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?

  • $\begingroup$ Do you have Peskin's book "an introduction to quantum field theory" ? If you have , you can read the part " 15.1 The Geometry of Gauge Invariance". $\endgroup$
    – Simon
    Apr 10, 2015 at 1:56
  • $\begingroup$ I do, this where that comes from. We know both that this overall phase convention is somehow physically superflous, but yet its mathematical realization implies that the phase convention is continuos and differentiable, which are usually requirements of physical states. $\endgroup$ Apr 10, 2015 at 2:38
  • $\begingroup$ I'm not quite sure about the question you ask , "the requirements of physical states" ? If you want to describe a physical state, you have to use mathematics, if you use the tool, you will get redundancy! But the physical law shouldn't change! That's my opinion! $\endgroup$
    – Simon
    Apr 10, 2015 at 3:14
  • $\begingroup$ Yes but if it s redundant then why cant $\psi(x)$ and $\psi(x+dx)$ differ by a finite phase difference? I mean, if it doesn't matter then why have the "phase field" be a continuous and differential object $\endgroup$ Apr 10, 2015 at 3:44
  • $\begingroup$ I get your point! But, hah! I don't know the answer. If you have just one discrete value, I mean $\alpha$ is constant, then it corresponds to Nother's charge. But if you have several discrete values , you might get a bunch of Dirac functions in order to compensate for the difference on the boundary of discrete points! I'm not sure! $\endgroup$
    – Simon
    Apr 10, 2015 at 4:34

1 Answer 1


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)

  • $\begingroup$ So in both you have a pure gauge (no curvature) but the second does not vanish. Thanks very much. Seems to be at odds with the ethos of gauge invariance, cool. $\endgroup$ Apr 10, 2015 at 21:45

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