$\DeclareMathOperator\SO{SO}\newcommand{\R}{\mathbb{R}}\newcommand{\S}{\mathbb{S}}$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 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?

$\DeclareMathOperator\SO{SO}\newcommand{\R}{\mathbb{R}}\newcommand{\S}{\mathbb{S}}$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?

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 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?

$\DeclareMathOperator\SO{SO}\newcommand{\R}{\mathbb{R}}\newcommand{\S}{\mathbb{S}}$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 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?