# Category Archives: Symmetry

## 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.

## 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!

## Landau Theory and the Ginzburg Criterion

The Landau theory of second order phase transitions has probably been one of the most influential theories in all of condensed matter. It classifies phases by defining an order parameter — something that shows up only below the transition temperature, such as the magnetization in a paramagnetic to ferromagnetic phase transition. Landau theory has framed the way physicists think about equilibrium phases of matter, i.e. in terms of broken symmetries. Much current research is focused on transitions to phases of matter that possess a topological index, and a major research question is how to think about these phases which exist outside the Landau paradigm.

Despite its far-reaching influence, Landau theory actually doesn’t work quantitatively in most cases near a continuous phase transition. By this, I mean that it fails to predict the correct critical exponents. This is because Landau theory implicitly assumes that all the particles interact in some kind of average way and does not adequately take into account the fluctuations near a phase transition. Quite amazingly, Landau theory itself predicts that it is going to fail near a phase transition in many situations!

Let me give an example of its failure before discussing how it predicts its own demise. Landau theory predicts that the specific heat should exhibit a discontinuity like so at a phase transition:

However, if one examines the specific heat anomaly in liquid helium-4, for example, it looks more like a divergence as seen below:

So it clearly doesn’t predict the right critical exponent in that case. The Ginzburg criterion tells us how close to the transition temperature Landau theory will fail. The Ginzburg argument essentially goes like so: since Landau theory neglects fluctuations, we can see how accurate Landau theory is going to be by calculating the ratio of the fluctuations to the order parameter:

$E_R = |G(R)|/\eta^2$

where $E_R$ is the error in Landau theory, $|G(R)|$ quantifies the fluctuations and $\eta$ is the order parameter. Basically, if the error is small, i.e. $E_R << 1$, then Landau theory will work. However, if it approaches $\sim 1$, Landau theory begins to fail. One can actually calculate both the order parameter and the fluctuation region (quantified by the two-point correlation function) within Landau theory itself and therefore use Landau theory to calculate whether or not it will fail.

If one does carry out the calculation, one gets that Landau theory will work when:

$t^{(4-d)/2} >> k_B/\Delta C \xi(1)^d \equiv t_{L}^{(4-d)/2}$

where $t$ is the reduced temperature, $d$ is the dimension, $\xi(1)$ is the dimensionless mean-field correlation length at $T = 2T_C$ (extrapolated from Landau theory) and $\Delta C/k_B$ is the change in specific heat in units of $k_B$, which is usually one per degree of freedom. In words, the formula essentially counts the number of degrees of freedom in a volume defined by  $\xi(1)^d$. If the number of degrees of freedom is large, then Landau theory, which averages the interactions from many particles, works well.

So that was a little bit of a mouthful, but the important thing is that these quantities can be estimated quite well for many phases of matter. For instance, in liquid helium-4, the particle interactions are very short-ranged because the helium atom is closed-shell (this is what enables helium to remain a liquid all the way down to zero temperatures at ambient conditions in the first place). Therefore, we can assume that $\xi(1) \sim 1\textrm{\AA}$, and hence $t_L \sim 1$ and deviations from Landau theory can be easily observed in experiment close to the transition temperature.

Despite the qualitative similarities between superfluid helium-4 and superconductivity, a topic I have addressed in the past, Landau theory works much better for superconductors. We can also use the Ginzburg criterion in this case to calculate how close to the transition temperature one has to be in order to observe deviations from Landau theory. In fact, the question as to why Ginzburg-Landau theory works so well for BCS superconductors is what awakened me to these issues in the first place. Anyway, we assume that $\xi(1)$ is on the order of the Cooper pair size, which for BCS superconductors is on the order of $1000 \textrm{\AA}$. There are about $10^8$ particles in this volume and correspondingly, $t_L \sim 10^{-16}$ and Landau theory fails so close to the transition temperature that this region is inaccessible to experiment. Landau theory is therefore considered to work well in this case.

For high-Tc superconductors, the Cooper pair size is of order $10\textrm{\AA}$ and therefore deviations from Landau theory can be observed in experiment. The last thing to note about these formulas and approximations is that two parameters determine whether Landau theory works in practice: the number of dimensions and the range of interactions.

*Much of this post has been unabashedly pilfered from N. Goldenfeld’s book Lectures on Phase Transitions and the Renormalization Group, which I heartily recommend for further discussion of these topics.

## Broken Symmetry and Degeneracy

Often times, when I understand a basic concept I had struggled to understand for a long time, I wonder, “Why in the world couldn’t someone have just said that?!” A while later, I will then return to a textbook or paper that actually says precisely what I wanted to hear. I will then realize that the concept just wouldn’t “stick” in my head and required some time of personal and thoughtful deliberation. It reminds me of a humorous quote by E. Rutherford:

All of physics is either impossible or trivial.  It is impossible until you understand it and then it becomes trivial.

I definitely experienced this when first studying the relationship between broken symmetry and degeneracy. As I usually do on this blog, I hope to illustrate the central points within a pretty rigorous, but mostly picture-based framework.

For the rest of this post, I’m going to follow P. W. Anderson’s article More is Different, where I think these ideas are best addressed without any equations. However, I’ll be adding a few details which I wished I had understood upon my first reading.

If you Google “ammonia molecule” and look at the images, you’ll get something that looks like this:

With the constraint that the nitrogen atom must sit on a line through the center formed by the triangular network of hydrogen atoms, we can approximate the potential to be one-dimensional. The potential along the line going through the center of the hydrogen triangle will look, in some crude approximation, something like this:

Notice that the molecule has inversion (or parity) symmetry about the triangular hydrogen atom network. For non-degenerate wavefunctions, the quantum stationary states must also be parity eigenstates. We expect, therefore, that the stationary states will look something like this for the ground state and first excited state respectively:

Ground State

First Excited State

The tetrahedral (pyramid-shaped) ammonia molecule in the image above is clearly not inversion symmetric, though. What does this mean? Well, it implies that the ammonia molecule in the image above cannot be an energy eigenstate. What has to happen, therefore, is that the ammonia molecule has to oscillate between the two configurations pictured below:

The oscillation between the two states can be thought of as the nitrogen atom tunneling from one valley to the other in the potential energy diagram above. The oscillation occurs about 24 billion times per second or with a frequency of 24 GHz.

To those familiar with quantum mechanics, this is a classic two-state problem and there’s nothing particularly new here. Indeed, the tetrahedral structures can be written as linear combinations of the symmetric and anti-symmetric states as so:

$| 1 \rangle = \frac{1}{\sqrt{2}} (e^{i \omega_S t}|S\rangle +e^{i \omega_A t}|A\rangle)$

$| 2 \rangle = \frac{1}{\sqrt{2}} (e^{i \omega_S t}|S\rangle -e^{i \omega_A t}|A\rangle)$

One can see that an oscillation frequency of $\omega_S-\omega_A$ will result from the interference between the symmetric and anti-symmetric states.

The interest in this problem, though, comes from examining a certain limit. First, consider what happens when one replaces the nitrogen atom with a phosphorus atom (PH3): the oscillation frequency decreases to about 0.14 MHz, about 200,000 times slower than NH3. If one were to do the same replacement with an arsenic atom instead (AsH3), the oscillation frequency slows down to 160 microHz, which is equivalent to about an oscillation every two years!

This slowing down can be simply modeled in the picture above by imagining the raising of the barrier height between the two valleys like so:

In the case of an amino acid or a sugar, which are both known to be chiral, the period of oscillation is thought to be greater than the age of the universe. Basically, the molecules never invert!

So what is happening here? Don’t worry, we aren’t violating any laws of quantum mechanics.

As the barrier height reaches infinity, the states in the well become degenerate. This degeneracy is key, because for degenerate states, the stationary states no longer have to be inversion-symmetric. Graphically, we can illustrate this as so:

Symmetric state, $E=E_0$

Anti-symmetric state, $E=E_0$

We can now write for the symmetric and anti-symmetric states:

$| 1 \rangle = e^{i\omega t} \frac{1}{\sqrt{2}} (|S\rangle + |A\rangle)$

$| 2 \rangle = e^{i\omega t} \frac{1}{\sqrt{2}} (|S\rangle - |A\rangle)$

These are now bona-fide stationary states. There is therefore a deep connection between degeneracy and the broken symmetry of a ground state, as this example so elegantly demonstrates.

When there is a degeneracy, the ground state no longer has to obey the symmetry of the Hamiltonian.

Technically, the barrier height never reaches infinity and there is never true degeneracy unless the number of particles in the problem (or the mass of the nitrogen atom) approaches infinity, but let’s leave that story for another day.