Can one encode topology into the position operator in quantum mechanics?

Question

We work with the concrete example of electromagnetism, but we intend to ask a question in broader scope at the end.

In classical field theory, the electromagnetic potential $\mathcal{A}$ lives on a principal bundle. Hence, part of the definition of the electromagnetic potential is the base manifold $M$. Concretely, $$\mathcal{A}: M \to \mathbb{R} \otimes T^*M,$$ so that $\mathcal{A}$ is an $\mathfrak{\mathbb{R}}$-valued $1$-form. We can then (at least traditionally) write a quantum mechanical Hamiltonian operator in terms of $\mathcal{A}$. A canonical example of this is the basic hydrogen atom Hamiltonian, which in the position basis might look like \begin{equation} H_{hyd} := -\frac{\hbar^2}{2m}\nabla^2 - q\mathcal{A}^0. \end{equation} We call the $\mathcal{A}$ that appears in the Hamiltonian the operatorized $\mathcal{A}$. Since we have included the potential $\mathcal{A}^0$, the Hamiltonian $H_{hyd}$ should carry the information of the base manifold encoded in $\mathcal{A}^0$.

Then, the question is: Given a classical field theory of a dynamic variable $\mathcal{A}$ over base manifold $M$, how does one encode the properties of the base manifold into the operatorized $\mathcal{A}$?

Answer 1

Mathematically the data of the topology is encoded in the structure of the operators themselves. I will focus on the simple case of the punctured plane $M=\mathbb{R}^2\setminus \{0\}$.

Suppose we are considering all $U(1)$-bundles with flat connections on $M$. From minimal-coupling, we have that $p_i = (D_A)_i$ where $A$ is our flat connection. $[p_i,p_j]=0$ and $[q_i, p_j]=i\delta_{ij}$. Thus we have some representation of the canonical commutation relations.

Note: The Stone-Von Neumann theorem does not apply here! In particular it is crucial that the Weyl relations, which are the relations actually relevant to Stone-Von-Neumann, may not necessarily hold thanks to the nontrivial topological structure of the problem

Call this representation $\pi_A$. One can prove that $\pi_A\cong \pi_{CCR}$ if and only if the holonomy $\int_{C_1} A \cdot dl \in 2\pi\mathbb{Z}$. In more generality, you can prove that $\pi_A \cong \pi_{A'}$ if and only if $A$ and $A'$ are gauge-equivalent.

A nice discussion of this framework is in Inequivalent Representations of Canonical Commutation and Anti-Commutation Relations by Asao Arai

Relating this to topology, notice that the relevant Hilbert space here is $L^2(\mathbb{R}^2\setminus \{0\})\cong L^2(\mathbb{R}^2)$. Thus the Hilbert space alone is completely blind to the distinction between the topologically trivial plane and the punctured plane. Instead it is the Hamiltonian itself having a different form (both in terms of their domains and their literal form).

To highlight how the domain itself plays a role, recall the relatively simple situation of the particle on an interval. Consider $H=-i\Delta$ on $C_c^\infty(0,1)$ in the Hilbert space $L^2(0,1)$. This operator admits many self-adjoint extensions determined entirely by fixing the domain. For example, if you set vanishing boundary conditions at the boundary, you get the classic infinite square well. If you set periodic boundary conditions, you get the particle-on-a-circle Hamiltonian, which you can then interpret as actually describing $M=S^1$.

So you can take a trivializing chart, and your Hilbert space will not see the difference. But the Hamiltonian is going to be some kind of differential operator, and this differential operator is sensitive to the boundary conditions at the boundaries of the domain that indicate whether or not a function is actually smooth, and this carries the information you had lost elsewhere.