The Kepler Problem (Part 10)

Azimuth 2025-08-06

The poet Blake wrote that you can

Today we’ll see a universe in an atom!

We’ll see that states of the hydrogen atom correspond to states of a massless spin-½ particle in the Einstein universe—a closed, static universe where space is a 3-sphere.

The rotational symmetries of the Einstein universe correspond to symmetries of the hydrogen atom. The energy eigenstates of the massless spin-½ particle correspond in a one-to-one way to energy eigenstates of the hydrogen atom.

Let’s dive in!

The ‘Einstein universe’ is a name for the manifold $\mathbb{R} \times S^3$ with Lorentzian metric $dt^2 - ds^2,$ where $dt^2$ is the usual Riemannian metric on $\mathbb{R}$ and $ds^2$ is the Riemannian metric on the unit sphere in 4-dimensional Euclidean speace. The Einstein universe has a lot of symmetry: the group $\mathbb{R} \times \text{SO}(4)$ acts as isometries. This group also acts on the bound states of the hydrogen atom, in a way that commutes with the Hamiltonian.

To describe a massless spin-½ particle on the Einstein universe, we’ll use the Weyl equation. This is a variant of the Dirac equation that describes massless spin-½ particles that are chiral, i.e., have an inherent handedness. We can trivialize the bundle of Weyl spinors over the Einstein universe, using right translations on the group

$\mathbb{R} \times \text{SU}(2) \cong \mathbb{R} \times S^3$

to identify every fiber of this bundle with the vector space $\mathbb{C}^2.$ Using this trivialization we can write the left-handed Weyl equation as

$\displaystyle{ \frac{\partial \psi}{\partial t} = -i\partial\!\!\!/ \psi }$

where

$\psi \colon \mathbb{R} \times S^3 \to \mathbb{C}^2$

is a spinor-valued function. (If the $-i$ in the Weyl equation looks weird, don’t worry: it’s there because in Part 8 we defined $\partial\!\!\!/$ to be a self-adjoint operator, which has an $i$ in it.)

The Weyl equation also comes in a right-handed form differing by a sign:

$\displaystyle{ \frac{\partial \psi}{\partial t} = i \partial\!\!\!/ \psi }$

We choose henceforth to work with the left-handed form. This is an arbitrary convention: if we used the right-handed Weyl equation, all our results would still hold, with suitable minus signs sprinkled in here and there. But it so happens that, back when people believed neutrinos were massless, we thought they obeyed the left-handed Weyl equation. The reason is that to a very good approximation, when a neutrino is moving in the direction of your thumb, it spins clockwise, like the fingers in your curled left hand.

Here’s how the Weyl equation works on the Einstein universe. We take $L^2(S^3) \otimes \mathbb{C}^2$ as the Hilbert space of solutions, and $\partial\!\!\!/$ as the Hamiltonian. Since we’ve defined things so that $\partial\!\!\!/$ is self-adjoint, this Hamiltonian generates a 1-parameter group of unitary operators

$U(t) = \exp(-it \partial\!\!\!/)$

Given any $\psi_0 \in L^2(S^3) \otimes \mathbb{C}^2,$ if we let

$\psi_t = U(t) \psi_0$

and define a function

$\psi \colon \mathbb{R} \times S^3 \to \mathbb{C}^2$

$\psi(t,x) = \psi_t (x)$

then this function will be a solution of the left-handed Weyl equation.

As we saw in Part 8, the hydrogen atom Hamiltonian $H$ is a function of the Hamiltonian $\partial\!\!\!/$ for the Weyl equation:

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

Thus, not only the Hilbert space but also the dynamics of the bound states of the hydrogen atom can be expressed in terms those for the Weyl equation on the Einstein universe. However, not all the symmetries we have found for the Hamiltonian $H$ are symmetries of $\partial\!\!\!/$ : this is possible because while $H$ is a function of $\partial\!\!\!/,$ $\partial\!\!\!/$ is not a function of $H.$ Let us make this precise.

In Part 7 we made $L^2(S^3) \otimes \mathbb{C}^2$ into a unitary representation of $\text{SU}(2)$ in three commuting ways: via left translations, via right translations, and via the spin-½ representation on $\mathbb{C}^2.$ The self-adjoint generators of these three representations are called $A_j,$ $B_j$ and $S_j,$ respectively, and these obey

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

Using some formulas from Part 8 we can write the Dirac operator in terms of these operators:

$\begin{array}{ccl} \partial\!\!\!/ &=& 4B_j S_j + \tfrac{3}{2}. \end{array}$

Using this and the commutation relations we just saw, we see $\partial\!\!\!/$ commutes with the operators $A_j.$ It does not commute with $B_j$ or $S_j$ separately, but it commutes with $B_j + S_j,$ since

$\begin{array}{ccl} [B_j + S_j, B_k S_k] &=& [B_j, B_k] S_k + B_k [S_j, S_k] \\ [2pt] &=& i \epsilon_{jk\ell} B_\ell S_k + i \epsilon_{jk\ell} B_k S_\ell \\ [2pt] &=& 0 \end{array}$

It follows that $\partial\!\!\!/$ commutes with the unitary representation $\rho$ of $\text{SU}(2) \times \text{SU}(2)$ on the Hilbert space $L^2(S^3) \otimes \mathbb{C}^2$ whose self-adjoint generators are $A_j$ and $B_j + S_k.$ If we think of this Hilbert space as consisting of functions $\psi \colon S^3 \to \mathbb{C}^2,$ this representation is given by

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

Geometrically, this representation arises from the natural way to lift the left and right translation actions of $\text{SU}(2)$ on $S^3$ to the spinor bundle of $S^3.$ The asymmetry between left and right here may seem puzzling. It has nothing to do with the fact that we’re studying the left-handed Weyl equation. Instead, it arises from how we arbitrarily chose to trivialize the spinor bundle of $S^3$ using the action of $\text{SU}(2)$ as left translations. Thus, the above action of $(g_1,1) \in \text{SU}(2) \times \text{SU}(2)$ on $\psi$ merely left translates $\psi,$ while the action of $(1,g_2)$ not only right translates $\psi$ but also acts on its value by $g_2.$

Summarizing, this is what we have seen so far. Made into representations of $\text{SU}(2) \times \text{SU}(2)$ as above, the Hilbert space of bound states of hydrogen atom and the Hilbert space for the left-handed Weyl equation on the Einstein universe are unitarily equivalent. Moreover, we can express the Hamiltonian for the hydrogen atom in terms of that for the left-handed Weyl equation!

The problem of negative energies

All this is fine mathematics, but there is a physical problem, noticed already by Dirac in a related context: the spectrum of $\partial\!\!\!/$ is unbounded below, giving states of arbitraily large negative energy! One widely accepted solution is to modify the complex structure on the Hilbert space, multiplying it by $-1$ on the negative-frequency solutions of the Weyl equation: that is, the subspace of $L^2(S^3) \otimes \mathbb{C}^2$ spanned by eigenvectors of the Dirac operator with negative eigenvalues. This is an updated version of Dirac’s original idea of treating antiparticles as ‘holes in the sea of negative-energy particles,’ or the later and still popular idea of switching annihilation and creation operators for negative-frequency solutions.

To modify the complex structure on $L^2(S^3) \otimes \mathbb{C}^2,$ we use the functional calculus to define an operator

$\displaystyle{ S = \frac{\partial\!\!\!/}{|\partial\!\!\!/|} }$

on this Hilbert space. This equals 1 on eigenvectors of $\partial\!\!\!/$ with positive eigenvalue and -1 on those with negative eigenvalue; we have seen that 0 is not an eigenvalue of $\partial\!\!\!/,$ so $S$ is well-defined. We then define an operator

$j = i S$

Since $S$ is both unitary and self-adjoint, it follows that $j$ is both unitary ( $jj^* = j^* j = 1$ ) and skew-adjoint ( $j^\ast = -j$ ), and thus a complex structure ( $j^2 = -1$ ). We henceforth use $\mathcal{H}$ to stand for $L^2(S^3) \otimes \mathbb{C}^2$ made into a complex Hilbert space with the same norm and this new complex structure $j.$

The operators $\partial\!\!\!/$ and $|\partial\!\!\!/|$ are still complex-linear on $\mathcal{H},$ despite the new complex structure, since they commute with $i$ and $S,$ and thus $j.$ The operator $\partial\!\!\!/$ is still self-adjoint on $\mathcal{H},$ since it has an orthonormal basis of eigenvectors with real eigenvalues. The operator $|\partial\!\!\!/|$ is not only self-adjoint but positive definite on $\mathcal{H},$ since

$\partial\!\!\!/ \psi = \lambda \psi \; \implies \; |\partial\!\!\!/| \psi = |\lambda| \psi$

In fact the operator $|\partial\!\!\!/|$ generates $U(t)$ as a one-parameter unitary group on $\mathcal{H},$ because

$\exp(jt|\partial\!\!\!/|) = \exp(it\partial\!\!\!/) = U(t) .$

Thus negative energy states have been eliminated, without changing the time evolution operators $U(t)$ at all, by changing the Hamiltonian from $\partial\!\!\!/$ to $|\partial\!\!\!/|$ and simultaneously changing the complex structure from $i$ to $j.$

Since all the operators $\rho(g_1,g_2)$ on $L^2(S^3) \otimes \mathbb{C}^2$ commute with $\partial\!\!\!/,$ they also commute with $S$ and thus with the new complex structure $j = iS.$ Thus $\rho,$ which began life as a unitary representation of $\text{SU}(2) \times \text{SU}(2)$ on $L^2(S^3) \otimes \mathbb{C}^2,$ gives a complex-linear representation of this group on $\mathcal{H},$ which we call $\rho_\mathcal{H}.$ This representation $\rho_\mathcal{H}$ is unitary, since the norm on $\mathcal{H}$ is the same as that on $L^2(S^3) \otimes \mathbb{C}^2,$ and any norm-preserving invertible linear operator on a Hilbert space is unitary.

Furthermore, the representation $\rho_\mathcal{H}$ is unitarily equivalent to $\rho.$ This is a nontrivial fact, because the unitary equivalence between them is not the identity operator! Indeed the map

$\begin{array}{rcl} I \colon L^2(S^3) \otimes \mathbb{C}^2 &\to &\mathcal{H} \\ \psi &\mapsto & \psi \end{array}$

is not even complex linear: it is complex linear on the +1 eigenspace of $S$ but conjugate-linear on the -1 eigenspace. To correct for this, we use a conjugate-linear map

$\begin{array}{c} C \colon L^2(S^3) \otimes \mathbb{C}^2 \to L^2(S^3) \otimes \mathbb{C}^2 \\ [4pt] (C \psi)(g) = \epsilon \, \overline{\psi}(g) \end{array}$

where we regard $\psi$ as a $\mathbb{C}^2$ -valued function on $S^3,$ let $\overline{\psi}$ denote its componentwise complex conjugate, and multiply $\overline{\psi}$ by

$\epsilon = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)$

The reason the map $C$ is important is that

$\begin{array}{ccl} C(g \psi) = \overline{g} C(\psi) \end{array}$

for all $g \in \text{SU}(2)$ and $\psi \in L^2(S^3) \otimes \mathbb{C}^2.$ Checking this is a little calculation with 2 × 2 matrices, but conceptually it says that $C$ is an equivalence between the spin-½ representation of $\text{SU}(2)$ and its conjugate representation. The desired unitary equivalence between $\rho$ and $\rho_\mathcal{H}$ is then the map

$\begin{array}{c} F \colon L^2(S^3) \otimes \mathbb{C}^2 \to \mathcal{H} \\ [4pt] F = I(p_+ + C p_-) \end{array}$

where

$p_+, p_- \colon L^2(S^3) \otimes \mathbb{C}^2 \to L^2(S^3) \otimes \mathbb{C}^2$

are the projections of $\psi$ to the +1 and -1 eigenspaces of $S,$ respectively. We include $I$ here because we should keep track of the difference between $L^2(S^3) \otimes \mathbb{C}^2$ and $\mathcal{H}.$

Theorem. The operator $F \colon L^2(S^3) \otimes \mathbb{C}^2 \to \mathcal{H}$ is a unitary equivalence between the representation $\rho$ of $\text{SU}(2) \times \text{SU}(2)$ on $L^2(S^3) \otimes \mathbb{C}^2$ and the representation $\rho_\mathcal{H}$ of this group on $\mathcal{H},$ and $F \partial\!\!\!/ = \partial\!\!\!/ F$ on the domain of $\partial\!\!\!/.$

Proof. For the proof, see the Appendix of my paper. █

Summary

Since changing the complex structure on a Hilbert space can be a bit bewildering, I should summarize what we’ve achieved here.

We have a unitary equivalence between the Hilbert space $L^2(S^3) \otimes \mathbb{C}^2$ of bound states of the hydrogen atom and the Hilbert space $\mathcal{H}$ of solutions of the left-handed Weyl equation on $\mathbb{R} \times S^3$ equipped with a complex structure that makes its Hamiltonian positive. The group $\text{SU}(2) \times \text{SU}(2)$ has equivalent unitary representations on these two Hilbert spaces. The Dirac operator $\partial\!\!\!/$ acts on both $L^2(S^3) \otimes \mathbb{C}^2$ and $\mathcal{H}$ in an manner compatible with their unitary equivalence. Finally, both the hydrogen atom Hamiltonian and the Hamiltonian for the left-handed Weyl equation can be expressed in terms of the Dirac operator: the former is

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

while the latter is just $|\partial\!\!\!/|.$

So, we’re seeing the universe in an atom—or at least, we’re seeing a massless spin-½ particle in the Einstein universe in the hydrogen atom.

But what about full-fledged quantum field theory? Can we understand the massless spin-½ quantum field in terms of atomic physics? Yes! But for that we’ll need to second quantize today’s story.