💼 Case: consider the Einstein model, lattice as many independent simple harmonic oscillators, where the centre atom vibrated under the influence of electrostatic forces of the other atoms but has no effect them
When a wave propagates in a real crystal atoms move in phase as longitudinal or transverse waves
left longitudinal and right transverse
If we consider $u_n$ the displacement of an atomic plane $n$
A visual demonstration of waves in a crystal
💼 Case: consider a string of atoms of mass $M$, of distance $a$, displaced with spring constant $K$
The forces experienced by the $n$th atoms are
1️⃣ Force to the left $F_l=K(u_n-u_{n-1})$
2️⃣ Force to the right $F_l=K(u_{n+1}-u_n)$
The equation of motion is then
$$ M\ddot u_n=F_r-F_f=K(u_{n+1}-2u_n+u_{n-1}) $$
If we assume all atoms have the same maximum amplitude $A$ the solution is
$$ u_n=A\exp[i( kx^0_n-\omega t)] $$
where $x^0_n=na$ is the equilibrium position of the $n$th atom
Subbing in we get
$$ \begin{aligned} -\omega^2 MAe^{i(kna-\omega t)}&=KA\left ( e^{i[k(n+1)a-\omega t]}-2e^{i(kna-\omega t)}+e^{i[k(n-1)a-\omega t]} \right ) \\ -\omega^2 M&=K(e^{ika}-2+e^{-ika})=2K[\cos(ka)-1] \end{aligned} $$
Thus we get the dispersion relation $\omega(k)$
$$ \omega^2M=4K \sin^2 \left ( \frac{ka}{2} \right ) \qquad \omega (k)=2\sqrt{\frac{K}{M}}\left | \sin\left ( \frac{ka}{2} \right ) \right | $$
🗒️ Note: The function $\omega(k)$ is periodic, with period $2\pi/a$
The range from $-\pi/a\le k\le \pi/a$ is known as the first Brillouin zone
For small $k$ we can approximate
$$ \omega= ka\sqrt{\frac{K}{M}} $$
with this approximation $\omega\propto k$ then as $k\to 0$, $\omega\to 0$, which gives
$$ \omega=v_p k \qquad k\to 0 \qquad v_p=a\sqrt{\frac{K}{M}} $$
where $v_p$ is the speed of sound
🗒️ Note: longitudinal is acoustic mode, with velocity independent of $\omega$ when $k=0$ in phase
For large $k$ we can approximate
$$ \frac{\text d\omega}{\text dk}\to 0 \qquad \text{as} \quad k\to\frac{\pi}{a} $$
the group velocity $v_g\to 0$
💎 Conclusion: at $k=\pm \pi/a$ we have a stationary wave, $\max\omega=2\sqrt{\frac KM}$ cut-off frequency
We can show that waves can be considered only in the first Brillouin zone (BZ)
💼 Case: Consider two chains of ions/atoms with masses $M$ and $m$ with spacing $a/2$
Consider only nearest-neighbour interactions we have
$$ \begin{aligned} M\ddot u_n&=K(u_{n+1}-2u_n+u_{n-1}) \\ m\ddot u_{n-1}&=K(u_n-2u_{n-1} +u_{n-2})
\end{aligned} $$
For atoms of mass $M$ and $m$ we assume solution
$$ \begin{aligned} \text{for }M: \quad \quad \; \; \,\, u_n&=A\exp[i(k x^0_n-\omega t)] \\ \text{for }m: \qquad u_{n-1}&=\alpha A\exp[i(kx^0_{n-1}-\omega t)]
\end{aligned} $$
where $x^0_n=na/2$ and $\alpha$ describes amplitude and phase for $m$ atoms relative to $M$ atoms
Substituting the equations we get
$$ \begin{aligned} \text{for }M: \quad -\omega^2M&=2K\left ( \alpha \cos \frac{ka}{2}-1 \right ) \\ \Rightarrow \quad \alpha &=\frac{2K-\omega^2 M}{2K\cos(ka/2)} \\ \text{for }m: \quad-\alpha \omega^2 m&=2K\left ( \cos \frac{ka}{2}-\alpha \right ) \\ \Rightarrow \quad \alpha &=\frac{2K\cos(ka/2)}{2K-\omega^2 m}
\end{aligned} $$