The Kepler Problem (Part 5)

Azimuth 2025-07-23

In Part 4 we saw how the classical Kepler problem is connected to a particle moving on a sphere in 4-dimensional space, and how this illuminates the secret 4-dimensional rotation symmetry of the Kepler problem. There are various ways to quantize the Kepler problem and obtain a description of the hydrogen atom’s bound states as wavefunctions on a sphere in 4-dimensional space. Here I’ll take a less systematic but quicker approach.

I’ll start with wavefunctions on the sphere in 4-dimensional space and describe operators on these arising from rotational symmetries. Combining these with standard facts about the hydrogen atom—which I’ll take as known, rather than derive—we can set up a correspondence between such wavefunctions and bound states of a hydrogen atom.

Here’s how. Chemists use a basis of bound states of the hydrogen called $|n, \ell, m \rangle$ where

• n tells us the energy of the state, • ℓ tells us its total angular momentum, and • m tells us its angular momentum around a chosen axis, say the z axis.

For now we’re ignoring the electron’s spin: we’ll bring that in later. We’ll show that wavefunctions on the sphere in 4-dimensional space also have a nice basis of states called $|n, \ell, m \rangle.$ This correspondence will let us show that the hydrogen atom has 4-dimensional rotation symmetry: all these rotations preserve the energy, which depends only on n.

So let’s get started!

We can identify the sphere in 4-dimensional space, $S^3,$ with the group $\text{SU}(2)$ of $2 \times 2$ unitary matrices with determinant 1. This group acts on itself in three important ways, which combine to give the mathematics we need:

• left multiplication by $g$ : $g$ maps $h$ to $gh,$

• right multiplication by $g^{-1}$ : $g$ maps $h$ to $hg^{-1},$

• conjugation by $g$ : $g$ maps $h$ to $ghg^{-1}.$

Left and right multiplication commute, so they combine to give a left action of $\text{SU}(2) \times \text{SU}(2)$ on $\text{SU}(2).$ We thus obtain a unitary representation $R$ of $\text{SU}(2) \times \text{SU}(2)$ on Hilbert space of wavefunctions on $\text{SU}(2),$ called $L^2(\text{SU}(2)),$ given by

$(R(g_1, g_2) \psi)(g) = \psi(g_1^{-1} g g_2)$

This is the key to everything in this game.

In what follows I’ll denote $\text{SU}(2)$ as $S^3$ when we regard it as the unit sphere in $\mathbb{R}^4,$ acted on by $\text{SU}(2) \times \text{SU}(2)$ as above. So I’ll write $L^2(S^3)$ instead of $L^2(\text{SU}(2)),$ but they’re really the same thing.

The Peter–Weyl theorem lets us decompose $L^2(S^3)$ into finite-dimensional irreducible unitary representations of $\text{SU}(2) \times \text{SU}(2)$ :

$\displaystyle{ L^2(S^3) \cong \bigoplus_{j} V_j \otimes V_j }$

Here $j = 0, \frac{1}{2}, 1, \frac{3}{2}, \dots$ and $V_j$ is the spin- $j$ representation of $\text{SU}(2),$ which is the irreducible unitary representation of dimension $2j + 1.$ If you don’t know the Peter–Weyl theorem, you should quit reading this article now and learn about that, because it’s a lot more important than anything I’m talking about! It’s a key tool throughout group representation theory and quantum mechanics.

The representation of $\text{SU}(2) \times \text{SU}(2)$ on $L^2(S^3)$ is generated by self-adjoint operators that correspond to familiar observables for bound states of the hydrogen atom! To see this, first note that the complexification of $\mathfrak{su}(2) \oplus \mathfrak{su}(2)$ has self-adjoint elements

$A_j = (\frac{1}{2} \sigma_j, 0), \qquad B_j = (0, \frac{1}{2} \sigma_j)$

where $\sigma_j$ are the Pauli matrices. We also use $A_j$ and $B_j$ to denote the corresponding self-adjoint operators on $L^2(S^3).$ They obey the following commutation relations, which are the quantum-mechanical analogues of the Poisson brackets we saw last time:

$\begin{array}{ccc} [A_j, A_k] &=& i\epsilon_{jk\ell} A_\ell \\ [2pt] [B_j, B_k] &=& i\epsilon_{jk\ell} B_\ell \\ [2pt] [A_j, B_k] &=& 0 \end{array}$

Geometrically speaking, the skew-adjoint operators $-iA_j$ acts on $L^2(S^3)$ as differentiation by vector fields on $S^3 \cong \text{SU}(2)$ that generate left translations, while the operators $-iB_j$ act by differentiation by vector fields that generate right translations. We also use $-iA_j$ and $-iB_j$ to stand for these vector fields. Beware: left-invariant-vector fields generate right translations and vice versa, the vector fields $-iA_j$ are right-invariant, while the $-iB_j$ are left-invariant!

As well known, we have

$\phi \in V_j \otimes V_j \implies A^2 \phi = B^2 \phi = j(j+1) \phi$

This implies that

$A^2 = B^2$

on all of $L^2(S^3).$ In terms of these we can define an operator on $L^2(S^3)$ that corresponds to the Hamiltonian for bound states of the hydrogen atom, namely

$\displaystyle{ H_0 \;=\; - \frac{1}{8(A^2 + \frac{1}{4})} \;=\;- \frac{1}{8(B^2 + \frac{1}{4})} }$

This is the quantum analogue of the classical Hamiltonian we saw last time! Later I’ll talk about the curious appearance of the number $\frac{1}{4}$ here. It holds deep mysteries, but we can see why it’s needed if we know the energy levels of the hydrogen atom. On the subspace $V_j \otimes V_j,$ the operator $4A^2 + 1$ acts as multiplication by

$4j(j+1) + 1 = 4j^2 + 4j + 1 = (2j + 1)^2.$

In atomic physics it is traditional to work with the dimension of $V_j,$

$n = 2j + 1$

rather than $j$ itself. Thus, we have

$\displaystyle{ \phi \in V_j \otimes V_j \implies H_0\phi = - \frac{1}{2n^2} \phi }$

Great! These are precisely the usual energy eigenvalues for the hydrogen atom, expressed in units with

$\hbar = e = 4 \pi \epsilon_0 = \mu = 1$

where:

• $\hbar$ is Planck’s constant,

• $-e$ is the charge of the electron (yes, some idiot made the electron charge negative),

• $\epsilon_0$ is the permittivity of the vacuum, and

• $\mu = m_e/(m_e + m_p)$ is the reduced mass of the electron.

So: we needed that $\frac{1}{4}$ in the Hamiltonian $H_0$ to get the usual energy eigenvalues for hydrogen.

We have not yet brought in the action of $\text{SU}(2)$ on $S^3$ by conjugation. This gives yet another important unitary representation of $\text{SU}(2)$ on $L^2(S^3).$ To conjugate by $g$ we both left multiply by $g$ and right multiply by $g^{-1}.$ Thus, the conjugation representation of $\text{SU}(2)$ on $L^2(S^3)$ has self-adjoint generators

$L_j = A_j + B_j$

and these obey

$[L_j, L_k] = i\epsilon_{jk\ell} L_\ell$

As we saw last time for the classical Kepler problem, the operators $L_j$ are the components of the angular momentum. Now we’re doing quantum mechanics, so this will be the angular momentum of a hydrogen atom—not counting the electron’s spin.

With respect to the conjugation representation, the subspace $V_j \otimes V_j \subset L^2(S^3)$ decomposes according to the usual rules for a tensor product of two representations of $\text{SU}(2)$ :

$\displaystyle{ V_j \otimes V_j \cong \bigoplus_{\ell = 0}^{2j} V_\ell }$

Here $\ell$ takes integer values going from $0$ to $2j = n - 1.$ Thus, with respect to the conjugation representation, we can decompose $L^2(S^3)$ into irreducible representations of $\text{SU}(2),$ like this:

$\displaystyle{ L^2(S^3) \cong \bigoplus_{n = 1}^\infty \bigoplus_{\ell = 0}^{n-1} V_\ell }$

The summand $V_\ell$ has a basis of eigenvectors for the $z$ component of angular momentum, $L_3,$ with eigenvalues taking integer values $m$ ranging from $-\ell$ to $\ell.$ Thus $L^2(S^3)$ has an orthonormal basis of states $|n , \ell , m\rangle$ where:

• $n$ ranges over positive integers;

• $\ell$ ranges from $0$ to $n-1$ in integer steps;

• $m$ ranges from $-\ell$ to $\ell$ in integer steps.

From the calculations thus far we have

$\begin{array}{ccl} A^2 |n, \ell, m \rangle &=& B^2 |n, \ell, m \rangle \; = \; \frac{1}{4}(n^2-1) |n, \ell, m \rangle \\ [8pt] H_0 |n, \ell, m \rangle &=& \displaystyle{ - \frac{1}{2n^2} \, |n, \ell, m \rangle } \\ [8pt] L^2 |n, \ell, m \rangle &=& \ell(\ell + 1) \, |n , \ell, m \rangle \\ [3pt] L_3 |n , \ell, m \rangle &=& m \, |n , \ell, m \rangle \end{array}$

The last three relations are familiar from work on the hydrogen atom. In this context

• $n$ is the ‘principal quantum number’,

• $\ell$ is the ‘azimuthal quantum number’,

• $m$ is the ‘magnetic quantum number’.

The operator $H_0$ corresponds to the Hamiltonian of the hydrogen atom, with the electron treated as spinless, and the states $|n, \ell, m\rangle$ are a well-known basis of the hydrogen atom’s bound states.

So, we’ve gotten to hydrogen, starting from just the space of wavefunctions on the 3-sphere, heavily using the fact that this sphere can be seen as the group $\text{SU}(2),$ which acts on itself in three ways. Pretty cool!

Next I’ll say more about that mysterious $\frac{1}{4}$ in the formula for the hydrogen atom Hamiltonian. Then I’ll bring in spin.