The diatomic chain

Exercises

Preliminary provocations

1. Verify that the expression for $\omega^2$ is always positive. Why is this important?

   We rewrite the equation for $\omega^2$ to the form of excercise 1. The term under the square root is always smaller than 1, from which we conclude that $\omega^2 \geq 0$. This is important, because a negative value would yield imaginary solutions for $\omega$, which does not correspond to an oscillating motion of the atoms (plug it into the Ansatz to see this)

2. When calculating the DOS, we only look at the first Brillouin zone. Why?

   Values of $k$ that differ by an integer multiple of $2\pi/a$ describe the same wave. Therefore, all information is already contained in the 1st Brillouin zone.

Exercise 1: analysing the diatomic vibrating chain

As we have shown, the normal modes of oscillation of a diatomic chain occur at frequencies:

$$\omega_\pm^2 = \frac{\kappa_1+\kappa_2}{m} \pm \sqrt{\frac{(\kappa_1+\kappa_2)^2 - 4\kappa_1\kappa_2\sin^2(ka/2)}{m^2}}$$

where the plus sign corresponds to the optical branch and the minus sign to the acoustic branch.

Hint

The final form of $\omega_\pm$ as given is not always the most useful: sometimes the complex exponential form - see 3.2.2 - can make life easier

1. Find the magnitude of the group velocity near $k=0$ for the acoustic branch.
2. Show that the group velocity at $k=0$ for the optical branch is zero.
3. Derive an expression of the density of states $g(ω)$ 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 of the Debye lecture.

Exercise 2: atomic chain with 3 different spring constants

Suppose we have a vibrating 1D atomic chain with 3 different spring constants alternating like $\kappa_ 1$, $\kappa_2$, $\kappa_3$, $\kappa_1$, etc. All the atoms in the chain have an equal mass $m$.

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?

1. Make a sketch of this chain and indicate the length of the unit cell $a$ in this sketch.

   The unit cell should contain exactly one spring of $\kappa_1$, $\kappa_2$ and $\kappa_3$ and exactly three atoms.

2. Derive the equations of motion for this chain.

   The equations of motion are

   $$\begin{align} m\ddot{u}_{n,1} &= -\kappa_1(u_{n,1} - u_{n,2}) - \kappa_3(u_{n,1} - u_{n-1,3})\\ m\ddot{u}_{n,2} &= -\kappa_2(u_{n,2} - u_{n,3}) - \kappa_1(u_{n,2}-u_{n,1})\\ m\ddot{u}_{n,3} &= -\kappa_3(u_{n,3} - u_{n+1,1}) - \kappa_2(u_{n,3}-u_{n,2}) \end{align}$$

3. By filling in the trial solutions into the equations of motion (which should be similar to those used in the dual spring constants case), show that the eigenvalue problem is $$\omega^2 \begin{pmatrix} A_1 \\ A_2 \\ A_3 \end{pmatrix} = \frac{1}{m} \begin{pmatrix} \kappa_1 + \kappa_ 3 & -\kappa_ 1 & -\kappa_ 3 e^{i k a} \\ -\kappa_ 1 & \kappa_1+\kappa_2 & -\kappa_ 2 \\ -\kappa_ 3 e^{-i k a} & -\kappa_2 & \kappa_2 + \kappa_ 3 \end{pmatrix} \begin{pmatrix} A_1 \\ A_2 \\ A_3 \end{pmatrix}$$

Substitute the following Ansatz into the equations of motion:

$$
\begin{pmatrix} u_{1,n} \\ u_{2,n} \\ u_{3,n} \end{pmatrix} = e^{i\omega t - ikx_n} \begin{pmatrix} A_1 \\ A_2 \\ A_3 \end{pmatrix}
$$

1. 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 $\kappa_1 = κ_2 = q$.

   Making use of the symmetry, we guess the form of the eigenvectors. As always, it is very useful to make a drawing. From a drawing of the unit cell, we expect an eigenvector $\mathbf{v_1} = [1 \quad 1 \quad 1]^\text{T}$ (all atoms moving in the same direction) with eigenfrequency equal to zero, an eigenvector $\mathbf{v_2} = [1 \quad a \quad 1]^\text{T}$ (outer atoms moving oppositely to the middle atom, with the middle atom having twice the amplitude), and an eigenvector $\mathbf{v_3} = [1 \quad 0 \quad -1]^\text{T}$

   We find the eigenvalues and the value of $a$ by multiplying the spring matrix by these eigenvectors, yielding $a=-2$ and

   $$\omega^2 = \begin{pmatrix} \omega_1^2 \\ \omega_2^2 \\ \omega_3^2 \end{pmatrix} = \frac{\kappa}{m} \begin{pmatrix} 0 \\ 3 \\ 1+2\kappa_3/\kappa \end{pmatrix}$$

   where we defined $\kappa = \kappa_1=\kappa_2$. More formally, we can use that because the spring matrix and the matrix $X$ commute, they share a common set of eigenvectors. The eigenvalues of the matrix $X$ are $\lambda = -1$ with an eigenvector $[1 \quad 0 \quad -1]^\text{T}$, and $\lambda = +1$ with eigenvectors $[ 1 \quad 0 \quad 1]^\text{T}$ and $[0 \quad 1 \quad 0 ]^\text{T}$.

   These eigenvectors can be used to calculate the eigenvalues of the spring matrix. However, be cautious! The eigenvalue $\lambda = +1$ is degenerate and to find the eigenvalue of the spring matrix we should take a linear combination of the two corresponding eigenvectors.

2. What will happen to the periodicity of the band structure if $\kappa_ 1 = \kappa_ 2 = \kappa_3$?

   If $\kappa_1 = \kappa_2 = \kappa_3$ then we have the uniform mono-atomic chain. This mono-atomic chain has unit cell of size $a/3$, so the Brillouin zone will span $-3\pi/a$ to $3\pi/a$. If we then make one of the spring constants slightly different, bandgaps will open at at $k=pi/a$ and $k=2\pi/a$, reflecting the new periodicity of the lattice.

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Last update: October 30, 2023

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