Derivation: Dispersion Relation of the Monatomic Lattice Vibration

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.

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Horizontal deflection of a chain results in a lattice vibration. The faded chains represent the rest position.

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 paralel (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.

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:

$$ \begin{align} F_n ~=~ \underset{z}{\boxed{+}} \, D_z \, (u_{n+z} - u_n) \end{align} $$

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 $z$.

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 $u$:

$$ \begin{align} m \, \frac{\text{d}^2 u_n}{\text{d} t^2} ~=~ \underset{z}{\boxed{+}} \, D_z \, (u_{n+z} - u_n) \end{align} $$

The solution of such a second order ordinary differential equation are harmonic functions. Let's make the following ansatz (exponential ansatz) for the deflection:

Exponential ansatz for the deflection

$$ \begin{align} u_{n+z} ~=~ C \, \mathrm{e}^{\mathrm{i}(kza - \omega t)} \end{align} $$

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 ($z=0$):

$$ \begin{align} u_{n} ~=~ C \, \mathrm{e}^{- \mathrm{i} \omega t)} \end{align} $$

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:

$$ \begin{align} -m \, C \, \omega^2 \mathrm{e}^{\mathrm{i} \omega t} ~=~ \underset{z}{\boxed{+}} \, D_z \, C \left( \mathrm{e}^{\mathrm{i}(kza-\omega t)} ~-~ \mathrm{e}^{-\mathrm{i}\omega t} \right) \end{align} $$

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:

$$ \begin{align} -m \, \omega^2 ~-~ \underset{z}{\boxed{+}} \, D_z \left( \mathrm{e}^{\mathrm{i} kza} ~-~ 1 \right) ~=~ 0 \end{align} $$

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:

$$ \begin{align} -m \, \omega^2 ~-~ \underset{z}{\boxed{+}} \, D_z \left( \mathrm{e}^{\mathrm{i} kza} ~+~ \mathrm{e}^{-\mathrm{i}kza} ~-~ 2 \right) ~=~ 0 \end{align} $$

Next, we rewrite complex exponential functions using Euler's formula, bringing cosine and sine into play:

$$ \begin{align} -m \, \omega^2 ~-~ \underset{z}{\boxed{+}} \, D_z \left( \cos(kza) + \mathrm{i}\sin(kza) + \cos(-kza) + \mathrm{i}\sin(-kza) ~-~ 2 \right) ~=~ 0 \end{align} $$

Exploit the symmetry and antisymmetry properties of cosine and sine, simplifying Eq. 8:

$$ \begin{align} -m \, \omega^2 ~-~ 2\,\underset{z}{\boxed{+}} \, D_z \left( \cos(kza) ~-~ 1 \right) ~=~ 0 \end{align} $$

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 $:

General dispersion relation for a monatomic chain

$$ \begin{align} \omega(k) ~=~ \sqrt{\frac{2}{m} \, \underset{z}{\boxed{+}} \, D_z \left( 1 ~-~ \cos(kza) \right)} \end{align} $$

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 $:

$$ \begin{align} \omega(k) ~=~ \sqrt{\frac{2 D}{m} \left( 1 ~-~ \cos(ka) \right)} \end{align} $$

Dispersion relation $\omega(k)$ reduced to the 1st Brillouin zone because all the other periods do not contain any additional information about the lattice vibration.

Using the double angle identity $ \sin^2(x) ~=~ \frac{1}{2}(1-\cos(2x)) $, Eq. 11 can also be written as follows:

Dispersion relation for a monatomic chain with neighbor interaction

$$ \begin{align} \omega(k) ~=~ \sqrt{\frac{4 D}{m} \sin^2\left(\frac{ka}{2}\right)} \end{align} $$

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 2).

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 $k$:

Group velocity of a monatomic chain

$$ \begin{align} v_{\text g} ~=~ \sqrt{\frac{D \, a^2}{m}} \, \cos\left(\frac{1}{2} \, k \, a\right) \end{align} $$