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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?

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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.

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  • $\begingroup$ Thank you very much, this has been very helpful ! $\endgroup$
    – emir sezik
    Nov 25 at 9:07
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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.

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  • $\begingroup$ 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 ? $\endgroup$
    – emir sezik
    Nov 25 at 0:11
  • $\begingroup$ 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. $\endgroup$
    – hft
    Nov 25 at 0:13

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