I am trying to understand how to see if a vibrational mode is Jahn-Teller active or not. According to the group theoretical description of the Jahn-Teller effect one needs to check if the symmetric part of the direct product the irreducible representation (=irrep) of the electronic state with itself contains the irrep of the distortional mode in its symmetric part. And this is where I have trouble to understand. I can understand that one can decompose tensors (tensor products) into a symmetric and an anti-symmetric part, but I fail to understand how to apply that to the the direct product of two irreps, as they are in most cases one-dimensional. Can someone explain to me what exactly means "symmetric, anti-symmetric and non-symmetric part" in this context and how to see that in specific cases?

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    $\begingroup$ When they are one-dimensional, there is nothing to decompose. $\endgroup$ – Ivan Neretin May 19 '18 at 12:08
  • $\begingroup$ @IvanNeretin: OK, so lets take the example E$_g$ in D$_{4h}$. $\endgroup$ – Rudi_Birnbaum May 19 '18 at 14:38
  • $\begingroup$ You should probably find some 'direct product' tables to speed up multiplying symmetry species. The symmetric stretch, say in CO2, is when CO bonds both stretch in phase, anti-symmetric when one stretches as the other compresses, $\pi/2$ out of phase. Anti-symmetric is somewhat subjective in a complex molecule and so this and anything else can be called asymmetric. (Asymmetric means 'no symmetry' which in not the case as all symmetry species are determined by the point group.) $\endgroup$ – porphyrin May 19 '18 at 14:44
  • $\begingroup$ In $D_{4h}$ product of E with anything but E is E, and $E \times E \to A_1+[A_2]+B_1+B_2$ where [] mean anti-symmetrized product of a degenerate representation with itself. $\endgroup$ – porphyrin May 19 '18 at 14:55
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    $\begingroup$ There seems to be a description of what you want, re symmetrized and anti - symmetrized direct products in section 5.14 of the 3rd edition of Atkins & Friedman 'Molecular Quantum Mechanics' publ OUP. $\endgroup$ – porphyrin May 19 '18 at 20:02

The symmetric and antisymmetric part of the direct product here refers to the fact that certain irreducible representations in the direct product are symmetric while others are antisymmetric (w.r.t some operations like $\sigma_v$, $C'_2$). So, for instance, when we write:

$$\Pi \times \Pi = \Sigma^+ + [\Sigma^-] + \Delta$$

the $[ \ ]$ simply means that the irreducible representation $\Sigma^-$ is the antisymmetric part of the direct product. It doesn't mean that you are somehow decomposing $\Sigma^-$ into a symmetric and antisymmetric part, and then selecting the antisymmetric one. This won't be possible since $\Sigma^-$ is a one-dimensional representation.

  • $\begingroup$ So what are the "defining" group elements then, that have to be checked, is that an arbitrary definition? One might suppose something like the highest order generator element.... $\endgroup$ – Rudi_Birnbaum Apr 22 '20 at 6:02

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