When the character in a point group table is listed as 1 or -1, I understand how an object/orbital etc. transforms to give either itself or the negative of itself.

When the character in a point group table is 2 or -2, what does this mean in terms of transformation? To understand this would be very helpful in terms of assigning the correct irreducible representations to atomic orbitals/LGOs to then determine which pairs mix.

Similarly, I do not understand how to interpret a character of sin or cos of an angle, nor a character of 0.

I am hoping there is a conceptual/visual way of explaining this as I am not up to speed on the matrix math, and the class that this is relevant too does not include that math either.

  • 1
    $\begingroup$ There are many sorts of characters needed to represent the symmetry behaviour; sine, cosine, complex numbers, complex exponential, square roots and you have to treat each the same way. The 2 and 3 indicates degeneracy, so is associated with E and T Mulliken symbols. Sometimes, as in the S4 group E appear as two rows so the degeneracy is clear. There is some information here chemistry.stackexchange.com/questions/58609/… $\endgroup$ – porphyrin Feb 10 at 8:48

When you observe that in the characer table of a symmetry group there are irreducible representations where the character of the identity operation (E) is 2 or higher, you may find that the character of some symmetry operations is 0, +2, -2, etc. This is due to the fact that this IR are degenerate.

To illustrate this idea, consider that you trying to determine the symmetry of the MO of bezene (D6h). You will find that some of these orbitals have the same energy (are degenerate) for example

benzene OM1 Benzene OM2

When you apply the symmetry operations of the D6h point group you can check that what you obtain is a linear combination of these orbitals. If you express this linear combinations in matrix form, the character of the matrix will give you the value that appears in the character table for the symmetry of that orbital. In this case these orbitals have a E1g symmetry.


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