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That is a follow-up to this question: Gauge symmetry is not a symmetry?

Ok, gauge symmetry is not a symmetry, but ...

... a redundancy in our description, by introducing fake degrees of freedom to facilitate calculations.

I want some simple and practical example for this.
So If I, say, take a simple $\phi^4$ theory, then I can gauge it by...

... introducing the proper fake degrees of freedom

Can I?

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Kostya
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1 Answers1

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A trivial example:

Take your original field $\phi$ to be a free real scalar field on $\mathbb{R}^n$. Double the number of fields by adding another free real scalar field $\chi$ to your list of fields

Now introduce a gauge symmetry by making the group of functions $g: \mathbb{R}^n \to \mathbb{R}$ act by $g: (\phi,\chi) \mapsto (\phi,\chi+ g)$. So your group of gauge transformations acts trivially on the space of $\phi$'s and freely and transitively on the space of $\chi$'s.

Now fix a gauge in your favorite way. You can grind through the BRST machinery, or you can just choose the gauge slice $\chi= 0$. Either way your original free scalar field is precisely equivalent to the new theory with two fields and the wacky gauge symmetry I described above.

user1504
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  • Could you explain what does gauge away mean? I came across this expression while reading about the Higgs mechanism but didn't understand what it means and what is the significance of doing it. I wanted to start a question on it but found it stated here, hence my question. (And by the way I'm trying to comprehend what you wrote here but this stuff is a bit heavy for me.) – user09876 Jun 22 '12 at 23:15
  • A gauge symmetry is a symmetry of the variables used to describe a physical system which isn't actually a symmetry of the physical system. Gauging away a variable means using a gauge symmetry to eliminate it from the description of the system. I was getting ahead of myself when I said 'gauge away' $\chi$, since I hadn't actually introduced the gauge symmetry or used it to eliminate $\chi$. Fixed that. – user1504 Jun 23 '12 at 02:08
  • When you say "not actually a symmetry" can you elaborate? Isn't U(1) symmetry of QED "due to" conservation of electric charge? I mean obviously since we're talking about a bundle over space-time the "translations" (phase changes) aren't actual translations but otherwise i'm not sure what you meant by this. – tachyonicbrane Jun 23 '12 at 02:50
  • Regarding gauge symmetry not being a real symmetry - see http://physics.stackexchange.com/questions/13870/gauge-symmetry-is-not-a-symmetry – DJBunk Jun 23 '12 at 03:32
  • Another way of doing this is adding a reality constraint to your field--by adding a term like $\alpha(\phi^{2}-\phi^{*}\phi)$, where $\alpha$ is taken to be a lagrange multipler. This makes the imaginary part of $\phi$ a gauge field. – Zo the Relativist Jun 23 '12 at 06:13
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    @JerrySchirmer: The imaginary part in your example in the comments is not a gauge field, it is just zero after the constraint. It isn't arbitrary, and you don't pick out a slice in the path integral. – Ron Maimon Jun 23 '12 at 11:14
  • @tachyonicdbrane: You're confusing the gauge group $U(1)$ with 'charge group', i.e., the symmetry group generated by the charge observable, which is also a copy of $U(1)$. The group of gauge transformations for QED is not the group $\mathcal{G}{big}$ of all functions on $\mathbb{R}^4$ valued in the gauge group; rather, it is the subgroup $\mathcal{G} \subset \mathcal{G}{big}$ consisting functions to $U(1)$ _which approach $1$ at $\infty$. The other subgroup, of constant $U(1)$-valued functions, is the charge group; this subgroup does not generate gauge transformations. – user1504 Jun 23 '12 at 13:36