@@ -30,32 +30,32 @@ _(based on chapters 10-11 of the book)_
- explain what happens with the band structure when the periodicity of the lattice is increased or reduced.
### More degrees of freedom per unit cell:
In the last lecture, we modeled phonons through a 1D homogeneous chain of atoms.
The model gave us insight into phonons and justified the crude approximations of the Debye model.
Despite the model's usefulness, we are not always able to model problems as a *homogeneous* chain.
Therefore, let us tackle a slightly more general problem.
In the last lecture, we modeled phonons through a 1D homogeneous chain of atoms.
The model gave us insight into phonons and justified the crude approximations of the Debye model.
Despite the model's usefulness, we are not always able to model problems as a *homogeneous* chain.
Therefore, let us tackle a slightly more general problem.
Consider a chain of atoms with two masses, $m_1$ and $m_2$.
In our chain, an atom with mass $m_1$ is always followed by an atom of mass $m_2$ and vice versa.
Similar to before, the atoms interact via a harmonic potential with spring constant $\kappa$.
Similar to before, the atoms interact via a harmonic potential with spring constant $κ$.
In the last lecture, we solved the homogenous chain problem by identifying its high level of symmetry.
In the last lecture, we solved the homogenous chain problem by identifying its high level of symmetry.
However, the problem here is not as symmetric.
We need to re-think and generalize our ansatz.
We need to re-think and generalize our ansatz.
First we identify a pattern that does repeat: an atom with mass $m_1$ followed by one with mass $m_2$.
This is called a **unit cell** (the smallest repeated element of the system)
We now label all degrees of freedom in a unit cell.
The two atoms in a unit cell have displacements $u_{1,n}$ and $u_{2,n}$, where the first index specifies the atom number within the unit cell and the second the unit cell number.
The two atoms in a unit cell have displacements $u_{1,n}$ and $u_{2,n}$, where the first index specifies the atom number within the unit cell and the second the unit cell number.
In our choice of unit cell, atom 1 has mass $m_1$ and atom 2 has mass $m_2$.
Having specified the degrees of freedom, let's write down the equations of motion:
Looking at the eigenvectors we see that for every $k$ there are now two values of $\omega$: one corresponding to in-phase motion (–) and anti-phase (+).
Looking at the eigenvectors we see that for every $k$ there are now two values of $ω$: one corresponding to in-phase motion (–) and anti-phase (+).
@@ -134,15 +134,15 @@ Unlike the simple atomic chain, the dispersion relation now has two branches (or
The reason an additional branch appears in the solution is due to the 2 degrees of freedom per unit cell.
If we had started with 3 different atoms, the eigenvalue problem would contain 3 equation, so that there would be three eigenfrequencies per $k$ and the dispersion relation would have 3 branches.
The lower branch is called _acoustic_ because its linear dispersion near $\omega=0$ matches the behavior of regular sound waves.
The lower branch is called _acoustic_ because its linear dispersion near $ω=0$ matches the behavior of regular sound waves.
The upper branch is the _optical branch_ because it can cross with the (linear) dispersion relation of photons, allowing these phonons to efficiently emit and absorb photons.
Like before, the phonon group velocity is $v=\textrm{d}\omega/\textrm{d}k$, and the density of states is $g(\omega)=\textrm{d}N/\textrm{d}ω = \frac{L}{2π} ∑| \textrm{d}k/\textrm{d} ω|$.
The sum goes over all states at a given energy.
Like before, the phonon group velocity is $v=\textrm{d}ω/\textrm{d}k$, and the density of states is $g(ω)=\textrm{d}N/\textrm{d}ω = \frac{L}{2π} ∑| \textrm{d}k/\textrm{d} ω|$.
The sum goes over all states at a given energy.
In this case, the sum ensures we include the contribution to the DOS from both the positive and negative momenta.
Since the energy of this system is symmetric with respect to momentum reversal and spin reversal, the sum only introduces a factor of 4.
An intuitive way to visualize the density of states $g(ω}$ is to consider it a histogram of the samples drawn from the dispersion relation $\omega(k)$:
An intuitive way to visualize the density of states $g(ω)$ is to consider it a histogram of the samples drawn from the dispersion relation $ω(k)$:
Note that $g(\omega)$ is generally plotted along the vertical axis and $\omega$ along the horizontal axis – the right plot above is just to demonstrate the relation between the dispersion and the DOS. The singularities in $g(\omega)$ at the bottom and top of each branch are called _van Hove singularities_.
Note that $g(ω)$ is generally plotted along the vertical axis and $ω$ along the horizontal axis – the right plot above is just to demonstrate the relation between the dispersion and the DOS. The singularities in $g(ω)$ at the bottom and top of each branch are called _van Hove singularities_.
### Consistency check with 1 atom per cell
To check if our result is consistent with the previous lecture, we examine what happens when we take $m_1\rightarrow m_2$.
@@ -216,7 +216,7 @@ In the 1D monatomic chain, $a$ was the distance between two neighbouring atoms,
Therefore we take the size of our unit cell in the diatomic case to be $2a$, so the physical systems are the same.
Looking at the graph we see that doubling the lattice constant "folds" the band structure on itself.
There are then $2$ possible values of $\omega$ for each $k$, creating the $2$ branches of the non-periodic system.
There are then $2$ possible values of $ω$ for each $k$, creating the $2$ branches of the non-periodic system.
We can look at the graph in two distinct ways:
We can consider the green line (monatomic chain solution). Its Brillouin zone extends from $0$ to $\pi / a$.
Alternatively, we consider the orange and blue lines together. In this case, the Brillouin zone is smaller ranging from $0$ to $\pi / 2a$ (because of the larger lattice constant).
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@@ -234,14 +234,14 @@ And although the band structures are different, working out the density of state
## Exercises
### Warm-up questions
1. Verify that the expression for $\omega^2$ is always positive. Why is this important?
2. Work out the expression of $\omega^2$ in the case $m_1 = m_2$. Compare this to the solution for the monatomic chain.
1. Verify that the expression for $ω^2$ is always positive. Why is this important?
2. Work out the expression of $ω^2$ in the case $m_1 = m_2$. Compare this to the solution for the monatomic chain.
3. When calculating the DOS, we only look at the first Brillouin zone. Why?
### Exercise 1: analyzing the diatomic vibrating chain
As we have derived, the eigenfreqencies of a diatomic vibrating chain with 2 different masses are:
where the plus sign corresponds to the optical branch and the minus sign to the acoustic branch.
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@@ -251,7 +251,7 @@ where the plus sign corresponds to the optical branch and the minus sign to the
Make use of a Taylor expansion.
2. Show that the group velocity at $k=0$ for the _optical_ branch is zero.
3. Derive an expression of the density of states $g(\omega)$ for the _acoustic_ branch and small $k$. Make use of your expression of the group velocity in 1.
3. Derive an expression of the density of states $g(ω)$ for the _acoustic_ branch and small $k$. Make use of your expression of the group velocity in 1.
Compare this expression with that of the derived density of states from [exercise 1](2_debye_model/#exercise-1-debye-model-concepts) of the Debye lecture.
### Exercise 2: the Peierls transition
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@@ -261,7 +261,11 @@ The spacing of the distorted chain alternates between two different distances an
Due to the alternating hopping energies, we must treat two consecutive atoms as two different orbitals ($\left| n,1 \right>$ and $\left| n,2 \right>$ in the figure) from the same unit cell. The corresponding LCAO of this chain is given by $$\left|\Psi \right> = \sum_n \left( \phi_n \left| n,1 \right> + \psi_n \left| n,2 \right>\right)$$ As usual, we assume that all these atomic orbitals are orthogonal to each other.
Due to the alternating hopping energies, we must treat two consecutive atoms as two different orbitals ($\left| n,1 \right⟩$ and $\left| n,2 \right⟩$ in the figure) from the same unit cell. The corresponding LCAO of this chain is given by
As usual, we assume that all these atomic orbitals are orthogonal to each other.
1. Indicate the length of the unit cell $a$ in the figure.
2. Using the Schrödinger equation, write the equations of motion of the electrons.
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@@ -281,13 +285,13 @@ Suppose we have a vibrating 1D atomic chain with 3 different spring constants al
2. Derive the equations of motion for this chain.
3. By filling in the trial solutions into the equations of motion (which should be similar to Ansazt used in the lecture), show that the eigenvalue problem is
4. In general, the eigenvalue problem above cannot be solved analytically, and can only be solved in specific cases. Find the eigenvalues $\omega^2$ when $k a = \pi$ and $\kappa_1 = \kappa_ 2 = q$.
4. In general, the eigenvalue problem above cannot be solved analytically, and can only be solved in specific cases. Find the eigenvalues $ω^2$ when $k a = \pi$ and $\kappa_1 = κ_2 = q$.
??? hint
To solve the eigenvalue problem quickly, make use of the fact that the mass-spring matrix in that case commutes with the matrix $$ X = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}. $$ What can be said about eigenvectors of two matrices that commute?
