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PHY.K02UF Molecular and Solid State Physics

## 1-d chain of atoms with two different masses

Consider a one-dimensional crystal with two atoms in the basis.

If the motion of the atoms is confined to one dimension, Newton's law yields the following equations of motion.

$$$M_1\frac{d^2u_l}{dt^2}=C(v_{l-1}-2u_l+v_l)\\ M_2\frac{d^2v_l}{dt^2}=C(u_{l}-2v_l+u_{l+1})$$$

Here $M_1$ is the mass of the heavy atoms, $M_2$ is the mass of the light atoms, $u_l$ is the displacement of the heavy atom in unit cell $l$ from its equilibrium position, $v_l$ is the displacement of the light atom in unit cell $l$ from its equilibrium position, and $C$ is the spring constant. To find the normal modes, we assume that all of the atoms move with the same frequency,

$$$u_l=u_k e^{i(kla-\omega t)},\\ v_l=v_ke^{i(kla-\omega t)}.$$$

Substituting this solution into the equations of motion yields,

$$$-\omega^2M_1u_k=-2Cu_k+Cv_k(1+\exp(-ika)),\\ -\omega^2M_2v_k=Cu_k(1+\exp(ika))-2Cv_k.$$$

These equations can be written in matrix form,

$$\left[ \begin{matrix} \omega^2M_1 -2C & C(1+\exp(-ika)) \\ C(1+\exp(ika)) & \omega^2M_2 -2C \end{matrix} \right]\left[ \begin{matrix} u_k \\ v_k \end{matrix} \right]=0.$$

The equations will have a solution when the determinant of the matrix equals zero.

$$$M_1M_2\omega^4-2C(M_1+M_2)\omega^2 +2C^2(1-\cos (ka)) = 0.$$$

This equation is quadratic in $\omega^2$ and can be solved using the quadratic equation.

$$\omega^2=\frac{C}{M_1M_2}\left(M_1+M_2 \pm \sqrt{M_1^2+M_2^2+2M_1M_2\cos (ka)}\right).$$

The phonon dispersion relation and an animation of the normal modes is shown below. There are two normal modes for every value of $k$. The modes with the lower frequency are called the acoustic modes and the modes with the higher frequency are called the optical modes.

 $\large \frac{\omega}{\sqrt{2C\left(\frac{1}{M_1}+\frac{1}{M_2}\right)}}$ $\frac{M_1}{M_2}=$1.1  $ka=$ $ka$

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The and buttons let you change the wavenumber $k$ to display different normal modes.

From the matrix equation above we can show that the amplitudes of the normal modes satisfy,

$$\frac{u_k}{v_k} = \frac{C(1+\exp(-ika))}{2C-\omega^2M_1} =\frac{2C-\omega^2M_2}{C(1+\exp(ika))}$$

Near $ka=0$, the ratio of the amplitude of the acoustic mode is $\frac{u_k}{v_k} \approx 1$ and the atoms oscillate in-phase whereas for the optical branch $\frac{u_k}{v_k} \approx -\frac{M_2}{M_1}$ and the atoms oscillate out-of-phase.

At the Brillouin zone boundaries $ka = \pm \pi$: for the acoustic branch the light atoms are still and the heavy atoms oscillate whereas for the optical branch, the heavy atoms are still and the light atoms oscillate.

To plot the normal modes we take the real parts of $u_l=u_k e^{i(kla-\omega t)}$ and $v_l=v_ke^{i(kla-\omega t)}$.

For the acoustic branch, the real part of the heavy atom motion is $u_l= \cos\omega_{ac} t$ and the real part of the light atom motion is $v_l= \frac{C}{2C-\omega_{ac}^2M_2}\left((1+\cos(ka))\cos(kla-\omega_{ac} t)-\sin(ka)\sin(kla-\omega_{ac} t)\right)$.

For the optical branch, the real part of the light atom motion is $v_l= \cos\omega_{op} t$ and the real part of the heavy atom motion is $u_l= \frac{C}{2C-\omega_{op}^2M_1}\left((1+\cos(ka))\cos(kla-\omega_{op} t)+\sin(ka)\sin(kla-\omega_{op} t)\right)$.