Question

In the first Brillouin zone of a rectangular lattice (lattice constants $a = 6$ \AA\ and $b = 4$ \AA), three incoming phonons with the same wave vector $\langle 1.2$ \AA$^{-1}, 0.6$ \AA$^{-1}\rangle$ interact to give one phonon. Which one of the following is the CORRECT wave vector of the resulting phonon?

• A. $\langle 2.56$ \AA$^{-1}, 0.23$ \AA$^{-1}\rangle$
• B. $\langle 3.60$ \AA$^{-1}, 1.80$ \AA$^{-1}\rangle$
• C. $\langle 0.48$ \AA$^{-1}, 0.23$ \AA$^{-1}\rangle$
• D. $\langle 3.60$ \AA$^{-1}, -0.80$ \AA$^{-1}\rangle$

Hint:

When three phonons interact in a lattice, the wave vector of the resulting phonon is the sum of the wave vectors of the incoming phonons. The result is then reduced into the first Brillouin zone.

Answer 1

The Correct Option is C

- The interaction of three phonons with the same wave vector leads to a phonon with a wave vector that is the sum of the individual wave vectors. Let the wave vector of each incoming phonon be $\vec{k_1} = \langle 1.2~\text{\AA}^{-1}, 0.6~\text{\AA}^{-1}\rangle$.
- The resulting wave vector of the new phonon is: \[ \vec{k_{\text{result}}} = 3 \times \vec{k_1} = 3 \times \langle 1.2~\text{\AA}^{-1}, 0.6~\text{\AA}^{-1}\rangle = \langle 3.6~\text{\AA}^{-1}, 1.8~\text{\AA}^{-1}\rangle. \] - However, since the phonon must lie within the first Brillouin zone, we must apply the concept of the reciprocal lattice and reduce the wave vector into the first Brillouin zone.
- For the reciprocal lattice vectors, the resulting phonon wave vector is reduced as: \[ \vec{k_{\text{result}}} = \langle 0.48~\text{\AA}^{-1}, 0.23~\text{\AA}^{-1}\rangle. \] Thus, the correct answer is (C) $\langle 0.48~\text{\AA}^{-1}, 0.23~\text{\AA}^{-1}\rangle$.