Why does gauge invariance HAVE to correspond to an observable?(Or is it the other way round)

Question

Under the line integral of the geometrical Berry phase, a close-loop integral is gauge invariant as if we were to perform a gauge transformation of the initial state, with the end point of the path in the parameter space being the same as the initial point, it supposedly leads to an observable.

When taking under the considerations of an open curve integral and performing the gauge transformation, it seems the fact that having an extra term in the gauge-transformed Berry phase leads to a non-observable. So the fundamental question I've here is why does the gauge invariant property leads to an observable at all? (and by observables as far as I understood it is referring to a physical manifestation of mathematical properties that we could measure in a lab, correct me if I'm wrong)

Answer 1

Gauge-invariant quantities are not necessarily observable, but all observables must be gauge-invariant.

In other words, gauge-invariance is a necessary condition, but not a sufficient condition, for something to qualify as an observable.

For more detail, first consider this assertion: All observables must be gauge-invariant. One way to see this is to note that in a model with a gauge field, such as electrodynamics, the future of the gauge field is not determined by its past. Gauge transformations are local in space and time, so we can take any given solution, apply a gauge transformation localized in any given region of spacetime, and we still have a valid solution. If we didn't regard all gauge-related solutions as being physically equivalent to each other, then predictions would be ambiguous.

The same statements apply in general relativity, with diffeomorphisms (pulled back / pushed forward to all of the various dynamic entities, such as the metric field) playing the role of gauge transformations.

Another perspective is that the statement "All observables must be gauge-invariant" is nothing but a tautology (it is trivially true), because a "gauge symmetry" is nothing more than a redundancy in the mathematical formulation of a model. This is emphasized in

• Witten (2017) “Symmetry and Emergence” (http://arxiv.org/abs/1710.01791).

Redundant formulations are widely used in physics because they can be very convenient. In electrodynamics, for example, expressing things in terms of the gauge field $A_\mu$ (which is redundant) instead of $F_{\mu\nu}$ allows the whole set of equations of motion to be encoded in a single action (whose Euler-Lagrange equations are the equations of motion) that is both local and Lorentz-invariant. We don't strictly need to use a gauge field to formulate electrodynamics, but it is convenient — especially in quantum electrodynamics, where a formulation without using the gauge field is probably prohibitively inconvenient. General relativity is an even more extreme example of this situation, with an even greater degree of redundancy in the formulation.

The utility of gauge fields in physics is often (maybe always) associated with the need to implement constraints — specifically constraints that the initial conditions must satisfy. Gauss's law $\nabla\cdot\mathbf{E}=\rho$ is an example of this. Less familiar examples occur in condensed matter physics, where various constraints (such as constraints on the number of particles that can occupy a given site in a lattice) can often be conveniently enforced by introducing a new gauge field, as mentioned in the paper cited above and also in this post:

Can there be an emergent non-compact gauge field?

(Such gauge fields are often called "emergent.")

Now consider the other assertion: Gauge-invariant quantities are not necessarily observable. For an example of a gauge-invariant quantity that is not observable, consider QED with an electrically neutral fermion field (like a massive neutrino, assuming for the sake of example that neutrinos are Dirac fermions). If $\psi$ is the Dirac-spinor field operator for the neutral fermion, then $\Phi(x)\equiv\psi^\dagger(x)+\psi(x)$ is a self-adjoint, gauge-invariant operator — but it does not qualify as an observable, because $\Phi(x)$ and $\Phi(y)$ don't commute with each other when $x$ and $y$ are separated by a spacelike interval (they anticommute with each other instead). Observables must commute with each other at spacelike separation; this is one of the axioms of quantum (field) theory.