Skip to main content

Derivation Dispersion Relation of the Diatomic Lattice Vibration

Dispersion Relation (Graph) of the Lattice Vibrations of a Diatomic Crystal Lattice
Level 4 (for physics pros)
Level 4 requires the knowledge of vector calculus, (multidimensional) differential and integral calculus. Suitable for advanced students.
Updated by Alexander Fufaev on

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 \( m_2 \).

Lattice vibration inside a crystal with diatomic basis
Hover the image! Get this illustration
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 \(m_1\)
Formula anchor
Differential equation for the deflection of the lattice plane with the masses \(m_2\)
Formula anchor

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:

First differential equation rearranged
Formula anchor
Second differential equation rearranged
Formula anchor

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

Exponential ansatz for the first deflection
Formula anchor
Exponential ansatz for the second deflection
Formula anchor

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:

Second derivative of the first deflection and ansatz for the second deflection
Formula anchor

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

First differential equation with exponential functions inserted
Formula anchor

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:

First differential equation with exponential functions inserted and simplified
Formula anchor

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:

First differential equation with exponential functions further simplified
Formula anchor

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:

First linear equation
Formula anchor

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:

Second derivative of the second deflection and ansatz for the first deflection
Formula anchor

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

Second differential equation with inserted exponential ansatz
Formula anchor

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:

Second linear equation
Formula anchor

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
Formula anchor

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

Linear system of equations in the matrix form
Formula anchor

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:

Determinant of the differential equation must be zero
Formula anchor
Why does the determinant have to be zero?

Let's abbreviate the matrix elements as \(a,b,c,d\):

General matrix equation with zero eigenvalue
Formula anchor

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

Linear system for a 2x2 matrix equation
Formula anchor

Rearrange the first equation for \(C_u\) and substitute it into the second equation of the linear system:

Equation for Cy
Formula anchor

Divide the equation by \(C_y\):

Determinant of the 2x2 matrix
Formula anchor

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:

Determinant for a crystal with diatomic base is zero
Formula anchor

Multiply out the parentheses in Eq. 22:

Determinant for a crystal with diatomic basis multiplied out
Formula anchor

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

Determinant for a crystal with diatomic basis using cosine
Formula anchor

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:

Dispersion relation for a diatomic crystal lattice with neighbor interaction using cosine
Formula anchor

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:

Formula anchor

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.

Dispersion Relation (Graph) of the Lattice Vibrations of a Diatomic Crystal Lattice
Dispersion relation for a diatomic basis. Two solutions are obtained: optical and acoustic branch.

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?