Skip to main content

All Questions

7 votes
2 answers
101 views

Point group of harmonic oscillator

The book Molecular Quantum Mechanics by Atkins and Friedman [1] says the point group of a harmonic oscillator is $C_\mathrm{s},$ composed by the identity operator $E$ and a reflection $\sigma_\mathrm{...
AlfredV's user avatar
  • 461
1 vote
1 answer
81 views

What does it mean that a state belongin to a given irrep transforms like $Rx$, $Ry$ or $Rz$

The present question is related to this other question I did few days ago. Given a point group and the list of the irreps (see for example here) the meaning of an irrep which transforms like $x$ or $x^...
Davide Sangalli's user avatar
7 votes
1 answer
412 views

How do I show that a transition is electric dipole allowed with group theory/symmetry?

This is actually a follow up of this question The follow-up is not because of the electric instead of magnetic dipole (this is trivial). It is because I'm interested in extra info. Suppose I have a ...
Davide Sangalli's user avatar
10 votes
0 answers
141 views

Using symmetry and group theory arguments to explain iron(II) in a tetrahedral crystal field

I am trying to figure out how to explain $1s \rightarrow 3d$ spectroscopic transitions for $\ce{Fe^{2+}}$ in $T_\mathrm{d}$ symmetry. These transitions make up the pre-edge region in K edge X-ray ...
Silverwilly's user avatar
2 votes
1 answer
91 views

How to find a symmetry group of a system if all the symmetry transformations do not obey closure and don't form a group?

For instance, consider a system with $p_x$ and $p_z$ orbitals at the vertices of a square (on xy-plane). A square by itself would have $D_4$ symmetry. However, because of the $p_x$ orbital; the $90^\...
Feynfan's user avatar
  • 21
8 votes
1 answer
3k views

Determination of +/- reflections in sigma molecular term symbols

This page, which depicts the molecular term symbols for the $\ce{O2}$ and $\ce{O2-}$ (Edit: Mistaken on $\ce{O2-}$) molecules, perhaps best summarizes the full scope of my questions. In general, I don'...
Blaise's user avatar
  • 1,648