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Introduction to 1D phonons

Original notebook: Phonon_1D.ipynb

Background theory

More on the background theory.

Tasks and exercises

1. Visualize the lattice vibration at $k=0$ and $k=\frac{\pi}{a}$. What characteristics do you observe ?

   Solution

   The atom indexed $n$ has a displacement given by $u=\exp(i(n\cdot ka-\omega t))$.

   At $k=0$, the displacement is given by $u=\exp(-i\omega t)$. Therefore, all atoms are displaced identically, which creates a translation. However, since $\omega=0$ at $k=0$, atoms stay static.

   At $k=\frac{\pi}{a}$, the displacement is given by $u=\exp(i(n\cdot\pi-\omega t))=\exp(i\cdot n\cdot\pi)\exp(-i\omega t)$. Therefore, the neighbouring atom at $n+1$ has an amplitude given by $u'=u\exp(i\pi)=-1$. Thus, neighbouring atoms oscillate out-of-phase.

   To make atoms oscillate out-of phase, a lot of energy is required since the phonon contains a lot of potential energy. As energy is directly linked to frequency with $E=h\nu$, the maximum of frequency is thus found at $k=\pm \frac{\pi}{a}$.

2. The first Brillouin zone is defined between $\frac{-\pi}{a}$ and $\frac{\pi}{a}$. Explain why the first Brillouin zone contains all the necessary information by looking at $k'=k+\frac{2\pi}{a}$

   Solution

   At $k'=k+2\frac{\pi}{a}$ the lattice vibration stays identical to the phonon with wave vector $k$. Therefore, the frequency will also be identical.

   From the equation of displacement, we get $u'=\exp(i(n\cdot k'a-\omega t))=\exp(i(n\cdot ka-\omega t))\exp(i\cdot n\cdot2\pi)=u$.

3. Visualize the phonon dispersion for a diatomic chain. What is the difference between the acoustic and optical branch at $k=0$ and $k=\pm\frac{\pi}{2a}$ ? How do the atoms displacements relate to frequency ?

   Solution

   At $k=0$, in the acoustic branch, as in the monoatomic chain, the atoms do not move. Instead, in the optical branch the two atoms are moving out-of-phase. The latter therefore has a high frequency.

   At $k=\pm\frac{\pi}{2a}$, in the acoustic branch, the light atoms stand still whearas the heavier atoms move. In the optical branch, the opposite is observed. Since frequency is proportional to $\sqrt{1/M}$, the frequency is higher when lighter atoms move.

4. Adjust the mass ratio of the molecules. What happens when the ratio is high ? How about when the ratio is close to unity ? How does the band dispersion change ?

   Solution

   As the ratio grows, the gap of forbidden frequencies grows as well. At $k=\pm\frac{\pi}{2a}$, the gap is proportional to $\sqrt{1/M_{gray}}-\sqrt{1/M_{red}}$.

   Furthermore, at $k=0$, we have $\frac{u_{red}}{u_{gray}}=\frac{-M_{gray}}{M_{red}}$, and thus the oscillation amplitude of the heavier atom decreases as it gets heavier.

5. Set the mass ratio to 1. How does it compare to the case of the monoatomic chain ?

   Solution

   When the ratio is 1, the chain is identical to the monoatomic one, except that the lattice parameter is twice as big, shrinking the Brillouin zone to half the size of the monoatomic case. One recognizes the dispersion relation of the monoatomic chain in the first BZ between $k=-\frac{\pi}{a}$ and $k=\frac{\pi}{a}$.

   The optical branch in the first BZ now represents the monoatomic dispersion relation but at $k'=k\pm\frac{\pi}{a}$. This is called band folding.

Interactive visualization

Using the interactive visualization

Phonon viewer

The bottom view shows an interactive animation of a lattice vibration.

• Use the Play / Pause button to start and stop the animation (it also stops while the view is off-screen).

• Rotate the surface by left clicking and dragging.

• Translate the surface by right clicking and dragging.

You can reset the view by clicking the corresponding ‘camera axis’ buttons.

Phonon

At the top are the parameters related to the atomic chain and the phonon dispersion plot.

Click on the phonon dispersion plot to get the corresponding lattice vibration.

Appearance

Under the view are parameters related to showing arrows along the oscillations or modifying the atoms appearance.