Can an observable be invariant under local $U(1)$ but not under global $U(1)$?

Question

Consider a quantum field theory with two fields, a complex scalar field $\phi$ and a $U(1)$ gauge field $A$. Both fields are dynamic fields, not background fields. Suppose that spacetime is topologically trivial. Suppose that the lagrangian is invariant under \begin{align*} \phi(x) &\to e^{-i\theta(x)}\phi(x) \\ A(x) &\to A(x)+d\theta(x), \tag{1} \end{align*} for all $\theta(x)$, and define two groups:

• $G$ is the group of all transformations of the form (1).

• $H$ is the subgroup with $\theta(x)\to 0$ as $|x|\to\infty$.

I'm deliberately leaving the lagrangian unspecified, but assume that if the field $A$ were omitted, then the global $U(1)$ symmetry of the remaining $\phi$-only model would not have an 't Hooft anomaly.

Question: Can an operator constructed from the fields $\phi$ and $A$ be invariant under $H$ but not under $G/H$?

Here's my attempt to construct an operator that is invariant under $H$ but not under $G/H$: $$ \phi(x)\exp\left(-i\sum_u \int_{P(x,u)} A\right), \tag{2} $$ where $P(x,u)$ is a path from $x$ to spacelike infinity, approaching spacelike infinity in the direction $u$. The sum over directions $u$ is an attempt to make (2) well-defined by "smearing out" the seemingly ill-defined behavior at spacelike infinity. Intuitively, this is a charged entity $\phi(x)$ "dressed" by an electromagnetic field (like the Coulomb field) as required by Gauss's law (= gauge invariance). If (2) really is well-defined, then it is invariant under $H$ but not under the transformations in $G$ with constant $\theta$.

But I don't know if (2) really is well-defined. Naively, operators like (2) seem to be needed so that we can create states with charged particles (including their electric fields) from the vacuum. However, states of different charge are normally considered to belong to different superselection sectors (different Hilbert-space representations of the algebra of observables), which suggests that such operators can't actually exist. And in this particular example, the model could be in the Higgs phase, with no unscreened charges. So I wouldn't be surprised if the example (2) is incurably ill-defined, but can this be turned into a compelling argument that no such operators exist?

The paper Lectures on the Infrared Structure of Gravity and Gauge Theory has things to say about the significance of $G/H$, but I didn't find an answer to my question there.

Maybe related: Why is $U(1)$ special when defining global charges?

Answer 1

The answer to your question is "yes".

The group G contains the global symmetry group $U(1)$, which is associated by Noether's theorem to electric charge. There are many observables which are not invariant under this symmetry.

Your operator 2 is close to correct. What you really want is a $\phi$-field dressed with a Wilson line going to spatial infinity. $$ \phi(x) e^{i\int_P A_\mu dx^\mu} $$ where $P$ is a path connecting $x$ to spatial infinity. This is well-defined operator, and it'll fail to be invariant under non-trivial transformations at infinity. It's a standard gauge-invariant construction of the charged state creation/annihilation operators.

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(Appendix added by OP based on user1504's comments)

One way to see that it's well-defined is to formulate the theory on a finite lattice. The factor $\phi(x)$ is associated with lattice site $x$, and the exponential factor is a product of "link variables" $U(x_1,x_2)$ along any path from $x$ to any point on the boundary. Each link variable is an element of $U(1)$ associated with a nearest-neighbor pair $(x_1,x_2)$ of sites. This is the lattice version of a $U(1)$ gauge field.

To help make the relationship bewteen the lattice formulation and the continuum picture more inuitive, the boundary can be structured as suggested in this comment (copied from user1504):

The lattice limits are easiest to deal with if you set things up so that the boundary of the lattice has no internal edges, only points.

Let $G$ be the group of transformations of the form \begin{align} \phi(x) &\to \phi(x)g(x) \\ U(x,y) &\to g^{-1}(x)U(x,y)g(y) \end{align} with $g(x)\in U(1)$ for each site. Define $H$ to be the subgroup of $G$ for which $g(x)=1$ on all boundary sites. Then the operator described above is invariant under $H$ but not under all of $G$. In particular, it is not invariant if $g(x)$ is independent of $x$ (a "global" $U(1)$ transformation).