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):

Jahn-Teller.png

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).

highSpin_Jahnteller

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.

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3 responses to “Jahn-Teller Distortion and Symmetry Breaking

  1. Is quasicrystal formation due Jahn Teller distortion? Asking out of ignorance

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  2. due to JT distortion , pl read as

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    • Not to my knowledge, no. There are many reasons solids can distort, and the Jahn-Teller effect is only one of them.

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