Group Theory and Energy Eigenstates in Quantum Mechanics

[praharmitra]

I have only recently begun to study group theory from Lie Algebra for Particle Physicists by Georgi. I am slightly confused about the language used by physicists.

What does it mean when the following is stated

"The energy eigenstates transform like irreducible representations of the group G"

(G is a transformation group that is a symmetry of a quantum mechanical system.

Does it mean that if u take irreducible representations of the group G (which are linear operators) and act them on the energy eigenstates, you get new eigenstates with the same energy?

 

[fresh_42]

The language physicists use in the area is quite a bit different from how mathematicians would say it.

The eigenstates are vectors, say of $V$. Then the group $G$ as well as its Lie algebra $\mathfrak{g}$ act on $\mathfrak{gl}(V)$ e.g. via the adjoint representations if $V=\mathfrak{g}$. I assume that we have the latter here, i.e. $\varphi\, : \,\mathfrak{g} \longrightarrow \mathfrak{gl}(V)$. So if we have any representation $(V,\varphi)$ of $\mathfrak{g}$, then, since $\mathfrak{g}$ is semisimple, $V=V^{(1)}\oplus \ldots V^{(m)}$ splits into irreducible subspaces $V^{(k)}$, i.e. $\{\,0\,\}$ and $V^{(k)}$ are the only $\varphi-$invariant subspaces. Nevertheless, the $V^{(k)}$ split into a direct sum $V^{(k)}=V^{(k)}_1\oplus \ldots \oplus V^{(k)}_{n_k}$ of subspaces which are invariant under the CSA of $\mathfrak{g}$ and the root spaces of $\mathfrak{g}$ shift vectors from on $V^{(k)}_i$ to another $V^{(k)}_j$. The energy eigenstates are represented by those irreducible components $V^{(k)}$.