction.
* 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 atom each, electron movement makes the atoms attract.
* If two atoms have one orbital and one electron each, they occupy the bonding orbital.
*[LCAO]: Linear Combination of Atomic Orbitals
## Exercises
### Warm-up questions
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.
@@ -173,17 +251,19 @@ $$Pd = [Kr] 5s^0 4d^{10}$$
$$Ag = [Kr] 5s^1 4d^{10}$$
$$Au = [Xe] 6s^1 4f^{14} 5d^{10}$$
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
Consider an electron moving in 1D between two delta-function shaped potential wells.
### 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:
$$
H = \frac{p^2}{2m}-V_0\delta(x_1-x)-V_0\delta(x_2-x),
\hat{H} = \frac{\hat{p}^2}{2m}-V_0\delta(x_1-x)-V_0\delta(x_2-x),
$$
with $V_0$ is the potential strength, $p$ the momentum of the electron, and $x_1$, $x_2$ the positions of the potential wells.
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.
??? hint "Properties of a $\delta$-function potential"
??? hint "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:
@@ -193,32 +273,42 @@ with $V_0$ is the potential strength, $p$ the momentum of the electron, and $x_1
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 the bound state has the energy $E_0$.
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:
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 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 $|2\rangle$:
$$|Ψ ⟩ = ϕ_1|1⟩ + ϕ_2|2\rangle.$$
The orbitals $|1\rangle$ and $|2\rangle$ correspond to the wave functions of the electron when only one of the nuclei is present:
$$|\psi⟩ = \varphi_1|1⟩ + \varphi_2|2\rangle.$$
The orbitals $|1\rangle$ and $|2\rangle$ correspond to the wave functions of the electron when there is only a single delta peak present:
$$H_1 |1\rangle = \epsilon_1 |1\rangle,$$
$$H_2 |2\rangle = \epsilon_2 |2\rangle,$$
$$H_2 |2\rangle = \epsilon_2 |2\rangle.$$
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 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 $\epsilon_2$.
Remember that you need to normalize the wave functions.
2. Construct the LCAO Hamiltonian. To simplify the calculations, assume that the orbitals are orthogonal.
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.
1. Find the expressions for the wave functions of the states $|1⟩$ and $|2⟩$: $\psi_1(x)$ and $\psi_2(x)$, as well as their energies $ε_1$ and $ε_2$.
Remember that you need to normalize the wave functions.
2. Find the LCAO Hamiltonian. To simplify the calculations, assume that the orbitals are orthogonal.
3. Diagonalize the effective Hamiltonian. Which are the expressions for the eigenenergies of the system?
### Exercise 3: Polarization 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
Consider two electrons in the system, each with charge $e$. The electrons are shared between the nuclei to form a covalent bond.
The LCAO Hamiltonian for each of the electrons is:
$$
H_{\textrm{eff}} = \begin{pmatrix}
E_0&&-t \\
@@ -226,16 +316,21 @@ H_{\textrm{eff}} = \begin{pmatrix}
\end{pmatrix}.
$$
1. Let us add an electric field $\mathcal{E}$ to the system. 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.
??? 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?
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 LCAO wave function of the ground state.
3. Diagonalize the modified LCAO Hamiltonian. Find the ground state wavefunction $\psi$.
4. Find the polarization $P$ of the molecule in the ground state.
??? hint "Reminder: polarization"
The polarization $P$ of a molecule with $n\leq 2$ electrons at its ground state $|\Psi\rangle$ is:
The polarization $P$ of a molecule with $n\leq 2$ electrons at its ground state $|\psi\rangle$ is:
$$
P = n e \langle\Psi|x|\Psi\rangle.
P = n e \langle\psi|x|\psi\rangle.
$$
Use that ground state you found in 3.2 is a linear superposition of two orthogonal orbitals centered at $-\frac{d}{2}$ and $+\frac{d}{2}$.