The Kepler Problem (Part 8)

Azimuth 2025-08-01

Now comes the really new stuff. I want to explain how the hydrogen atom is in a certain sense equivalent to a massless spin-½ particle in the ‘Einstein universe’. This is the universe Einstein believed in before Hubble said the universe was expanding! It has a 3-sphere $S^3$ for space, and this sphere stays the same size as time passes.

Today I’ll just lay the groundwork. To study relativistic spin-½ quantum particles, we need to understand the Dirac operator. So, we need to bring out the geometrical content of what we’ve already done.

The main trick is that we can see the 3-sphere as the group $\text{SU}(2)$ , which acts on itself in two ways, via left and right translations. We get all the rotational symmetries of the 3-sphere this way. In Part 4 we studied operators called $A_i$ and $B_i$ on $L^2(S^3),$ which are the self-adjoint generators of these left and right translations. We saw that

$A^2 = B^2$

Today we’ll see that $A^2 = B^2$ is proportional to the Laplacian on the unit 3-sphere!

But then I want to look at spinor fields on the 3-sphere. We can think of elements of $L^2(S^3) \otimes \mathbb{C}^2$ as spinor fields on the 3-sphere if we trivialize the spinor bundle using the action of $\text{SU}(2)$ as right translations on $S^3 \cong \text{SU}(2).$ You could use left translations, but you have to pick one or the other, and we’ll use right translations.

In some notes from a course he gave at Harvard, Peter Kronheimer used this trick to study the Dirac operator $\partial\!\!\!/$ on these spinor fields:

• Peter Kronheimer, Bott periodicity and harmonic theory on the 3-sphere.

I’ll explain some of what he did, and show that the hydrogen atom Hamiltonian, thought of as an operator on $L^2(S^3) \otimes \mathbb{C}^2,$ is

$\displaystyle{ H = - \frac{1}{2 (\partial\!\!\!/ - \frac{1}{2})^2} }$

Next time we’ll use this to relate hydrogen to the massless relativistic spin-½ particle on the Einstein universe.

Okay, on to business!

The Laplacian on the 3-sphere

Let’s start with the Laplacian on the 3-sphere. From what we’ve already seen, the operators $-iB_j$ are a basis of left-invariant vector fields on $S^3.$ Each vector field $-iB_j$ gives a tangent vector at the identity of $\text{SU}(2),$ namely

$-\frac{i}{2} \sigma_j \in \mathfrak{su}(2)$

What is the length of this vector if we give $\text{SU}(2)$ the usual Riemannian metric on the unit 3-sphere? Exponentiating this vector we get

$\exp(-\frac{i}{2} \sigma_j t)$

which is the identity precisely when $t$ is an integer times $4\pi.$ Since a great circle on the unit sphere has circumference $2\pi,$ this vector must have length ½. It follows that the vector fields

$X_j = -2i B_j$

have unit length everywhere, and one can check that they form an orthonormal basis of vector fields on $S^3.$ We thus define the (positive definite) Laplacian on $S^3$ to be differential operator

$\displaystyle{ \Delta = - \sum_{i = 1}^3 X_j^2 = 4B^2 }$

In Part 5 we saw that

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

where $V_j$ is the spin-j representation of $\text{SU}(2).$ We also saw that

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

It follows that

$\phi \in V_j \otimes V_j \implies \Delta \phi = 4j(j+1) \phi$

But chemists like to work with $n = 2j+1$ instead: they call this the ‘principal quantum number’ for a state of hydrogen atom. Since

$4j(j+1) = 4j^2 + 4j = n^2 - 1$

it follows that

$\phi \in V_j \otimes V_j \implies \Delta \phi = (n^2 - 1) \phi$

so the eigenvalues of the Laplacian on the unit 3-sphere are $n^2 - 1$ where $n$ ranges over all positive integers.

Tensoring $\Delta$ with the identity we obtain a differential operator on $L^2(S^3) \otimes \mathbb{C}^2,$ which by abuse of notation we again call $\Delta.$ We know from Part 7 that the hydrogen atom Hamiltonian is

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

but now we know $B^2 = \Delta,$ so

$\displaystyle{ H = - \frac{1}{2(\Delta + 1)} }$

The Dirac operator on the 3-sphere

Next we turn to the Dirac operator.

Up to isomorphism there is only one choice of spin structure on $S^3,$ namely the trivial bundle. To get this can trivialize the tangent bundle of $S^3 \cong \text{SU}(2)$ using left translations. This lets us identify the oriented orthonormal frame bundle of $S^3$ with the trivial bundle

$S^3 \times \text{SO}(3) \to S^3$

This gives a way to identify the spin bundle on $S^3$ with the trivial bundle

$S^3 \times \text{SU}(2) \to S^3$

This in turn lets us identify spinor fields on $S^3$ with $\mathbb{C}^2$ -valued functions.

There are at least two important connections on the tangent bundle of $S^3:$

• One is the Cartan connection: a vector field is covariantly constant with respect to this connection if and only if it is invariant under left translations on $S^3 \cong \text{SU}(2).$

• The other is the Levi–Civita connection: the unique torsion-free connection for which parallel translation preserves the metric.

Parallel translation with respect to the Cartan connection also preserves the metric, but the Cartan connection is flat and has torsion, while the Levi–Civita connection is curved and torsion-free.

Each of these connections lifts uniquely to a connection on the spin bundle which then gives a Dirac-like operator. The Cartan connection gives covariant derivative operators $\nabla^c_j$ on $L^2(S^3) \otimes \mathbb{C}^2$ with

$\nabla^c_j = X_j \otimes 1$

while the Levi–Civita connection gives covariant derivative operators $\nabla_j$ with

$\nabla_j = X_j \otimes 1 - \tfrac{i}{2}(1 \otimes \sigma_j)$

We can define a Dirac operator $\partial\!\!\!/$ on $L^2(S^3) \otimes \mathbb{C}^2$ using the Levi–Civita connection:

$\partial\!\!\!/ = i \nabla_j (1 \otimes \sigma_j)$

I should warn you that this operator has an $i$ in it, to make it self-adjoint! This may be nonstandard, but it will make our life easier.

On the other hand, Kronheimer defined a Dirac-like operator $D$ using the Cartan connection:

$D = i \nabla^c_j (1 \otimes \sigma_j)$

An easy calculation shows how $\partial\!\!\!/$ and $D$ are related:

$\begin{array}{ccl} \partial\!\!\!/ &=& i \nabla_j (1 \otimes \sigma_j) \\ [3pt] &=& i \left( \nabla^c_j - \frac{i}{2}(1 \otimes \sigma_j)\right)(1 \otimes \sigma_j) \\ [3pt] &=& D + \frac{3}{2} \end{array}$

where we use $\sigma_j^2 = 1$ and the 3-dimensionality of space.

Let us compute $D^2.$ Using the identities

$\sigma_j \sigma_k = \delta_{jk} + i \epsilon_{jk\ell} \sigma_\ell , \qquad \epsilon_{jk\ell} B_j B_k = iB_\ell$

we obtain

$\begin{array}{ccl} D^2 &=& -X_j X_k \otimes \sigma_j \sigma_k \\ [2pt] &=& 4 B_j B_k \otimes (\delta_{jk} + i \epsilon_{jk\ell} \sigma_\ell ) \\ [2pt] &=& 4B^2 \otimes 1 - 4 B_\ell \otimes \sigma_\ell \\ [2pt] &=& \Delta - 2 D \end{array}$

It follows that $\Delta = D(D+2),$ so

$\Delta + 1 = (D+1)^2 = (\partial\!\!\!/ - \frac{1}{2})^2$

Combining this with our earlier formula for the hydrogen atom Hamiltonian:

$\displaystyle{ H = - \frac{1}{2(\Delta + 1)} }$

we can now express the Hamiltonian for the hydrogen atom in terms of the Dirac operator on the 3-sphere:

$\displaystyle{ H \; = \; - \frac{1}{2(\partial\!\!\!/ - \frac{1}{2})^2} }$

This is pretty cool. We will exploit this next time.

Further details

That was the main result we’ll need, but when working on this I got interested in understanding the eigenvectors and eigenvalues of the Dirac operator in more detail. Here are some facts about those.

Since $\partial\!\!\!/$ maps each finite-dimensional subspace

$V_j \otimes V_j \otimes \mathbb{C}^2 \subset L^2(S^3) \otimes \mathbb{C}^2$

to itself, and it is self-adjoint on these subspaces, each of these subspaces has an orthonormal basis of eigenvectors. So, consider an eigenvector: suppose $\psi \in V_j \otimes V_j \otimes \mathbb{C}^2$ has

$\partial\!\!\!\psi = \lambda \psi.$

Then we must have

$(\lambda - \frac{1}{2})^2 \psi = (\partial\!\!\!/ - \frac{1}{2})^2 \psi = (\Delta + 1)\psi = n^2 \psi$

where $n = 2j+1$ as usual, so we must have $\lambda - \frac{1}{2} = \pm n.$ Thus, the only possible eigenvalues of $\partial\!\!\!/$ on the subspace $V_j \otimes V_j \otimes \mathbb{C}^2$ are $\pm n + \frac{1}{2},$ or in other words:

$\psi \in V_j \otimes V_j \otimes \mathbb{C}^2 \implies \partial\!\!\!/ \psi = \lambda \psi \text{ for } \lambda = \pm (2j+1) + \frac{1}{2}$

To go further, we can use some results from Kronheimer. First, he shows the spectrum of $\partial\!\!\!/$ is symmetric about the origin. To do this he identifies $\mathbb{C}^2$ with the quaternions, and thus $L^2(S^3) \otimes \mathbb{C}^2$ with a space of quaternion-valued functions on the 3-sphere. Then quaternionic conjugation gives a conjugate-linear operator

$\dagger \colon L^2(S^3) \otimes \mathbb{C}^2 \to L^2(S^3) \otimes \mathbb{C}^2$

with $\dagger^2 = 1.$ He then proves a result, using his Dirac-like operator $D,$ that implies

$\partial\!\!\!/ \psi = \lambda \psi \; \iff \; \partial\!\!\!/ \psi^\dagger = -\lambda \psi^\dagger$

Thus the Dirac operator $\partial\!\!\!/$ has as a negative eigenvalue for each positive eigenvalue, and their multiplicities are the same!

Second, he proved a result which implies that the eigenspace

$F_\lambda = \{ \psi \in L^2(S^3) \otimes \mathbb{C}^2 : \; \partial\!\!\!/ \psi = \lambda \psi \}$

has dimension

$\text{dim}(F_\lambda) = (\lambda+\frac{1}{2})(\lambda-\frac{1}{2})$

when $\lambda \in \mathbb{Z} + \frac{1}{2}$ and zero otherwise. Thus every number that’s an integer plus $\frac{1}{2}$ is an eigenvalue of the Dirac operator on the 3-sphere—except $\pm\frac{1}{2}.$ Also, while we’ve already seen that

$V_j \otimes V_j \otimes \mathbb{C}^2 = F_{n+\frac{1}{2}} \oplus F_{-n +\frac{1}{2}}$

where $n = 2j+1,$ this additional result implies that these two summands have different dimensions, namely $n(n+1)$ and $n(n-1),$ respectively. Their total dimension is $2n^2,$ as we already knew! We knew it because this is the number of electron states in the shell with principal quantum number $n.$

So, a bit more than half the electron states in the nth shell are positive eigenvectors of the Dirac operator on the 3-sphere, while a bit fewer than half are negative eigenvectors. Weird but true!

For an explicit basis of eigenvectors of the Dirac operator on $S^3,$ see:

• Fabio Di Cosmo and Alessandro Zampini, Some notes on Dirac operators on the $S^2$ and $S^3$ spheres.