# Guage symmetry - invariance under arbitrary phase change

1. Jan 18, 2012

### arlesterc

As I understand it the heart of gauge symmetry is that I can change the phase at different points different amounts and the Lagrangian/action is unchanged. What I am not clear on is whether the changes I can make are completely arbitrary - I can make any change I want at any point - or whether the changes are a specific function evaluated at different points which lead to different amounts but the amounts are connected by the fact that they are formed by the same function. By way of crude example, let's say I have a point A with values x=1,y=2 and another point B with values x=5,y=6. By arbitrary I mean that I change the phase at point A by some amount and then completely at random - spin a wheel/toss the dice - choose to change the phase at B by some other amount. Or is it the case that the changes at each point are calculated by some function of the points. Let's say the function is 2x+3y so that at point A I change the phase by 2+6=8 whereas at B the function changes by 10+18=28.

Any clarification is appreciated in advance.

2. Jan 18, 2012

### The_Duck

I think we probably want the phase change we make to be a continuous function of the spacetime coordinates. But other than that it can be any function whatsoever. So infinitesimally close points should have similar phase changes, but points separated by a finite distance can have completely different phase changes.

3. Jan 24, 2012

### arlesterc

Thanks for the response. You write 'I think we probably want the phase change we make to be a continuous function of the spacetime coordinates.' But it doesn't have to be? What's the physical implication if it isn't as opposed to when it is?

I thought the point was that we could have different observers at each point picking their starting point completely unaware of anybody else's starting point and we could still reconcile using a gauge transformation.

4. Jan 24, 2012

### Lapidus

If you pick one function then you have fixed a gauge. Before that the phase is a completly arbitrary scalar space-time function, i.e. it is not specified.

Say you have not fixed a gauge, i.e. you have not specified a space-time dependent phase function. You can pick then any phase value at some space-time point A. But consider that there are infinitely many different functions thinkable that have your chosen phase value at point A. All of them give completly different phase values at other space-time points.

Not so when you pick one phase function, or as physicists say fix a specific gauge. Then phase values at different points in space-time are connected.

Last edited: Jan 24, 2012
5. Jan 24, 2012

### The_Duck

I meant I don't think it makes mathematical sense to choose a discontinuous function. I was thinking of the fact that when you perform a gauge transformation, in addition to rotating the wave function phase at each point by some function f(x), you must change the vector potential:

$A_\mu \to A_\mu + \partial_\mu f$

So really I meant that f needs be a _differentiable_ function of the spacetime coordinates or else the transformation of A doesn't make sense.

I'm not clear what you mean by this?

At its most basic level, gauge invariance is the fact that the same physical state can be described by infinitely many equivalent configurations of the wave function and A field, with all equivalent configurations related to each other by gauge transformations. So it is just the fact that there is a lot of redundancy in the variables you are using to describe the physics.

6. Jan 24, 2012

### strangerep

Typically we want to be able to perform Fourier transforms, and we still want to be able to do so after the gauge transformation. So one constraint is that the gauge function (and/or at least some of its derivatives should be Fourier-transformable). But we might well be working with generalized functions (distributions), so some of the constraints should be interpreted in that context. E.g., a "derivative" of a generalized function might need to be interpreted as a "weak derivative": http://en.wikipedia.org/wiki/Weak_derivative

Other constraints are that we usually want the fields to vanish at spatial infinity, so gauge transformations must not spoil that.

In a Lagrangian context, the field values at the start and end points of the action integral are considered fixed, so gauge transformations must not affect those either. Hence the gauge function should vanish at those points. (This is just a special case of the principle that variations must vanish at the endpoints.)

An observer can't make any measurement which is gauge-dependent. That's kinda the whole point of gauge transformations -- they just reflect redundant degree(s) of freedom in the way we've set up the theory. They're not physical.

7. Jan 30, 2012

### arlesterc

Thanks for the explanations. I will read over and try to absorb.

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