$\DeclareMathOperator\SO{SO}\newcommand{\R}{\mathbb{R}}\newcommand{\S}{\mathbb{S}}$The periodic table of elements has row lengths $2, 8, 8, 18, 18, 32, \ldots $, i.e., perfect squares doubled. The group theoretic explanation for this that I know (forgive me if it is an oversimplification) is that the state space of the hydrogen atom is made of functions on $\R^3$, which we can decompose as functions on $\S^2$ times functions on $\R^+$. Then $\SO(3)$ acts on the functions on $\S^2$ and commutes with the action of the Hamiltonian, so we can find pure states inside irreducible representations of $\SO(3)$. The orbital lengths depend on how these representations line up by energy, which is a function on $\R^+$. It happens that the spaces with the same energy have the form $(V_0 \oplus V_2 \oplus V_4 \oplus \cdots ) \otimes W$, where $V_i$ is the irreducible representation of $\SO(3)$ of dimension $i+1$, and $W$ is a $2$-dimensional space that represents the spin. So the orbital length is the dimension of $V_0 \oplus V_2 \oplus V_4 \oplus \cdots$ is the sum of the first $k$ odd numbers, which is a perfect square, and you double it because of the spin.
So, in short, the perfect squares arise as the sums of the first $k$ odd numbers, and the invariant subspaces arrange themselves into energy levels that way because… well, here I get stuck. Factoring out the spin, which explains the doubling, can anyone suggest a more conceptual (symmetry-based?) explanation for why perfect squares arise here?
Best Answer
In this answer, I'm going to crib from this presentation by @JohnBaez and the paper On the Regularization of the Kepler Problem. Milnor's paper includes a lot of the same information.
First, I'm going to state a few facts without proof. One can compose the stereographic projection of $\mathbb{R}^3$ to $S^3$ with the symplectomorphism swapping $p$s and $q$s on $T^*(\mathbb{R}^3)$ to get a symplectomorphism from the punctured $T^*(S^3)$ to $T^*(\mathbb{R}^3)$.
Furthermore, one can show that the Hamiltonian flow of $p^2$ on a a constant energy surface in $T^*(S^3)$ maps to the Hamiltonian flow of the Kepler potential on $T^*(\mathbb{R}^3)$. This maps the constant energy classical mechanics of a negative energy state in the Kepler potential to a free particle on $S^3$ with fixed energy. This also exhibits the $SO(4)$ symmetry as rotations on $S^3$. Thus (and I'm still undecided if there's some handwaving here), the energy eigenstates in the quantum theory should be irreps of $SO(4)$. You can also exhibit the $SO(4)$ symmetry directly in the quantum theory, so any handwaving isn't a problem.
To see what the representations are, the $SO(4)$ action on $S^3$ can be exhibited by the two $SU(2)$ factors in $\operatorname{Spin}(4)$ acting on both sides of $S^3 \cong SU(2)$. The element $(-I,-I) \in SU(2) \times SU(2)$ acts trivially, so you get an $SO(4)$ action.
With this, we can decompose a la the Peter–Weyl theorem: $$ L^2(S^3) \cong \bigoplus_i \rho_i \otimes \rho_i^\star $$ Each $\rho_i$ is an irrep of $SU(2)$, and those irreps can be labelled by an integer $n$. Thus, the problem decomposes into $n^2$-dimensional irreps of $SO(4)$, which explains the question asked.
[N.B. -- I'd be interested in understanding if this can be done "all at once" as opposed to working with constant energy surfaces and arguing by scaling as I see in the references. If I have time, I'd also want to show that the different irreps of SO(4) have different energies, or maybe I'm missing something obvious. This can all be done by looking at the symmetry explicitly in the quantum theory, I'm sure, but it would be nice to see it geometrically.]