Categories: Physics.

# Lyddane-Sachs-Teller relation

While the Lorentz oscillator model originally studied the electric dipole formed by an electron and its nucleus, it can also be applied to the nuclei of polar crystals, i.e. crystals held together by polar bonds between ions. When an electromagnetic wave passes by, its electric field $\vb{E}(t)$ exerts a force on the ions, leading to an optical response.

We are talking about light waves (photons) creating lattice vibrations (phonons), i.e. a photon-phonon conversion, where the total energy and momentum must be conserved. If the photon has frequency $\omega$ and wavenumber $k$, and the phonon $\Omega$ and $K$, then:

$\begin{aligned} \hbar \omega = \hbar \Omega \qquad \qquad \hbar k = \hbar K \end{aligned}$In other words, such a conversion can only take place
at intersections of the dispersion relations $\omega(k)$ and $\Omega(K)$.
The latter consists of two branches:
low-frequency *acoustic* modes and higher-frequency *optical* modes.
Meanwhile, the photon dispersion is simply $\omega = c k / n$,
where $n$ is the medium’s refractive index.

For acoustic phonons, the dispersions only intersect at $k = K = 0$,
which is simply a static solid in a static electric field.
For optical phonons, the intersection is at a nonzero $k$.
In addition, light is a transverse wave,
so it can only interact with transverse phonons,
meaning that we must only consider **transverse optical (TO) phonons**.

A wave’s group velocity is the slope of its dispersion,
so $\ipdv{\omega}{k}$ and $\ipdv{\Omega}{K}$ in this case.
Clearly, light is much faster than sound,
so $\omega(k)$ is much steeper than $\Omega(K)$,
meaning that the photon-phonon conversion
will happen at relatively low $k$.
In practice, the intersection is in the infrared (IR),
hence TO phonons are sometimes called **IR active**.

We consider a 1D chain of unit cells along the $z$-axis, each containing a positive and a negative ion oscillating transversely along the $x$-axis. For optical phonon modes, the ions always move in opposite directions. Let the ions have masses $m_{-}$ and $m_{+}$, then the Lorentz oscillator model tells us that the displacements $\vb{x}_{+}(t)$ and $\vb{x}_{-}(t)$ are governed by:

$\begin{aligned} m_{+} \dvn{2}{\vb{x}_{+}}{t} &= - \kappa (\vb{x}_{+} - \vb{x}_{-}) + q \vb{E} \\ m_{-} \dvn{2}{\vb{x}_{-}}{t} &= - \kappa (\vb{x}_{-} - \vb{x}_{+}) - q \vb{E} \end{aligned}$Where $\vb{E}(t) = \vb{E}_0 e^{- i \omega t}$ represents the light, and $\kappa$ is the spring constant of the polar bonds’ restoring force. Note that the latter depends on the displacement between the ions, instead of from their equilibrium position, so we need to write $\vb{x}_{+} - \vb{x}_{-}$ instead of $\vb{x}_{+}$.

Respectively dividing the equations by $m_{+}$ and $m_{-}$ and subtracting the latter from the former, we arrive at the following combined equation, where $m$ is the reduced mass:

$\begin{aligned} \dvn{2}{}{t} (\vb{x}_{+} - \vb{x}_{-}) = - \frac{\kappa}{m} (\vb{x}_{+} - \vb{x}_{-}) + \frac{q}{m} \vb{E} \end{aligned}$Defining the relative displacement $\vb{x} \equiv \vb{x}_{+} \!-\! \vb{x}_{-}$, and recognizing that $\kappa / m$ is the TO phonons’ natural resonance frequency $\Omega_\mathrm{TO}^2$:

$\begin{aligned} \dvn{2}{\vb{x}}{t} + \Omega_\mathrm{TO}^2 \vb{x} = \frac{q}{m} \vb{E} \end{aligned}$Note that $\Omega_\mathrm{TO}$ is the phonon frequency for $K = 0$. This is because IR light waves are much larger than the crystal’s unit cell, so we are ignoring all spatial variation in $\vb{E}$ (i.e. the electric dipole approximation). This is equivalent to assuming that $K \approx 0$.

For the sake of generality, we also introduce an empirical damping rate $\gamma$, like in the original Lorentz oscillator model:

$\begin{aligned} \dvn{2}{\vb{x}}{t} + \gamma \dv{\vb{x}}{t} + \Omega_\mathrm{TO}^2 \vb{x} = \frac{q}{m} \vb{E} \end{aligned}$Inserting the ansatz $\vb{x}(t) = \vb{x}_0 e^{- i \omega t}$ and isolating for the amplitude $\vb{x}_0$, we find:

$\begin{aligned} \vb{x}_0 = \frac{q \vb{E}_0}{m (\Omega_\mathrm{TO}^2 - \omega^2 - i \gamma \omega)} \end{aligned}$The induced polarization density $\vb{P}$ is then the sum of the electrons’ and ions’ contributions $\vb{P}_e$ and $\vb{P}_i$. The former is described by a background susceptibility $\chi$, and the latter by each unit cell’s dipole moment $\vb{p} = q \vb{x}$ multiplied by the number of cells per unit volume $N$:

$\begin{aligned} \vb{P} \approx \varepsilon_0 \chi \vb{E} + N q \vb{x} = \bigg( \varepsilon_0 \chi + \frac{N q^2}{m (\Omega_\mathrm{TO}^2 - \omega^2 - i \gamma \omega)} \bigg) \vb{E} \end{aligned}$Note that we are neglecting how each dipole shields its neighbors. This approximation can be improved afterwards by using the Clausius-Mossotti relation.

With our expression for $\vb{P}$, we can find the dielectric function $\varepsilon_r(\omega)$ using the definition of the electric displacement field $\vb{D} = \varepsilon_0 \vb{E} + \vb{P} = \varepsilon_0 \varepsilon_r \vb{E}$, yielding:

$\begin{aligned} \boxed{ \varepsilon_r(\omega) = 1 + \chi(\omega) + \frac{N q^2}{\varepsilon_0 m (\Omega_\mathrm{TO}^2 - \omega^2 - i \gamma \omega)} } \end{aligned}$In the limits of low and high frequencies $\omega$, we see that $\varepsilon_r$ is higher in the former:

$\begin{aligned} \varepsilon_{\mathrm{low}} &= \, \lim_{\omega \to 0} \, \varepsilon_r(\omega) = 1 + \chi_\mathrm{low} + \frac{N q^2}{\varepsilon_0 m \Omega_\mathrm{TO}^2} \\ \varepsilon_{\mathrm{high}} &= \lim_{\omega \to \infty} \varepsilon_r(\omega) = 1 + \chi_\mathrm{high} \end{aligned}$We can use these quantities to rewrite the relative permittivity $\varepsilon_r$ as follows:

$\begin{aligned} \varepsilon_r(\omega) = \varepsilon_{\mathrm{high}} + (\varepsilon_{\mathrm{low}} - \varepsilon_{\mathrm{high}}) \frac{\Omega_\mathrm{TO}^2}{\Omega_\mathrm{TO}^2 - \omega^2 - i \gamma \omega} \end{aligned}$For weak damping $\gamma \approx 0$, there exists a frequency, which we will call $\Omega_\mathrm{LO}$ in anticipation, where the dielectric function is zero:

$\begin{aligned} 0 = \varepsilon_r(\Omega_\mathrm{LO}) = \varepsilon_{\mathrm{high}} + (\varepsilon_{\mathrm{low}} - \varepsilon_{\mathrm{high}}) \frac{\Omega_\mathrm{TO}^2}{\Omega_\mathrm{TO}^2 - \Omega_\mathrm{LO}^2} \end{aligned}$The physical significance of $\varepsilon_r = 0$ can be seen from Gauss’ law, under the assumption that there is no net charge density:

$\begin{aligned} \nabla \cdot \vb{D} = \varepsilon_0 \varepsilon_r \nabla \cdot \vb{E} = 0 \end{aligned}$If $\varepsilon_r \neq 0$, then $\nabla \cdot \vec{E} = 0$,
corresponding to a transverse light wave as usual.
However, if $\varepsilon_r = 0$, then $\nabla \cdot \vec{E} \neq 0$,
representing a longitudinal electric wave, like a plasmon in metal.
Rearranging the equation for $\Omega_\mathrm{LO}$
gives us the **Lyddane-Sachs-Teller (LST) relation**:

$\Omega_\mathrm{LO}$ is the natural frequency
of such **longitudinal optical (LO) phonons** for $K = 0$.
Recall that only transverse phonons interact with light:
the significance of this result is that we can measure
$\varepsilon_\mathrm{low}$, $\varepsilon_\mathrm{high}$,
and $\Omega_\mathrm{TO}$ with light,
and use that to calculate a quantity for an effect
that we cannot interact with directly.
The caveat is that this is only valid for simple polar crystals.

For $\omega$-values between $\Omega_\mathrm{TO}$ and $\Omega_\mathrm{LO}$, the permittivity $\varepsilon_r$ is negative, meaning the reflectivity $R$ equals $1$, i.e. the material becomes a perfect reflector:

$\begin{aligned} R = \bigg| \frac{i \sqrt{-\varepsilon_r} - 1}{i \sqrt{-\varepsilon_r} + 1} \bigg|^2 = \frac{\varepsilon_r^2 + 1^2}{\varepsilon_r^2 + 1^2} = 1 \end{aligned}$This region of 100% reflectivity is called the **Reststrahlen band**.
In practice, real materials have $\gamma > 0$, which reduces $R$ somewhat.

Because the photons and TO phonons interact so strongly
for $\omega \approx \Omega_\mathrm{TO}$,
they can be treated as a single **phonon polariton** there,
with a dispersion relation given by:

Earlier, when treating the photon and phonon separately,
we wanted the intersection between $\omega(k)$ and $\Omega(K)$.
But now, for $\omega_\mathrm{pp}(K)$, there is none! This is a good example
of the typical *anti-crossing* behavior of strongly coupled systems.

## References

- M. Fox,
*Optical properties of solids*, 2nd edition, Oxford.