* Postulated the dispersion relation of free electrons and phonons
* Calculated the heat capacity of free electrons and phonons
* Calculated the heat capacity of free electrons and phonons
As a result we:
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@@ -48,7 +48,7 @@ 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?
* 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)
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@@ -199,13 +199,13 @@ Because $ψ_- = 0$ between the two atoms, and $ψ_+$ is not, we conclude that $E
### Bonding vs antibonding
If we decrease the interatomic distance, the two atoms get closer and their atomic orbitals have more overlap.
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.
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.
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@@ -218,7 +218,7 @@ On the other hand, if each atom has 0 or 2 electrons in the outermost shell, the
### 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 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.
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@@ -260,7 +260,7 @@ where $V_0>0$ is the potential strength, $\hat{p}$ the momentum of the electron,
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 $E<0$.
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:
$$
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@@ -271,20 +271,20 @@ where $V_0>0$ is the potential strength, $\hat{p}$ the momentum of the electron,
Let us apply the LCAO model to solve this problem. Consider the trial wave function for the ground state to be a linear combination of two orbitals $|1⟩$ and $|2⟩$:
$$|ψ⟩ = φ_1|1⟩ + φ_2|2⟩.$$
The orbitals $|1⟩$ and $|2⟩$ correspond to the wave functions of the electron when there is only a single delta peak present:
The orbitals $|1⟩$ and $|2⟩$ correspond to the wave functions of the electron when there is only a single delta peak present:
$$H_1 |1⟩ = \epsilon_1 |1⟩,$$
$$H_2 |2⟩ = \epsilon_2 |2⟩.$$
We start of by calculating the wavefunction of an electron bound to a single delta-peak.
To do so, you first need to set up the Schrödinger equation of a single electron bound to a single delta-peak.
To do so, you first need to set up the Schrödinger equation of a single electron bound to a single delta-peak.
You do not have to solve the Schrödinger equation twice—you can use the symmetry of the system to calculate the wavefunction of the other electron bound to the second delta-peak.
1. Find the expressions for the wave functions of the states $|1⟩$ and $|2⟩$: $ψ_1(x)$ and $ψ_2(x)$.
Also find an expression for their energies $\epsilon_1$ and $\epsilon_2$.
Remember that you need to normalize the wave functions.
2. Construct the LCAO Hamiltonian. To simplify the calculations, assume that the orbitals are orthogonal.
3. Diagonalize the LCAO Hamiltonian and find an expression for the eigenenergies of the system.
2. Construct the LCAO Hamiltonian. To simplify the calculations, assume that the orbitals are orthogonal.
3. Diagonalize the LCAO Hamiltonian and find an expression for the eigenenergies of the system.
It was previously mentioned that $V_0>0$.
Using this, determine which energy corresponds to the bonding energy.
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@@ -300,7 +300,7 @@ $$
H_{\textrm{eff}} = \begin{pmatrix}
E_0&&-t \\
-t&&E_0
\end{pmatrix}.
\end{pmatrix}.
$$
1. Let us add an electric field $\mathcal{E} \hat{\bf{x}}$ to the system.
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@@ -324,4 +324,6 @@ $$
$$
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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[^1]:See the book exercise 6.5 for relaxing the orthogonality assumption.
