# Tag Archives: Polarization

## An Undergraduate Optics Problem – The Brewster Angle

Recently, a lab-mate of mine asked me if there was an intuitive way to understand Brewster’s angle. After trying to remember how Brewster’s angle was explained to me from Griffiths’ E&M book, I realized that I did not have a simple picture in my mind at all! Griffiths’ E&M book uses the rather opaque Fresnel equations to obtain the Brewster angle. So I did a little bit of thinking and came up with a picture I think is quite easy to grasp.

First, let me briefly remind you what Brewster’s angle is, since many of you have probably not thought of the concept for a long time! Suppose my incident light beam has both components, s– and p-polarization. (In case you don’t remember, p-polarization is parallel to the plane of incidence, while s-polarization is perpendicular to the plane of incidence, as shown below.) If unpolarized light is incident on a medium, say water or glass, there is an angle, the Brewster angle, at which the light comes out perfectly s-polarized.

An addendum to this statement is that if the incident beam was perfectly p-polarized to begin with, there is no reflection at the Brewster angle at all! A quick example of this is shown in this YouTube video:

So after that little introduction, let me give you the “intuitive explanation” as to why these weird polarization effects happen at the Brewster angle. First of all, it is important to note one important fact: at the Brewster angle, the refracted beam and the reflected beam are at 90 degrees with respect to each other. This is shown in the image below:

Why is this important? Well, you can think of the reflected beam as light arising from the electrons jiggling in the medium (i.e. the incident light comes in, strikes the electrons in the medium and these electrons re-radiate the light).

However, radiation from an oscillating charge only gets emitted in directions perpendicular to the axis of motion. Therefore, when the light is purely p-polarized, there is no light to reflect when the reflected and refracted rays are orthogonal — the reflected beam can’t have the polarization in the same direction as the light ray! This is shown in the right image above and is what gives rise to the reflectionless beam in the YouTube video.

This visual aid enables one to use Snell’s law to obtain the celebrated Brewster angle equation:

$n_1 \textrm{sin}(\theta_B) = n_2 \textrm{sin}(\theta_2)$

and

$\theta_B + \theta_2 = 90^o$

to obtain:

$\textrm{tan}(\theta_B) = n_2/n_1$.

The equations also suggest one more thing: when the incident light has an s-polarization component, the reflected beam must come out perfectly polarized at the Brewster angle. This is because only the s-polarized light jiggles the electrons in a way that they can re-radiate in the direction of the outgoing beam. The image below shows the effect a polarizing filter can therefore have when looking at water near the Brewster angle, which is around 53 degrees for water.

To me, this is a much simpler way to think about the Brewster angle than dealing with the Fresnel equations.

## LST Relation – The Physical Picture

In 1941, Lydanne, Sachs and Teller wrote a paper entitled “On the Polar Vibrations of Alkali Halides”, where they derived a result now known as the Lydanne-Sachs-Teller (LST) relation. It has wide applicability for polar insulators. I reproduce the relation below:

$\frac{\omega_{LO}^2}{\omega_{TO}^2} = \frac{\epsilon(o)}{\epsilon(\infty)}$

In the equation above, $\omega_{LO}$ and $\omega_{TO}$ refer to the frequencies of the longitudinal and transverse optical phonons respectively. $\epsilon(0)$ and $\epsilon(\infty)$ refer to the static and high frequency (above the phonon frequencies, but below any electronic energy scale) dielectric constants. All these quantities are understood to be the values in the long-wavelength limit (i.e. $q \approx 0$).

The beautiful thing about the LST result is that it is independent of any microscopic description, which is quite unusual in solid-state physics. Therefore, the result can be derived from classical electrodynamics, without resorting to any quantum mechanics. It is an interesting question as to whether or not quantum mechanics plays a role in the long-wavelength optical response in general.

Regardless, it turns out that all quantities in the LST relation are experimentally accessible! I find this relation quite remarkable and deep. Not only that, the agreement with experiment in many polar semiconductors is excellent. Take a look at the table below to get an idea of how well this relation holds for a few materials (reproduced from Mark Fox’s textbook Optical Properties of Solids):

I have found textbook derivations don’t give a good intuition of why this relation holds, so here is my attempt to rectify this situation. First, let me state an important assumption that goes into the LST relation:

The phonons are assumed to be in the harmonic limit (i.e. no phonon anharmonicity) and as a result, the dielectric constant has the following form:

$\epsilon(\omega) = \epsilon(\infty) + \frac{C}{\omega_{TO}^2-\omega^2}$

where $C$ is a constant. This form of the dielectric constant can be arrived at using either classical electrodynamics or quantum mechanics (see e.g. Ashcroft and Mermin, Kittel or Ziman).

Now, with this result under our belts, it turns out that it is quite simple to understand why the LST relation holds. In a simple polar semiconductor, we have two atoms per unit cell that are oppositely charged like so:

Therefore, for the longitudinal optical phonon we have an extra polarization effect due to the long-range nature of the Coulomb interaction. This extra polarization results in an extra restoring force (in addition to the springy restoring force between the ions), yielding a higher longitudinal phonon frequency compared to the transverse optical phonon. I have discussed this a little more extensively in a previous post. This extra restoring force (which is only present for the longitudinal oscillation) is pictured below:

The longitudinal optical phonon is at a higher energy because of the extra Coulombic polarization effect

More precisely, we can write the following when including this extra restoring force:

$\omega_{LO}^2 = \omega_{TO}^2 + \frac{C}{\epsilon(\infty)}$

There is an $\epsilon(\infty)$ in the formula above because this polarization will necessarily be screened by higher energy (electronic) processes. Dividing both sides by $\omega_{TO}^2$, we can write the above equation suggestively as:

$\frac{\omega_{LO}^2}{\omega_{TO}^2} = \frac{\epsilon(\infty)+C/\omega_{TO}^2}{\epsilon(\infty)}$

Looking at the equation for the dielectric constant from earlier, this is precisely the LST relation! In effect, the same extra restoring due to the long-range Coulomb interaction leads to the extra screening in the static limit, yielding, in my mind, a delightful little result.

Using the LST relation, we can deduce a property of ferroelectric materials. Namely, we know that at the transition temperature between the normal state and a ferroelectric ground state, the static dielectric constant, $\epsilon(0)$, diverges. Therefore, we can surmise from the LST relation that a zone center transverse optical phonon must go to zero energy (soften) at the transition temperature (see here for PbTiO3). This is a totally non-trivial consequence of the LST relation, demonstrating again its far-reaching utility.

Did I mention that I think this result is pretty excellent?

I’d like to acknowledge Zhanybek Alpichshev for enlightening some aspects regarding this topic.

## An Integral from the SSH Model

A while ago, I was solving the Su-Schrieffer-Heeger (SSH) model for polyacetylene and came across an integral which I immediately thought was pretty cool. Here is the integral along with the answer:

$\int_{0}^{2\pi} \frac{\delta(1+\mathrm{tan}^2(x))}{1+\delta^2\mathrm{tan}^2(x)}\frac{dx}{2\pi} = \mathrm{sgn}(\delta)$

Just looking at the integral, it is difficult to see why no matter what the value of $\delta$, the integral will always give +1 or -1, which only depends on the sign of $\delta$. This means that if $\delta=1,000,000$ or if $\delta=0.00001$, you would get the same result, in this case +1, as the answer to the integral! I’ll leave it to you to figure out why this is the case. (Hint: you can use contour integration, but you don’t have to.)

It turns out that the result actually has some interesting topological implications for the SSH model, as there are fractional statistics associated with the domain wall solitons. I guess it’s not so surprising that an integral that possesses topological properties would show up in a physical system with topological characteristics! But I thought the integral was pretty amusing anyhow, so I thought I’d share it.

Aside: For those who are interested in how I arrived at this integral in the SSH model, here are some of my notes. (Sorry if there are any errors and please let me know!) Also, the idea of solitons in the SSH model actually bears a strong qualitative resemblance to the excellent zipper analogy that Brian Skinner used on his blog.

## Modern Theory of Polarization

It is quite curious that the simple concept of polarization in a solid was not understood until the early to mid-90s. The solution to the problem actually came from the computational physics community because of their inability to calculate accurately the polarization in solids. Prior to the solution, there were papers seriously discussing whether polarization was a bulk phenomenon or whether it was a property of a crystal surface. The solution to the problem is commonly associated with these papers by Resta (pdf!) and King-Smith and Vanderbilt (pdf!) .

There are a few startling realizations that arose from the modern theory of polarization:

1. Polarization is not a well-defined quantity in that it is multi-valued.
2. Only polarization difference has any physical meaning.
3. Experiments only measure differences in polarization.
4. Polarization is deeply rooted in the concept of the Berry phase.
5. (not startling) Polarization is a bulk phenomenon.

There is a fantastic pedagogical introduction to the subject by Spaldin entitled A beginner’s guide to the modern theory of polarization.

As just a little preview of the Spaldin paper, let me outline the issues as she does. The problem with defining polarization as the dipole moment per unit cell can easily be understood using the picture of the one-dimensional chain below:

Using the box on the left, one would calculate the dipole moment per unit length as:

$p =\frac{e}{a}\sum_i q_id_i = \frac{e}{a}(\frac{a}{4}\times -1 + \frac{3a}{4}\times 1) = \frac{e}{2}$

whereas the box on the right gives:

$p =\frac{e}{a}\sum_i q_id_i = \frac{e}{a}(\frac{a}{4}\times 1 + \frac{3a}{4}\times -1) = \frac{-e}{2}$.

That these don’t match has to do precisely with the fact that polarization in a solid is multi-valued, as alluded to earlier. Now consider the following one-dimensional lattice with a distortion:

The polarization now for the distorted lattice (second row in the image above) for the left and right boxes are respectively:

Left Box: $p = \frac{e}{2} +\frac{ed}{a}$

Right Box: $p = \frac{-e}{2} +\frac{ed}{a}$

We can therefore see that for both boxes, the change in polarization is $\delta p = \frac{ed}{a}$, which is a single-valued and experimentally well-defined quantity.

While this illustration was classical, when one includes the wavefunction of the electrons, one is forced to consider the Berry phase of the Bloch electrons. While I have known about this result for some time now, I still find it quite surprising that the simple concept of polarization in a solid has any relationship to the Berry phase at all. I strongly recommend Spaldin’s eminently readable article as an excellent introduction to the subject.