Derivation: Dispersion Relation of the Diatomic Lattice Vibration

The goal is to derive a dispersion relation $ \omega_{\pm}(k) $ for an infinitely extended crystal with a diatomic basis. The basis contains two atoms with mass $ m_1 $ and mass $ m_2 $.

Hover the image!

A lattice plane $ n $ contains two subplanes (chains) with atoms of different mass. Planes shown in pale represent the equilibrium position.

Just as in deriving the dispersion relation for a monatomic basis, we make the approximation that a deflection of the $n$ lattice plane has an effect only on the neighboring lattice planes - that is, only an effect on the $n+1$ and $n-1$ lattice planes, but not, for example, on the $n+2$ lattice plane and so on. Außerdem sollen die Auslenkungen orthogonal zur jeweiligen Netzebene sein, wie in der Illustration 1 gezeigt.

With $ u_n $ we denote the deflection of the $n$th plane of the lattice from the equilibrium position. In this plane there are the masses $m_1$.
With $ y_n $ we denote the deflection of the $n$-th lattice plane from the rest position. In this second $n$-th plane there are the masses $m_2$.

In this problem, we need to set up two differential equations for the deflections $y_n$ and $u_n$. For this we use the Hook's spring law and equate it with Newton's 2nd law of motion:

Differential equation for the deflection of the lattice plane with the masses $m_1$

$$ \begin{align} m_1 \, \frac{\partial^2 u_{n}}{\partial t^2} ~&=~ D*(y_n ~-~ u_n) ~+~ D*(y_{n-1} ~-~ u_n) \end{align} $$

Differential equation for the deflection of the lattice plane with the masses $m_2$

$$ \begin{align} m_2 \, \frac{\partial^2 y_{n}}{\partial t^2} ~=~ D*(u_n ~-~ y_n) ~+~ D*(u_{n+1} ~-~ y_n) \end{align} $$

Here, $ D $ is the spring constant that couples two adjacent lattice planes. For Eqs. 1 and 2, let's first multiply out the parentheses on the right-hand side and then factor out $D$. Then we bring everything to the left side:

$$ \begin{align} m_1 \, \frac{\partial^2 u_{n}}{\partial t^2} ~+~ D*(2u_n ~-~ y_n ~-~ y_{n-1}) ~=~ 0 \end{align} $$

$$ \begin{align} m_2 \, \frac{\partial^2 y_{n}}{\partial t^2} ~+~ D*(2y_n ~-~ u_n ~-~ u_{n+1}) ~=~ 0 \end{align} $$

As a solution ansatz for the two differential equations 3 and 4 we take the exponential ansatz:

Exponential ansatz for the first deflection

$$ \begin{align} u_n(k) ~=~ \frac{1}{\sqrt{m_1}} \, C_u \mathrm{e}^{\mathrm{i}(kna - \omega t)} \end{align} $$

Exponential ansatz for the second deflection

$$ \begin{align} y_n(k) ~=~ \frac{1}{\sqrt{m_2}} \, C_y \mathrm{e}^{\mathrm{i}(kna - \omega t)} \end{align} $$

Here $ k $ is the wave number and $ \omega $ is the frequency of the oscillation of the lattice plane in the crystal. $C_u $ are $ C_y$ are unknown constants.

Insert exponential ansatzes into the first differential equation:
Before we can insert the exponential ansatz 5 into the first differential equation 3, we have to differentiate it twice with respect to time $t$. And since in the differential equation also the deflection $y_{n-1}$ occurs, we adapt the exponential ansatz 6 for it:

$$ \begin{align} \frac{\partial^2 u_{n}}{\partial t^2} ~&=~ \frac{-\omega^2 \, C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} \\\\
y_{n-1} ~&=~ \frac{C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}ka - \mathrm{i}\omega t} \end{align} $$

Now we can substitute the derivative and exponential functions 5, 6, and 7 into the first differential equation 3:

$$ \begin{align} m_1 \, \frac{-\omega^2 \, C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} & ~+~ 2D\, \frac{C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} \\\\
~-~ \frac{D \, C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} & ~-~ \frac{D \, C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}ka - \mathrm{i}\omega t} ~=~ 0 \end{align} $$

Now we only have to simplify the equation 8. To do this, factor out $ \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} $ and divide the equation by this factor:

$$ \begin{align} m_1 \, \frac{-\omega^2 \, C_u}{\sqrt{m_1}} ~+~ \frac{2D\, C_u}{\sqrt{m_1}} ~-~ \frac{D \, C_y}{\sqrt{m_2}} ~-~ \frac{D \, C_y}{\sqrt{m_2}} \, \mathrm{e}^{-\mathrm{i}ka} ~=~ 0 \end{align} $$

Factor out $\frac{C_u}{\sqrt{m_1}}$ for the first two summands and $\frac{D \, C_y}{\sqrt{m_2}}$ for the last two summands:

$$ \begin{align} \frac{C_u}{\sqrt{m_1}} \, (2D - \omega^2 \, m_1) ~-~ \frac{D \, C_y}{\sqrt{m_2}} \, (1+\mathrm{e}^{\mathrm{i}ka}) ~=~ 0 \end{align} $$

Next, we multiply 10 by $ 1/\sqrt{m_1} $ and rearrange the equation so that the $C_u$ and $C_y$ are preceded by the coefficients. This will be our first linear equation:

$$ \begin{align} \left(\frac{2D}{m_1} - \omega^2\right) \, C_u ~-~ \frac{D}{\sqrt{m_2 \, m_1}} \, \left( 1+\mathrm{e}^{\mathrm{i}ka} \right) \, C_y ~=~ 0 \end{align} $$

Insert exponential functions into the second differential equation:
We proceed analogously with the second differential equation. For this we first differentiate $y_n$ twice with respect to $t$ and adjust the exponential ansatz 5 for the deflection $u_{n+1}$, because it also occurs in the second differential equation:

$$ \begin{align} \frac{\partial^2 y_{n}}{\partial t^2} ~&=~ \frac{-\omega^2 \, C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} \\\\
u_{n+1} ~&=~ \frac{C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t}\,\mathrm{e}^{ika} \end{align} $$

Now we can substitute the second derivative and the exponentials 12, 5, and 6 into the second differential equation 4:

$$ \begin{align} m_2 \, \frac{-\omega^2 \, C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} & ~+~ \frac{2D\,C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} \\\\
~-~ \frac{D \, C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} & ~-~ \frac{D \, C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t}\,\mathrm{e}^{ika} ~=~ 0 \end{align} $$

Let's bracket out $ \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} $ and then divide by this factor. Then we multiply the equation by $ 1/\sqrt{m_2} $ and rearrange it so that before the $C_y$ and $C_u$ we have the coefficients:

$$ \begin{align} \left( \frac{2D}{m_2} - \omega^2 \right) \, C_y ~-~ \frac{D}{\sqrt{m_2 \, m_1}} \left( 1+\mathrm{e}^{\mathrm{i}ka} \right) \, C_u ~=~ 0 \end{align} $$

Combine result:
By setting up the differential equations and plugging in the exponential functions (solutions), we set up a system of linear equations:

Linear system of equations

$$ \begin{align} \left(\frac{2D}{m_1} - \omega^2\right) \, C_u ~-~ \frac{D}{\sqrt{m_2 \, m_1}} \, \left( 1+\mathrm{e}^{\mathrm{i}ka} \right) \, C_y ~&=~ 0 \\\\
\left( \frac{2D}{m_2} - \omega^2 \right) \, C_y ~-~ \frac{D}{\sqrt{m_2 \, m_1}} \left( 1+\mathrm{e}^{\mathrm{i}ka} \right) \, C_u ~&=~ 0 \end{align} $$

We can express this linear system of equations in matrix notation:

$$ \begin{align} \begin{bmatrix} \frac{2D}{m_1}-\omega^2 & -\frac{D(1+\mathrm{e}^{-\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} \\ -\frac{D(1+\mathrm{e}^{\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} & \frac{2D}{m_2}-\omega^2 \end{bmatrix} \begin{bmatrix} C_u\\\\ C_y\end{bmatrix} ~=~ \begin{bmatrix} 0 \\\\ 0 \end{bmatrix} \end{align} $$

From mathematics we know that for any $ C_y$ and $C_u $ the determinant of the above matrix must be zero. Only then the eigenvalue equation 16 with eigenvalue $ 0 $ is satisfied:

$$ \begin{align} \det\left(\begin{array}{c} \frac{2D}{m_1}-\omega^2 & -\frac{D(1+\mathrm{e}^{-\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} \\ -\frac{D(1+\mathrm{e}^{\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} & \frac{2D}{m_2}-\omega^2 \end{array}\right) ~\stackrel{!}{=}~ 0 \end{align} $$

Why does the determinant have to be zero?

Let's abbreviate the matrix elements as $a,b,c,d$:

$$ \begin{align} \begin{bmatrix} a & b\\ c & d \end{bmatrix} \begin{bmatrix} C_u\\ C_y \end{bmatrix} ~=~ \begin{bmatrix} 0 \\ 0 \end{bmatrix} \end{align} $$

The matrix equation corresponds to the following linear system of equations that we have to solve:

$$ \begin{align} a \, C_u ~+~ b \, C_y ~&=~ 0 \\
c \, C_u ~+~ d \, C_y ~&=~ 0 \end{align} $$

Rearrange the first equation for $C_u$ and substitute it into the second equation of the linear system:

$$ \begin{align} \frac{-c\,b}{a}\,C_y ~+~ d\,C_y ~=~ 0 \end{align} $$

Divide the equation by $C_y$:

$$ \begin{align} -c\,b ~+~ d \, a ~=~ 0 \end{align} $$

Using Laplace's expansion, you easily find that the determinant of this 2x2 matrix is equal to $-c\,b ~+~ d\,a $. And we found out that the determinant must be zero.

Determinant of the matrix 17 is easy to determine with the Laplace expansion. If you don't know how to do that, look at equations 18 and 21. The determinant found, as we have already reasoned, must be zero:

$$ \begin{align} \left( \frac{2D}{m_1} - \omega^2 \right) \, \left( \frac{2D}{m_2} - \omega^2 \right) ~-~ \frac{D \, (1+\mathrm{e}^{\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} \, \frac{D \, (1+\mathrm{e}^{-\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} ~=~ 0 \end{align} $$

Multiply out the parentheses in Eq. 22:

$$ \begin{align} \frac{4D^2}{m_1 \, m_2} ~-~ \frac{2D}{m_1} \, \omega^2 ~-~ \frac{2D}{m_2} \, \omega^2 ~+~ \omega^4 ~-~ \frac{D^2}{m_1 \, m_2}(2+\mathrm{e}^{\mathrm{i}ka} + \mathrm{e}^{\mathrm{i}ka}) ~=~ 0 \end{align} $$

Sort eq. 23 by the power of the frequency $ \omega $. Also, you can rewrite the exponential functions to cosine using Euler's formula:

$$ \begin{align} \omega^4 ~-~ 2D \, \left( \frac{1}{m_1} + \frac{1}{m_2} \right) \, \omega^2 ~+~ \frac{2D^2}{m_1 \, m_2} \, \left( 1-\cos(ka) \right)~=~ 0 \end{align} $$

As you can see, it is a quartic equation for the frequency $\omega$. We have to rearrange this for $\omega$, because we want to determine the dispersion relation. Substitute $ \omega^4 := \omega_{\pm}^2 $ to make it a quadratic equation. The two solutions of the quadratic equation give you the quadratic formula:

$$ \begin{align} \omega_{\pm}^2 ~=~ D \, \left( \frac{1}{m_1} + \frac{1}{m_2} \right) ~\pm~ D \, \sqrt{\left(\frac{1}{m_1} + \frac{1}{m_2}\right)^2 ~-~ \frac{2}{m_1 \, m_2}\,\left( 1-\cos(ka) \right) } \end{align} $$

Thus, we have found out the dispersion relation for a crystal with a diatomic basis. We can rewrite the solution 25 (if we want) with the trigonometric relation $ \sin^2(x) ~=~ \frac{1}{2}(1-\cos(2x)) $ like this:

Dispersion relation for a diatomic crystal lattice with neighbor interaction

$$ \begin{align} \omega_{\pm}^2 ~=~ D \, \left( \frac{1}{m_1} + \frac{1}{m_2} \right) ~\pm~ D \, \sqrt{\left(\frac{1}{m_1} + \frac{1}{m_2}\right)^2 ~-~ \frac{4}{m_1 \, m_2}\,\sin^2\left(\frac{k\,a}{2}\right) } \end{align} $$

Note that this dispersion relation is only valid for a crystal in which the lattice planes oscillate purely longitudinally (or transversely) and the deflection of one lattice plane only has an influence on the neighboring lattice planes.

Only positive frequencies are physically meaningful, that means: If you take the root of 26, you get two solutions:

Solution $ \omega_- $ is called acoustic dispersion branch.
Solution $ \omega_+ $ is called optical dispersion branch.

Why are lattice vibrations called optical and acoustic?