## 1. Perturbations

As already mentioned, the Jahn-Teller effect has its roots in group theory. The essence of the argument is that the energy of the compound is stabilised upon distortion to a lower-symmetry point group. This distortion may be considered to be a normal mode of vibration, with the corresponding vibrational coordinate $q$ labelling the "extent of distortion". There is one condition on the vibrational mode: it cannot transform as the totally symmetric irreducible representation of the molecular point group, as such a vibrational mode cannot bring about any distortion in the molecular geometry.
$\require{begingroup} \begingroup
\newcommand{\ket}[1]{| #1 \rangle}
\newcommand{\odiff}[2]{\frac{\md #1}{\md #2}}$

In the undistorted geometry (i.e. $q = 0$), the electronic Hamiltonian is denoted $H_0$. The corresponding unperturbed electronic wavefunction is $\ket{\n{0}}$, and the electronic energy is $\En{0}$. We therefore have

$$H_0 \ket{\n{0}} = \En{0}\ket{\n{0}} \tag{1}$$

Here, the Hamiltonian, wavefunction, and energy are all functions of $q$. We can expand them as Taylor series about $q = 0$:

H &= H_0 + q \left(\odiff{H}{q}\right) + \frac{q^2}{2}\left(\frac{\md^2 H}{\md q^2}\right) + \cdots \tag{2} \\
\ket{n} &= \ket{\n{0}} + q\ket{\n{1}} + \frac{q^2}{2}\ket{\n{2}} + \cdots \tag{3} \\
E_n &= \En{0} + q\En{1} + \frac{q^2}{2}\En{2} + \cdots \tag{4}

In the new geometry (i.e. $q \neq 0$), the Schrodinger equation must still be obeyed and therefore

$$H\ket{n} = E_n \ket{n} \tag{5}$$

By substituting in equations $(2)$ through $(4)$ into equation $(5)$, one can compare coefficients of $q$ to reach the results:

\En{1} &= \left< \n{0} \middle| \odiff{H}{q} \middle| \n{0} \right> \tag{6} \\
\En{2} &= \left< \n{0} \middle| \frac{\md^2 H}{\md q^2} \middle| \n{0} \right> + 2\sum_{m \neq n}\frac{\left|\left<m^{(0)} \middle|(\md H/\md q)\middle|\n{0} \right>\right|^2}{\En{0} - E_m^{(0)}} \tag{7}

The derivation of equations $(6)$ and $(7)$ will not be discussed further here.<sup>1</sup>

Distortions that arise from a negative value of $\En{1}$ are called **first-order** Jahn-Teller distortions, and distortions that arise from a negative value of $\En{2}$ are called **second-order** Jahn-Teller distortions.


## 2. The first-order Jahn-Teller effect

Recall that

$$E_n = \En{0} + q\En{1} + \cdots \tag{8}$$

Therefore, if $\En{1} > 0$, then stabilisation may be attained with a negative value of $q$; if $\En{1} < 0$, then stabilisation may be attained with a positive value of $q$. These simply represent distortions in opposite directions along a vibrational coordinate. A well-known example is the distortion of octahedral $\ce{Cu^2+}$: there are two possible choices, one involving axial compression, and one involving axial elongation. These two distortions arise from movement along the same vibrational coordinate, except that one has $q > 0$ and the other has $q < 0$.

Therefore, in order for there to be a first-order Jahn-Teller distortion, we require that $\En{1} \neq 0$. Within group theory, the condition for the integral to be nonzero is that the integrand must contain a component that transforms as the totally symmetric irreducible representation (TSIR). In other words

$$\Gamma_{\text{TSIR}} \in \Gamma_n \otimes \Gamma_{(\md H/\md q)} \otimes \Gamma_n$$

We can simplify this slightly by noting that the Hamiltonian, $H$, itself transforms as the TSIR. Therefore, $\md H/\md q$ transforms as $\Gamma_q$, and the requirement is that

$$\Gamma_{\text{TSIR}} \in \Gamma_n \otimes \Gamma_q \otimes \Gamma_n$$

In all point groups, for any non-degenerate irrep $\Gamma_n$, $\Gamma_n \otimes \Gamma_n = \Gamma_{\text{TSIR}}$. Therefore, if $\Gamma_n$ is non-degenerate, then

$$\Gamma_n \otimes \Gamma_q \otimes \Gamma_n = \Gamma_q \neq \Gamma_{\text{TSIR}}$$

and the molecule is stable against a first-order Jahn-Teller distortion. Therefore, all closed-shell molecules ($\Gamma_n = \Gamma_{\text{TSIR}}$) do not undergo first-order Jahn-Teller distortions.

However, what happens if $\Gamma_n$ is non-degenerate? Now, the product $\Gamma_n \otimes \Gamma_n$ contains other irreps apart from the TSIR.<sup>2</sup> If the molecule possesses a vibrational mode that transforms as one of these irreps, then the direct product $\Gamma_n \otimes \Gamma_q \otimes \Gamma_n$ will contain the TSIR.

In a rather inelegant article,<sup>3</sup> Hermann Jahn and Edward Teller did the maths for every important point group and found that:

> stability and degeneracy are not possible simultaneously unless the molecule is a linear one...

In other words, if a **non-linear molecule** has a **degenerate ground state**, then it is susceptible towards a (first-order) Jahn-Teller distortion.

Take, for example, octahedral $\ce{Cu^2+}$. This has a $\mathrm{^2E_g}$ term symbol (see [this question](https://chemistry.stackexchange.com/questions/64458)) - which is doubly degenerate. The symmetric direct product $\mathrm{E_g \otimes E_g = A_{1g} + E_g}$. Therefore, if we have a vibrational mode of symmetry $\mathrm{E_g}$ (and we do), then distortion along this vibrational coordinate will occur to give a more stable compound.

What does an $\mathrm{e_g}$ vibrational mode look like? Here is a diagram:<sup>4</sup>

[![Vibrational modes of Oh complex][1]][1]

It's an axial elongation, just as expected! However, there is a slight catch: the vibrational mode is doubly degenerate (the other $\mathrm{e_g}$ mode is not shown), and any linear combination of these two degenerate vibrational modes also transforms as $\mathrm{e_g}$. Therefore, the exact form of the distortion can be any linear combination of these two degenerate modes.

On top of that, there's also no indication of how *much* distortion there is. That depends on (amongst other things) the value of $\En{1}$, and all we have said is that it is *nonzero* - we have not said *how large* it is.

This is what is meant by "impossible to predict the extent or the exact form of the distortion".


## 3. The second-order Jahn-Teller effect

Pearson has written an article on second-order Jahn-Teller effects.<sup>5</sup> **To be continued another time**


## Notes and references

(1) For more details, look up perturbation theory in your quantum mechanics book of choice. In such treatments, the perturbation is usually formulated slightly differently: e.g. $H$ is taken as $H_0 + \lambda V$, and the eigenstates and eigenvalues are expanded as a power series instead of a Taylor series. Notwithstanding that, the principles remain the same.

(2) There is a subtlety in that the symmetric direct product must be taken. For example, in the $D_\mathrm{\infty h}$ point group, we have $\Pi \otimes \Pi = \Sigma^+ + [\Sigma^-] + \Delta$. The antisymmetric direct product $\Sigma^-$ has to be discarded.

(3) Jahn, H. A.; Teller, E. Stability of Polyatomic Molecules in Degenerate Electronic States. I. Orbital Degeneracy. *Proc. R. Soc. A* **1937,** *161* (905), 220–235. [DOI: 10.1098/rspa.1937.0142](http://doi.org/10.1098/rspa.1937.0142).

(4) Albright, T. A.; Burdett, J. K.; Whangbo, M.-H. *Orbital Interactions in Chemistry,* 2nd ed.; Wiley: Hoboken, NJ, 2013.

(5) Pearson, R. G. The second-order Jahn-Teller effect. *J. Mol. Struct.: THEOCHEM* **1983,** *103,* 25–34. [DOI: 10.1016/0166-1280(83)85006-4](http://doi.org/10.1016/0166-1280(83)85006-4).

  [1]: https://i.stack.imgur.com/sF9VRm.jpg