Independent Phases in Gauge Theory Excuse my naivety.
When we postulate a local gauge invariance we say that we allow the overall phase of the field variables $\psi(x)$ can be changed and that this overall phase can vary from point to point. We can do this given that a change $\psi(x)\to e^{i\alpha (x)}\psi(x)$ is accompanied by $A_\mu \to A_\mu +\partial_\mu \alpha (x)$.
this seems to assume that these unimportant phases are stiched together continuously and differentiably. What is the reason for this? If the overall local phase is just another description then what is the physical reason for why we describe this redundancy like this?
 A: Suppose you have an abelian gauge theory (forget about $\psi$ for now) in a pure gauge state, that is using a gauge trasformation $A_\mu\rightarrow A_\mu + \partial_\mu\alpha$, you can go to $A_\mu = 0$ (call this gauge 1), i.e. $A_\mu = \partial_\mu \alpha$ (call this gauge 2). Now let's assume that $\alpha$ is not differentiable and see what goes wrong.
Let's find the lagrangian density in gauge 1, its proportional to
$$
F_{\mu\nu}F^{\mu\nu} = \partial_{[\mu}A_{\nu]}\partial^{[\mu}A^{\nu]} = 0
$$
in gauge 2
$$
F_{\mu\nu}F^{\mu\nu} = \partial_{[\mu}A_{\nu]}\partial^{[\mu}A^{\nu]} =\partial_{[\mu}\partial_{\nu]}\alpha\partial^{[\mu}\partial^{\nu]} \alpha
$$
This expression need not vanish if alpha is not differentiable according to Schwarz's theorem. This would mean that a gauge transformation with a non differentiable gauge parameter could actually change the energy of the system and therefore doesn't correspond to a redundancy in the physical description but rather to two physically inequivalent physical states.
(see http://en.wikipedia.org/wiki/Symmetry_of_second_derivatives for an example of a non differentiable function where the commutator is indeed not zero)
