The Kepler Problem (Part 11)

Azimuth 2025-08-09

In Part 9 we saw, loosely speaking, that the theory of a hydrogen atom is equivalent to the theory of a massless left-handed spin-½ particle in the Einstein universe—a static universe where space is a 3-sphere. Today we’ll ‘second quantize’ both of these equivalent theories and get new theories that again are equivalent.

‘Second quantization’ is a funny term. It’s a completely systematic way to get a new quantum theory from an old one. When you second quantize the theory of some particle, you get a theory that describes noninteracting collections of that kind of particle. So, when we second quantize the theory of a hydrogen atom, we get a theory that describes collections of electrons orbiting the nucleus—but in a simplified way, where the electrons do not interact. When we second quantize the theory of a massless left-handed spin-½ particle in the Einstein universe, we get a theory of noninteracting collections of such particles. This is called a free quantum field theory: it’s the easiest sort of quantum field theory to understand.

So, there’s a relationship between multi-electron atoms and a free quantum field theory on the Einstein universe! Next time we’ll think a bit harder about how electrons in an atom actually do interact, and how that affects the structure of the periodic table. But for now we’ll ignore that.

To get started, let’s recall how to build the fermionic Fock space on an arbitrary Hilbert space $\mathcal{H}.$ We start with the exterior algebra

$\displaystyle{ \Lambda \mathcal{H} = \bigoplus_{n = 0}^\infty \Lambda^n \mathcal{H} }$

and give it the inner product such that if $e_j$ is any orthonormal basis for $\mathcal{H},$ the wedge products $e_{i_1} \wedge \cdots \wedge e_{i_n}$ with $i_1 < \cdots < i_n$ form an orthonormal basis for $\Lambda^n \mathcal{H},$ and the different subspaces $\Lambda^n \mathcal{H}$ are orthogonal. Completing $\Lambda \mathcal{H}$ with respect to the norm coming from this inner product, we obtain a Hilbert space we call $\mathbf{\Lambda} \mathcal{H}.$

If $\mathcal{H}$ is the Hilbert space for a single particle of some sort, $\Lambda^n \mathcal{H}$ is the Hilbert space of states of a collection of n particles of this sort, treated as fermions, and $\mathbf{\Lambda} \mathcal{H}$ is the Hilbert space for arbitrary finite collections of such particles. We call $\mathbf{\Lambda} \mathcal{H}$ the fermionic Fock space on $\mathcal{H},$ and call $\Lambda^n \mathcal{H}$ the n-particle subspace.

To define observables on the Fock space, recall how any self-adjoint operator on $\mathcal{H}$ gives rise to one on $\mathbf{\Lambda} \mathcal{H}.$ First, any unitary operator

$U \colon \mathcal{H} \to \mathcal{H}$

gives rise to a unitary operator

$\mathbf{\Lambda}(U) \colon \mathbf{\Lambda} \mathcal{H} \to \mathbf{\Lambda} \mathcal{H}$

determined by the property that

$\mathbf{\Lambda}(U) (\psi_1 \wedge \cdots \wedge \psi_n) = U(\psi_1) \wedge \cdots \wedge U(\psi_n)$

for any vectors $\psi_i \in \mathcal{H}.$ Note that

$\mathbf{\Lambda}(UV) = \mathbf{\Lambda}(U) \mathbf{\Lambda}(V)$

Let $\text{U}(\mathcal{H})$ be the group of unitary operators on $\mathcal{H},$ and $\text{U}(\mathbf{\Lambda} \mathcal{H})$ the group of unitary operators on $\mathbf{\Lambda}(\mathcal{H}).$ If $G$ is any topological group, any strongly continuous unitary representation

$R \colon G \to \text{U}(\mathcal{H})$

gives rise to a strongly continuous unitary representation

$\mathbf{\Lambda} R \colon G \to \text{U}(\mathbf{\Lambda} \mathcal{H})$

defined by

$(\mathbf{\Lambda} R)(g) = \mathbf{\Lambda} (R(g))$

In particular, any self-adjoint operator $A$ on $\mathcal{H}$ gives a strongly continuous unitary one-parameter group $\exp(itA)$ on $\mathcal{H},$ and thus a strongly continuous unitary one-parameter group $\mathbf{\Lambda}(\exp(itA))$ on $\mathbf{\Lambda}\mathcal{H}.$ Stone’s theorem says the latter is generated by a unique self-adjoint operator on $\mathbf{\Lambda}\mathcal{H},$ which we call $d\mathbf{\Lambda}(A).$ We thus have

$\exp(it d\mathbf{\Lambda}(A)) = \mathbf{\Lambda}(\exp(itA))$

for all $t \in \mathbb{R}.$ If the vectors $\psi_1, \dots, \psi_n$ are in the domain of $A,$ we can differentiate both sides of the above formula applied to $\psi_1 \wedge \cdots \wedge \psi_n$ and set $t = 0,$ obtaining

$d\mathbf{\Lambda}(A) (\psi_1 \wedge \cdots \wedge \psi_n) \displaystyle{ = \sum_{i = 1}^n \psi_1 \wedge \cdots \wedge A \psi_i \wedge \cdots \wedge \psi_n }$

In particular if all $\psi_i$ are eigenvectors of $A,$ then their wedge product is an eigenvector of $d\mathbf{\Lambda}(A):$

$A \psi_i = \lambda_i \psi_i \implies$

$d \mathbf{\Lambda}(A)(\psi_1 \wedge \cdots \wedge \psi_n) = (\lambda_1 + \cdots + \lambda_n) (\psi_1 \wedge \cdots \wedge \psi_n)$

We can apply all this mathematics in two ways:

1) We can $\mathcal{H}$ to be the Hilbert space of bound states of a hydrogen atom:

$\mathcal{H} = L^2(S^3) \otimes \mathbb{C}^2$

Then $\mathbf{\Lambda} \mathcal{H}$ is the Hilbert space for an arbitrary finite collection of electrons occupying such states. In particular, if $H$ is the hydrogen atom Hamiltonian, then $d\mathbf{\Lambda}(H)$ restricted to n-particle subspace $\Lambda^n \mathcal{H}$ is the Hamiltonian for an idealized atom with n noninteracting electrons. Since in fact electrons do interact, the lowest-energy eigenstate in the n-particle space gives a very crude approximation to the nth element in the periodic table. To do better we must modify the Hamiltonian! I’ll talk about this next time.

2) Alternatively we can start with $\mathsf{H},$ the Hilbert space of a single left-handed massless spin-½ partice in the Einstein universe. We have seen that the Hamiltonian for such a particle is $|\partial\!\!\!/|.$ Then $\mathbf{\Lambda} \mathsf{H}$ is the Hilbert space for an arbitrary collection of left-handed massless spin-½ particles, treated as fermions. If these particles are noninteracting, their Hamiltonian is $d\mathbf{\Lambda}(|\partial\!\!\!/|),$ and we have a free quantum field theory. This is called the left-handed massless spin-½ quantum field! Back when we thought neutrinos were massless, this would be a pretty good approximation to how they work. To get a better approximation, we’d need to add some interactions with other fields.

How are these related? Very closely!

Any unitary operator between Hilbert spaces, for example the unitary operator $F \colon \mathcal{H} \to \mathsf{H}$ that we saw in Part 9, induces a unitary operator between their fermionic Fock spaces, like this:

$\begin{array}{c} \mathbf{\Lambda}(F) \colon \mathbf{\Lambda} \mathcal{H} \to \mathbf{\Lambda} \mathsf{H} \\ [3pt] \mathbf{\Lambda}(F) (\psi_1 \wedge \cdots \wedge \psi_n) = F \psi_1 \wedge \cdots \wedge F \psi_n \end{array}$

for all $\psi_1, \dots, \psi_n \in \mathcal{H}.$ Thus, an equivalence between two theories at the single-particle level induces an equivalence between their second quantized versions! And with some extra work we get this:

Theorem 2. The map $\mathbf{\Lambda}(F) \colon \mathbf{\Lambda} \mathcal{H} \to \mathbf{\Lambda} \mathsf{H}$ is a unitary equivalence between the representation of the group $\text{SU}(2) \times \text{SU}(2)$ on the fermionic Fock space $\mathbf{\Lambda} \mathcal{H}$ for bound states of the hydrogen atom Hamiltonian and the representation of this group on the fermionic Fock space $\mathbf{\Lambda} \mathsf{H}$ for left-handed massless spin-½ particles on the Einstein universe. That is,

$\mathbf{\Lambda}(F) \mathbf{\Lambda}(\rho(g_1,g_2)) = \mathbf{\Lambda}(\rho_\mathsf{H}(g_1,g_2)) \mathbf{\Lambda}(F)$

for all $(g_1,g_2) \in \text{SU}(2) \times \text{SU}(2).$ Moreover

$\mathbf{\Lambda}(F) d\mathbf{\Lambda}(\partial\!\!\!/) = d\mathbf{\Lambda}(\partial\!\!\!/) \mathbf{\Lambda}(F)$

on the domain of $d\mathbf{\Lambda}(\partial\!\!\!/).$

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

Next time we’ll see whether we can get the periodic table from our second quantized theory of the hydrogen atom.