Dispersion Relation of the Lattice Vibration for Mono/Diatomic Basis

We want to derive a dispersion relation \(\omega(k)\) for a 1-atom, infinitely extended, one-dimensional chain (one-dimensional lattice planes). The chains are a distance \( a \) apart and contain atoms of mass \(m\). We can then use the dispersion relation to describe the oscillation of the atomic chain classically. Here \( \omega \) is the frequency and \( k \) is the wavenumber of the oscillation. For example, for an electromagnetic wave, the dispersion relation is linear: \( \omega(k) = c \, k \), with \( c \) as the speed of light. We want to calculate such a relation for vibrations in a monatomic crystal.

We numerate the chains with a natural number \(n \in \mathbb{N} \). To get to other chains we use an integer \(z \in \mathbb{Z} \) which we can add to or subtract from \(n\). If now the \((n+z)\)-th chain is orthogonally deflected by the amount \( u_{n+z} \), then this deflection has an effect on the \(n\)-th chain, which thereby experiences a deflection \( u_n \). Here we want to describe the acting force \( F_n \) on the \(n\)th chain by Hook's spring law: 1 $$F_n ~=~ \underset{z}{\boxed{+}} \, D_z \, (u_{n+z} - u_n)$$

Here, \( D_z \) is a spring constant that describes the strength of the coupling between the \((n+z)\)-th and \( n \)-th chains. Since we have many chains that can be coupled to the \(n\)th chain, we sum over \(z\).

If one of the atoms is deflected orthogonally (longitudinally) to the chain, then all the other atoms feel a force that also deflects them from their neutral position. However, they are not deflected in any direction, but also orthogonally to the chains. Why only orthogonally? If you project the forces acting on two opposite atoms within a chain onto the chain, then the projections cancel each other out. Only the orthogonal component of the force remains. An analogous consideration is also valid for purely parallel (transversal) deflection of an atom from its equilibrium position. Of course, this is not true for an arbitrary deflection of the atom, so we consider only the simple case of a longitudinal lattice vibration.

Equate Eq. 1 with Newton's 2nd law of motion \( m \, \frac{\text{d}^2 u_n}{\text{d} t^2} \) to obtain a differential equation for the deflection \(u\): 2 $$m \, \frac{\text{d}^2 u_n}{\text{d} t^2} ~=~ \underset{z}{\boxed{+}} \, D_z \, (u_{n+z} - u_n)$$

The solution of such a second order ordinary differential equation are harmonic functions. Let's make the following ansatz (exponential ansatz) for the deflection: 3 $$u_{n+z} ~=~ C \, \mathrm{e}^{\mathrm{i}(kza - \omega t)}$$

Here \(k\) is a wave number and \( \omega \) is the frequency of the wave, which results from the oscillation of the chains. And \(C\) is an unknown constant.

According to the solution ansatz, the \(n\)th deflection is (\(z=0\)): 4 $$u_{n} ~=~ C \, \mathrm{e}^{- \mathrm{i} \omega t)}$$

We have to differentiate the deflection 4 twice with respect to \(t\). Then we can insert the solution 3, 4 and the derivative into the differential equation 2: 5 $$-m \, C \, \omega^2 \mathrm{e}^{\mathrm{i} \omega t} ~=~ \underset{z}{\boxed{+}} \, D_z \, C \left( \mathrm{e}^{\mathrm{i}(kza-\omega t)} ~-~ \mathrm{e}^{-\mathrm{i}\omega t} \right)$$

Thereby the factor \( C \, \mathrm{e}^{\mathrm{i} \omega t} \) gets eliminated and with it the unknown constant \(C\). Let's bring everything to the left-hand side: 6 $$-m \, \omega^2 ~-~ \underset{z}{\boxed{+}} \, D_z \left( \mathrm{e}^{\mathrm{i} kza} ~-~ 1 \right) ~=~ 0$$

Since the chain is symmetric, the spring constant relation \( D_z = D_{-z} \) is valid. That is, both the \( n+z \) chain and the \( n-z \) chain are equally coupled with the \( n \) chain. With this information, Eq. 6 can be simplified: 7 $$-m \, \omega^2 ~-~ \underset{z}{\boxed{+}} \, D_z \left( \mathrm{e}^{\mathrm{i} kza} ~+~ \mathrm{e}^{-\mathrm{i}kza} ~-~ 2 \right) ~=~ 0$$

Next, we rewrite complex exponential functions using Euler's formula, bringing cosine and sine into play: 8 $$-m \, \omega^2 ~-~ \underset{z}{\boxed{+}} \, D_z \left( \cos(kza) + \mathrm{i}\sin(kza) + \cos(-kza) + \mathrm{i}\sin(-kza) ~-~ 2 \right) ~=~ 0$$

Exploit the symmetry and antisymmetry properties of cosine and sine, simplifying Eq. 8: 9 $$-m \, \omega^2 ~-~ 2\,\underset{z}{\boxed{+}} \, D_z \left( \cos(kza) ~-~ 1 \right) ~=~ 0$$

Now we got an equation which does not contain any deflection and depends only on constants, the angular frequency \(\omega\) and wavenumber \(k\). Since we are looking for the dispersion relation \( \omega(k) \), we need to rearrange Eq. 9 for the angular frequency \( \omega \): 10 $$\omega(k) ~=~ \sqrt{\frac{2}{m} \, \underset{z}{\boxed{+}} \, D_z \left( 1 ~-~ \cos(kza) \right)}$$

The dispersion relation 10 also takes into account the interaction between all the distant chains. If you consider only the interaction between adjacentchains, all coupling constants \( D_z \) with \( z \neq 1 \) are dropped. That means a deflection of the chain \( q=n+2\) has no effect on the chain \( n \). Only coupling constant \( D_1 \) remains, which we simply call \( D \): 11 $$\omega(k) ~=~ \sqrt{\frac{2 D}{m} \left( 1 ~-~ \cos(ka) \right)}$$

Using the double angle identity \( \sin^2(x) ~=~ \frac{1}{2}(1-\cos(2x)) \), Eq. 11 can also be written as follows: 12 $$\omega(k) ~=~ \sqrt{\frac{4 D}{m} \sin^2\left(\frac{ka}{2}\right)}$$

The dispersion relation 12 applies to a monatomic chain interacting only with its right and left neighboring chains and being deflected only longitudinally (or transversely). From the formula 12 for the dispersion relation we see that it is symmetrical and independent of the direction of propagation of the oscillation because of \( \sin^2 \): \( \omega(k) = \omega(-k) \) (see illustration).

We can also calculate from the dispersion relation, for example, the group velocity \(v_{\text g}\) if we differentiate Eq. 12 with respect to the wavenumber \(k\): 13 $$v_{\text g} ~=~ \sqrt{\frac{D \, a^2}{m}} \, \cos\left(\frac{1}{2} \, k \, a\right)$$

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 \( m_2 \).

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. In addition, the deflections should be orthogonal to the respective lattice planes, as shown in Illustration.

• 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: 1 $$m_1 \, \frac{\partial^2 u_{n}}{\partial t^2} ~=~ D*(y_n ~-~ u_n) ~+~ D*(y_{n-1} ~-~ u_n)$$ 2 $$m_2 \, \frac{\partial^2 y_{n}}{\partial t^2} ~=~ D*(u_n ~-~ y_n) ~+~ D*(u_{n+1} ~-~ y_n)$$

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: 3 $$m_1 \, \frac{\partial^2 u_{n}}{\partial t^2} ~+~ D*(2u_n ~-~ y_n ~-~ y_{n-1}) ~=~ 0$$ 4 $$m_2 \, \frac{\partial^2 y_{n}}{\partial t^2} ~+~ D*(2y_n ~-~ u_n ~-~ u_{n+1}) ~=~ 0$$

As a solution ansatz for the two differential equations 3 and 4 we take the exponential ansatz: 5 $$u_n(k) ~=~ \frac{1}{\sqrt{m_1}} \, C_u \mathrm{e}^{\mathrm{i}(kna - \omega t)}$$ 6 $$y_n(k) ~=~ \frac{1}{\sqrt{m_2}} \, C_y \mathrm{e}^{\mathrm{i}(kna - \omega t)}$$

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: 7 $$\begin{align}\frac{\partial^2 u_{n}}{\partial t^2} ~&=~ \frac{-\omega^2 \, C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} \\\\ y_{n-1} ~&=~ \frac{C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}ka - \mathrm{i}\omega t}\end{align}$$

Now we can substitute the derivative and exponential functions 5, 6, and 7 into the first differential equation 3: 8 $$\begin{align}m_1 \, \frac{-\omega^2 \, C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} & ~+~ 2D\, \frac{C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} \\\\ ~-~ \frac{D \, C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} & ~-~ \frac{D \, C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}ka - \mathrm{i}\omega t} ~=~ 0\end{align}$$

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: 9 $$m_1 \, \frac{-\omega^2 \, C_u}{\sqrt{m_1}} ~+~ \frac{2D\, C_u}{\sqrt{m_1}} ~-~ \frac{D \, C_y}{\sqrt{m_2}} ~-~ \frac{D \, C_y}{\sqrt{m_2}} \, \mathrm{e}^{-\mathrm{i}ka} ~=~ 0$$

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: 10 $$\frac{C_u}{\sqrt{m_1}} \, (2D - \omega^2 \, m_1) ~-~ \frac{D \, C_y}{\sqrt{m_2}} \, (1+\mathrm{e}^{\mathrm{i}ka}) ~=~ 0$$

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: 11 $$\left(\frac{2D}{m_1} - \omega^2\right) \, C_u ~-~ \frac{D}{\sqrt{m_2 \, m_1}} \, \left( 1+\mathrm{e}^{\mathrm{i}ka} \right) \, C_y ~=~ 0$$ 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: 12 $$\begin{align}\frac{\partial^2 y_{n}}{\partial t^2} ~&=~ \frac{-\omega^2 \, C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} \\\\ u_{n+1} ~&=~ \frac{C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t}\,\mathrm{e}^{ika}\end{align}$$

Now we can substitute the second derivative and the exponentials 12, 5, and 6 into the second differential equation 4: 13 $$\begin{align}m_2 \, \frac{-\omega^2 \, C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} & ~+~ \frac{2D\,C_y}{\sqrt{m_2}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} \\\\ ~-~ \frac{D \, C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t} & ~-~ \frac{D \, C_u}{\sqrt{m_1}} \, \mathrm{e}^{\mathrm{i}kna - \mathrm{i}\omega t}\,\mathrm{e}^{ika} ~=~ 0\end{align}$$

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: 14 $$\left( \frac{2D}{m_2} - \omega^2 \right) \, C_y ~-~ \frac{D}{\sqrt{m_2 \, m_1}} \left( 1+\mathrm{e}^{\mathrm{i}ka} \right) \, C_u ~=~ 0$$

Combine result:
By setting up the differential equations and plugging in the exponential functions (solutions), we set up a system of linear equations: 15 $$\begin{align}\left(\frac{2D}{m_1} - \omega^2\right) \, C_u ~-~ \frac{D}{\sqrt{m_2 \, m_1}} \, \left( 1+\mathrm{e}^{\mathrm{i}ka} \right) \, C_y ~&=~ 0 \\\\ \left( \frac{2D}{m_2} - \omega^2 \right) \, C_y ~-~ \frac{D}{\sqrt{m_2 \, m_1}} \left( 1+\mathrm{e}^{\mathrm{i}ka} \right) \, C_u ~&=~ 0\end{align}$$

We can express this linear system of equations in matrix notation: $$\begin{bmatrix} \frac{2D}{m_1}-\omega^2 & -\frac{D(1+\mathrm{e}^{-\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} \\ -\frac{D(1+\mathrm{e}^{\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} & \frac{2D}{m_2}-\omega^2 \end{bmatrix} \begin{bmatrix} C_u\\\\ C_y\end{bmatrix} ~=~ \begin{bmatrix} 0 \\\\ 0 \end{bmatrix}$$

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: 17 $$\det\left(\begin{array}{c} \frac{2D}{m_1}-\omega^2 & -\frac{D(1+\mathrm{e}^{-\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} \\ -\frac{D(1+\mathrm{e}^{\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} & \frac{2D}{m_2}-\omega^2 \end{array}\right) ~\stackrel{!}{=}~ 0$$

Determinant of the matrix 17 is easy to determine with the Laplace expansion. If you don't know how to do that, then go back to this section. The determinant found, as we have already reasoned, must be zero: 18 $$\left( \frac{2D}{m_1} - \omega^2 \right) \, \left( \frac{2D}{m_2} - \omega^2 \right) ~-~ \frac{D \, (1+\mathrm{e}^{\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} \, \frac{D \, (1+\mathrm{e}^{-\mathrm{i}ka})}{\sqrt{m_2 \, m_1}} ~=~ 0$$

Multiply out the parentheses in Eq. 18: $$\frac{4D^2}{m_1 \, m_2} ~-~ \frac{2D}{m_1} \, \omega^2 ~-~ \frac{2D}{m_2} \, \omega^2 ~+~ \omega^4 ~-~ \frac{D^2}{m_1 \, m_2}(2+\mathrm{e}^{\mathrm{i}ka} + \mathrm{e}^{\mathrm{i}ka}) ~=~ 0$$

Sort equation by the power of the frequency \( \omega \). Also, you can rewrite the exponential functions to cosine using Euler's formula: $$\omega^4 ~-~ 2D \, \left( \frac{1}{m_1} + \frac{1}{m_2} \right) \, \omega^2 ~+~ \frac{2D^2}{m_1 \, m_2} \, \left( 1-\cos(ka) \right)~=~ 0$$

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

Thus, we have found out the dispersion relation for a crystal with a diatomic basis. We can rewrite the solution (if we want) with the trigonometric relation \( \sin^2(x) ~=~ \frac{1}{2}(1-\cos(2x)) \) like this: 26 $$\omega_{\pm}^2 ~=~ D \, \left( \frac{1}{m_1} + \frac{1}{m_2} \right) ~\pm~ D \, \sqrt{\left(\frac{1}{m_1} + \frac{1}{m_2}\right)^2 ~-~ \frac{4}{m_1 \, m_2}\,\sin^2\left(\frac{k\,a}{2}\right) }$$

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?

The dispersion relation \( \omega(k) \) of a one-dimensional crystal lattice with a diatomic basis has two branches. These branches result from the solution of two differential equations for this problem of diatomic chains.

The one branch (one solution) \(\omega_-(k) \) is called acoustic because in this case the lattice planes oscillate in phase, as is the case of acoustic waves.

The other branch (the other solution) \(\omega_+(k) \) is called optical because this solution of the respective differential equation gives an opposite-phase oscillation of the lattice planes (with \(m_1\) and \(m_2\)). For example, if the atoms are ions, then they are electrically charged. An antiphase oscillation creates electric dipole moments in the crystal, which affects the optical properties of the crystal.