# Selection Rules using Group Theory

I was learning about the applications of Group Theory and one important application is looking at the selection rules in a weak EM field. We essentially want to see whether the matrix element $$\langle i|H'|f\rangle$$ vanishes due to the reasons of symmetry. Here $$i$$ denotes the initial state, $$f$$ denotes the final state and $$H'$$ denotes the perturbed part of the hamiltonian. Dresselhaus (link: http://web.mit.edu/course/6/6.734j/www/group-full02.pdf, page 139, equation 7.29) states that the perturbed Hamiltonian can be expanded in terms of the irreducible representations of the group of the Hamiltonian.

How can we have a right to do this as the perturbed hamiltonian is not necessarily in the group of the unperturbed Hamiltonian? I understand that by Maschke's theorem we can expand any representation of group $$G$$ by its irreducible representations. But can an arbitrary reducible representation of a group $$H$$, be expanded by the irreducible representations of $$G$$? Furthermore, even though the irreps (irreducible representations) of $$G$$ form on orthogonal basis, how can we make sure that the perturbed Hamiltonian spans a space that has the same dimensionality as the irreps of the Group of the Hamiltonian?

A useful way to see through this is write the perturbation $$H’$$ as \begin{align} H’=\sum_{ij} h_{ij}\vert i\rangle\langle j\vert\, ,\qquad h_{ij}=\langle i\vert H’\vert j\rangle. \end{align} If $$\vert j\rangle$$ is in the irrep $$\Gamma_1$$, and $$\vert i\rangle$$ in $$\Gamma_2$$, then $$\vert i\rangle\langle j\vert$$ is in $$\Gamma_2^*\otimes\Gamma_1$$, where $$\Gamma_2^*$$ is the irrep conjugate to $$\Gamma_2$$. Since the $$h_{ij}$$ are just numbers, this shows that one can indeed expand $$H’$$ in terms of irreps of the group $$G$$ for the unperturbed Hamiltonian.

This is not a constructive way of expanding, since you would need to know the $$h_{ij}\ne 0$$, but it does show that such an expansion is possible if $$\langle i\vert H’\vert j\rangle\ne 0$$.

Now, if contrariwise you assume $$H’$$ cannot be so expanded, then $$H’\vert j\rangle$$ can given you a vector $$\vert k\rangle$$ that is NOT in an irrep of $$G$$, the symmetry group of your unperturbed Hamiltonian. You then have a different problem, which is to understand how $$\vert k\rangle$$ appears in your unperturbed Hilbert space, i.e. your unperturbed states do not span the entire Hilbert space of your problem.

• Thank you very much, this has been very helpful ! Nov 25, 2021 at 9:07

We essentially want to see whether the matrix element <i|H'|f> vanishes due to the reasons of symmetry

How can we have a right to do this as the perturbed hamiltonian is not necessarily in the group of the unperturbed Hamiltonian ?

The states "|i>" and "|f>" are typically taken to be states of the unperturbed Hamiltonian. Otherwise you can't get very far, in general.

For example, if i and f are single particle atomic states (states for a spherically symmetric potential) and the perturbation is the $$\vec{r} \cdot \vec E$$, you can expand i and f in terms of $$Y_{lm}$$ functions. You can also write $$\vec r \cdot \vec E$$ as a linear combo of the $$Y_{1m}$$ functions. Then you get selected rules pretty straightforwardly (Clebsch Gordan coefficients and whatnot).

Here, it is the spherically symmetric potential of the unperturbed Hamiltonian that allows us to use the $$Y_{lm}$$ functions to expand the unperturbed eigenstates in a helpful way.

• I understand that but what is stopping us from having a perturbed potential that is not in the group of the hamiltonian and cannot be expanded in terms of the irreps of the group of the hamiltonian ? Nov 25, 2021 at 0:11
• Nothing is stopping us. The group theory helps us when it can. If we had to include a ton of terms in the expansion of the perturbation, or if we couldn't do an expansion, then group theory would not be of much help.
– hft
Nov 25, 2021 at 0:13