# Tag Archives: Toy Model

## Jahn-Teller Distortion and Symmetry Breaking

The Jahn-Teller effect occurs in molecular systems, as well as solid state systems, where a molecular complex distorts, resulting in a lower symmetry. As a consequence, the energy of certain occupied molecular states is reduced. Let me first describe the phenomenon before giving you a little cartoon of the effect.

First, consider, just as an example, a manganese atom with valence $3d^4$, surrounded by an octahedral cage of oxygen atoms like so (image taken from this thesis):

The electrons are arranged such that the lower triplet of orbital states each contain a single “up-spin”, while the higher doublet of orbitals only contains a single “up-spin”, as shown on the image to the left. This scenario is ripe for a Jahn-Teller distortion, because the electronic energy can be lowered by splitting both the doublet and the triplet as shown on the image on the right.

There is a very simple, but quite elegant problem one can solve to describe this phenomenon at a cartoon level. This is the problem of a two-dimensional square well with adjustable walls. By solving the Schrodinger equation, it is known that the energy of the two-dimensional infinite well has solutions of the form:

$E_{i,j} = \frac{h}{8m}(i^2/a^2 + j^2/b^2)$                where $i,j$ are integers.

Here, $a$ and $b$ denote the lengths of the sides of the 2D well. Since it is only the quantity in the brackets that determine the energy levels, let me factor out a factor of $\gamma = a/b$ and write the energy dependence in the following way:

$E \sim i^2/\gamma + \gamma j^2$

Note that $\gamma$ is effectively an anisotropy parameter, giving a measure of the “squareness of the well”. Now, let’s consider filling up the levels with spinless electrons that obey the Pauli principle. These electrons will fill up in a “one-per-level” fashion in accordance with the fermionic statistics. We can therefore write the total energy of the $N-$fermion problem as so:

$E_{tot} \sim \alpha^2/ \gamma + \gamma \beta^2$

where $\alpha$ and $\beta$ parameterize the energy levels of the $N$ electrons.

Now, all of this has been pretty simple so far, and all that’s really been done is to re-write the 2D well problem in a different way. However, let’s just systematically look at what happens when we fill up the levels. At first, we fill up the $E_{1,1}$ level, where $\alpha^2 = \beta^2 = 1^2$. In this case, if we take the derivative of $E_{1,1}$ with respect to $\gamma$, we get that $\gamma_{min} = 1$ and the well is a square.

For two electrons, however, the well is no longer a square! The next electron will fill up the $E_{2,1}$ level and the total energy will therefore be:

$E_{tot} \sim 1/\gamma (1+4) + \gamma (1+1)$,

which gives a $\gamma_{min} = \sqrt{5/2}$!

Why did this breaking of square symmetry occur? In fact, this is very closely related to the Jahn-Teller effect. Since the level is two-fold degenerate (i.e. $E_{2,1} = E_{1,2}$), it is favorable for the 2D well to distort to lower its electronic energy.

Notice that when we add the third electron, we get that:

$E_{tot} \sim 1/\gamma (1+4+1) + \gamma (1+1+4)$

and $\gamma_{min} = 1$ again, and we return to the system with square symmetry! This is also quite similar to the Jahn-Teller problem, where, when all the states of the degenerate levels are filled up, there is no longer an energy to be gained from the symmetry-broken geometry.

This analogy is made more complete when looking at the following level scheme for different $d$-electron valence configurations, shown below (image taken from here).

The black configurations are Jahn-Teller active (i.e. prone to distortions of the oxygen octahedra), while the red are not.

In condensed matter physics, we usually think about spontaneous symmetry breaking in the context of the thermodynamic limit. What saves us here, though, is that the well will actually oscillate between the two rectangular configurations (i.e. horizontal vs. vertical), preserving the original symmetry! This is analogous to the case of the ammonia ($NH_3$) molecule I discussed in this post.

## Zener’s Electrical Breakdown Model

In my previous post about electric field induced metal-insulator transitions, I mentioned the notion of Zener breakdown. Since the idea is not likely to be familiar to everyone, I thought I’d use this post to explain the concept a little further.

Simply stated, Zener breakdown occurs when a DC electric field applied to an insulator is large enough such that the insulator becomes conducting due to interband tunneling. Usually, when we imagine electrical conduction in a solid, we think of the mobile electrons moving only within one or more partially filled bands. Modelling electrical transport within a single band can already get quite complicated, so it was a major accomplishment that C. Zener was able to come up with a relatively simple and solvable model for interband tunneling.

To make the problem tractable, Zener came up with a hybrid real-space / reciprocal-space model where he could use the formalism of a 1D quantum mechanical barrier:

In Zener’s model, the barrier height is set by the band gap energy, $E_{g}$, between the valence and conduction bands in the insulator, while the barrier width is set by the length scale relevant to the problem. In this case, we can say that the particle can gain enough kinetic energy to surpass the barrier if $e\mathcal{E}d = E_{g}$, in which case our barrier width would be:

$d = E_{g} / e\mathcal{E}$,

where $\mathcal{E}$ is the applied electric field and $e$ is the electron charge.

Now, how do we solve this tunneling problem? If we were to use the WKB formalism, like Zener, we get that the transmission probability is:

$P_T = e^{-2\gamma}$             where           $\gamma = \int_0^d{k(x) dx}$.

Here, $k(x)$ is the wavenumber. So, really, all that needs to be done is to obtain the correct funtional form for the wavenumber and (hopefully) solve the integral. This turns out not to be too difficult — we just have to make sure that we include both bands in the calculation. This can be done in similar way to the nearly free electron problem.

Quickly, the nearly-free electron problem considers the following $E-k$ relations in the extended zone scheme:

Near the zone boundary, one needs to apply degenerate perturbation theory due to Bragg diffraction of the electrons (or degeneracy of the bands from the next zone, or however you want to think about it). So if one now zooms into the hatched area in the figure above, one gets that a gap opens up by solving the following determinant and obtaining $\epsilon(k)$:

$\left( \begin{array}{cc} \lambda_k - \epsilon & E_g/2 \\ E_g/2 & \lambda_{k-G} - \epsilon \end{array} \right)$,

where $\lambda_k$ is $\hbar^2k^2/2m$ in this problem, and the hatched area becomes gapped like so:

In the Zener model problem, we take a similar approach. Instead of solving for $\epsilon(k)$, we solve for $k(\epsilon)$. To focus on the zone boundary, we first let $k \rightarrow k_0 + \kappa$ and $\epsilon \rightarrow \epsilon_0 + \epsilon_1$, where $k_0 = \pi/a$ (the zone boundary) and $\epsilon_0 = \hbar^2k_0^2/2m$, under the assumption that $\kappa$ and $\epsilon_1$ are small. All this does is shift our reference point to the hatched region in previous figure above.

The trick now is to solve for  $k(\epsilon)$ to see if imaginary solutions are possible. Indeed, they are! I get that:

$\kappa^2 = \frac{2m}{\hbar^2} (\frac{\epsilon_1^2 - E_g^2/4}{4\epsilon_0})$,

so as long as $\epsilon_1^2 - E_g^2/4 < 0$, we get imaginary solutions for $\kappa$.

Although we have a function $\kappa(\epsilon_1)$, we still need to do a little work to obtain $\kappa(x)$, which is required for the WKB exponent. Here, Zener just assumed the simplest thing that he could, that $\epsilon_1$ is related to the tunneling distance, $x$, linearly. The image I’ve drawn above (that shows the potential profile) and the fact that work done by the electric field is $e\mathcal{E}x$ demonstrates that this assumption is very reasonable.

Plugging all the numbers in and doing the integral, one gets that:

$P_T = \exp-\left(\pi^2 E_g^2/(4 \epsilon_0 e \mathcal{E} a)\right)$.

If you’re like me in any way, you’ll find the final answer to the problem pretty intuitive, but Zener’s methodology towards obtaining it pretty strange. To me, the solution is quite bizarre in how it moves between momentum space and real space, and I don’t have a good physical picture of how this happens in the problem. In particular, there is seemingly a contradiction between the assumption of the lattice periodicity and the application of the electric field, which tilts the lattice, that pervades the problem. I am apparently not the only one that is uncomfortable with this solution, seeing that it was controversial for a long time.

Nonetheless, it is a great achievement that with a couple simple physical pictures (albeit that, taken at face value, seem inconsistent), Zener was able to qualitatively explain one mechanism of electrical breakdown in insulators (there are a few others such as avalanche breakdown, etc.).

## Nonlinear Response and Harmonics

Because we are so often solving problems in quantum mechanics, it is sometimes easy to forget that certain effects also show up in classical physics and are not “mysterious quantum phenomena”. One of these is the problem of avoided crossings or level repulsion, which can be much more easily understood in the classical realm. I would argue that the Fano resonance also represents a case where a classical model is more helpful in grasping the main idea. Perhaps not too surprisingly, a variant of the classical harmonic oscillator problem is used to illustrate the respective concepts in both cases.

There is also another cute example that illustrates why overtones of the natural harmonic frequency components result when subject to slightly nonlinear oscillations. The solution to this problem therefore shows why harmonic distortions often affect speakers; sometimes speakers emit frequencies not present in the original electrical signal. Furthermore, it shows why second harmonic generation can result when intense light is incident on a material.

First, imagine a perfectly harmonic oscillator with a potential of the form $V(x) = \frac{1}{2} kx^2$. We know that such an oscillator, if displaced from its original position, will result in oscillations at the natural frequency of the oscillator $\omega_o = \sqrt{k/m}$ with the position varying as $x(t) = A \textrm{cos}(\omega_o t + \phi)$. The potential and the position of the oscillator as a function of time are shown below:

(Left) Harmonic potential as a function of position. (Right) Variation of the position of the oscillator with time

Now imagine that in addition to the harmonic part of the potential, we also have a small additional component such that $V(x) = \frac{1}{2} kx^2 + \frac{1}{3}\epsilon x^3$, so that the potential now looks like so:

The equation of motion is now nonlinear:

$\ddot{x} = -c_1x - c_2x^2$

where $c_1$ and $c_2$ are constants. It is easy to see that if the amplitude of oscillations is small enough, there will be very little difference between this case and the case of the perfectly harmonic potential.

However, if the amplitude of the oscillations gets a little larger, there will clearly be deviations from the pure sinusoid. So then what does the position of the oscillator look like as a function of time? Perhaps not too surprisingly, considering the title, is that not only are there oscillations at $\omega_0$, but there is also an introduction of a harmonic component with $2\omega_o$.

While the differential equation can’t be solved exactly without resorting to numerical methods, that the harmonic component is introduced can be seen within the framework of perturbation theory. In this context, all we need to do is plug the solution to the simple harmonic oscillator, $x(t) = A\textrm{cos}(\omega_0t +\phi)$ into the nonlinear equation above. If we do this, the last term becomes:

$-c_2A^2\textrm{cos}^2(\omega_0t+\phi) = -c_2 \frac{A^2}{2}(1 + \textrm{cos}(2\omega_0t+2\phi))$,

showing that we get oscillatory components at twice the natural frequency. Although this explanation is a little crude — one can already start to see why nonlinearity often leads to higher frequency harmonics.

With respect to optical second harmonic generation, there is also one important ingredient that should not be overlooked in this simplified model. This is the fact that frequency doubling is possible only when there is an $x^3$ component in the potential. This means that the potential needs to be inversion asymmetric. Indeed, second harmonic generation is only possible in inversion asymmetric materials (which is why ferroelectric materials are often used to produce second harmonic optical signals).

Because of its conceptual simplicity, it is often helpful to think about physical problems in terms of the classical harmonic oscillator. It would be interesting to count how many Nobel Prizes have been given out for problems that have been reduced to some variant of the harmonic oscillator!

## 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.

## Lunar Eclipse and the 22 Degree Halo

The beautiful thing about atmospheric optics is that (almost) everyone can look up at the sky and see stunning optical phenomena from the sun, moon or some other celestial object. In this post I’ll focus on two particularly striking phenomena where the physical essence can be captured with relatively simple explanations.

The 22 degree halo is a ring around the sun or moon, which is often observed on cold days. Here are a couple images of the 22 degree halo around the sun and moon respectively:

22 degree halo around the sun

22 degree halo around the moon

Note that the 22 degree halo is distinct from the coronae, which occur due to different reasons. While the coronae arise due to the presence of water droplets, the 22 degree halo arises specifically due to the presence of hexagonal ice crystals in the earth’s atmosphere. So why 22 degrees? Well, it turns out that one can answer the question using rather simple undergraduate-level physics. One of the most famous questions in undergraduate optics is that of light refraction through a prism, illustrated below:

Fig. 1: The Snell’s Law Prism Problem

But if there were hexagonal ice crystals in the atmosphere, the problem is exactly the same, as one can see below. This is so because a hexagon is just an equilateral triangle with its ends chopped off. So as long as the light enters and exits on two sides of the hexagon that are spaced one side apart, the analysis is the same as for the triangle.

Equilateral triangle with ends chopped off, making a hexagon

It turns out that $\theta_4$ in Fig. 1 can be solved as a function of $\theta_1$ with Snell’s law and some simple trigonometry to yield (under the assumption that $n_1 =1$):

$\theta_4 = \textrm{sin}^{-1}(n_2 \times \textrm{sin}(60-\textrm{sin}^{-1}(\textrm{sin}(\theta_1)/n_2)))$

It is then pretty straightforward to obtain $\delta$, the difference in angle between the incident and refracted beam as a function of $\theta_1$. I have plotted this below for the index of refraction of ice crystals for three different colors of light, red, green and blue ($n_2 =$ 1.306, 1.311 and 1.317 respectively):

The important thing to note in the plot above is that there is a minimum angle below which there is no refracted beam, and this angle is precisely 21.54, 21.92 and 22.37 degrees for red, green and blue light respectively. Because there is no refracted beam below 22 degrees, this region appears darker, and then there is a sudden appearance of the refracted beam at the angles listed above. This is what gives rise to the 22 degree halo and also to the reddish hue on the inside rim of the halo.

Another rather spectacular celestial occurrence is the lunar eclipse, where the earth completely obscures the moon from direct sunlight. This is the geometry for the lunar eclipse:

Geometry of the lunar eclipse

The question I wanted to address is the reddish hue of the moon, despite it lying in the earth’s shadow. It would naively seem like the moon should not be observable at all. However, there is a similar effect occurring here as with the halo. In this case, the earth’s atmosphere is the refracting medium. So just as light incident on the prism was going upward and then exited going downward, the sun’s rays similarly enter the atmosphere on a trajectory that would miss the moon, but then are bent towards the moon after interacting with the earth’s atmosphere.

But why red? Well, this has the same origins as the reddish hue of the sunset. Because light scatters from atmospheric particles as $1/\lambda^4$, blue light gets scattered away much more easily than red light. Hence, the only color of light left by the time the light reaches the moon is primarily of red color.

It is interesting to imagine what the earth looks like from the moon during a lunar eclipse — it likely looks completely dark apart from a spectacular red halo around the earth. Anyway, one should realize that Snell’s law was first formulated in 984 by Arab scientist Ibn Sahl, and so it was possible to come to these conclusions more than a thousand years ago. Nothing new here!