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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 $T^*(S^3)$ a constant energy surface 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.]

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.]

I haven't worked through the details, but since nobody has answered this fully, I'm going to crib from this presentation by @JohnBaez to hopefully expand on the comments above.

First, I'm going to state a few facts without proof. The space of bound state solutions to the Kepler problem in three dimensions is the (punctured) cotangent bundle to $S^3$. Thus, the classical question is equivalent to a free particle on $S^3$. I believe this is due to Moser. The $SO(4)$ action arises from the two $SU(2)$ factors in $\operatorname{Spin}(4)$ acting on both sides of $S^3 \cong SU(2)$.

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 $\rho$ is an irrep of $SU(2)$, and the irreps can be labelled by an integer $n$, so the problem decomposes into $n^2$-dimensional irreps of $SO(4)$, which explains the question asked.

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 $T^*(S^3)$ a constant energy surface 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.]

I haven't worked through the details, but since nobody has answered this fully, I'm going to crib from this presentation by @JohnBaez to hopefully expand on the comments above.

First, I'm going to state a few facts without proof. The space of bound state solutions to the Kepler problem in three dimensions is the (punctured) cotangent bundle to $S^3$. Thus, the classical question is equivalent to a free particle on $S^3$. I believe this is due to Moser. The $SO(4)$ action arises from the two $SU(2)$ factors in $Spin(4)$ acting on both sides of $S^3 \cong SU(2)$.

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 $\rho$ is an irrep of $SU(2)$, and the irreps can be labelled by an integer $n$, so the problem decomposes into $n^2$ dimensional irreps of $SO(4)$, which explains the question asked.

I haven't worked through the details, but since nobody has answered this fully, I'm going to crib from this presentation by @JohnBaez to hopefully expand on the comments above.

First, I'm going to state a few facts without proof. The space of bound state solutions to the Kepler problem in three dimensions is the (punctured) cotangent bundle to $S^3$. Thus, the classical question is equivalent to a free particle on $S^3$. I believe this is due to Moser. The $SO(4)$ action arises from the two $SU(2)$ factors in $\operatorname{Spin}(4)$ acting on both sides of $S^3 \cong SU(2)$.

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 $\rho$ is an irrep of $SU(2)$, and the irreps can be labelled by an integer $n$, so the problem decomposes into $n^2$-dimensional irreps of $SO(4)$, which explains the question asked.