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: 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.
A: 
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.
