# I Gauge Symmetry

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1. Jan 13, 2018 at 5:06 PM

### arlesterc

I have reviewed the various posts on gauge symmetry in particular this one which is now closed. In this post there is the following link:http://www.vttoth.com/CMS/physics-notes/124-the-principle-of-gauge-invariance.

This is a good read. However, there is some clarification I need.

The article has the following:
"By far the simplest gauge theory is electromagnetism. And by far the simplest way to present electromagnetism as a gauge theory is through the non-relativistic Schrödinger equation of a particle moving in empty space:

iℏ∂ψ∂t=−ℏ22m∇2ψ.

Although the equation contains the wave function ψ, we know that the actual probability of finding a particle in some state is a function of |ψ|. In other words, the phase of the complex function ψ can be changed without altering the outcome of physical experiments. In other words, all physical experiments will produce the same result if we perform the following substitution:

ψ→eip(x,t)ψ,

where p(x,t) is an arbitrary smooth function of space and time coordinates."

Then it goes on to derive the Schrödinger substituting in this varying function and after a lot of steps shows:

"iℏ∂ψ∂t=−ℏ22m{[∇+i∇p(x,t)]2−2mℏ∂p(x,t)∂t}ψ.
which is not the original Schrödinger equation. "

My question/line of thinking: Is the Schrodinger equation supposed to produce the same 'answer' for the wave function at every point in space? In other words two different observers at two different points in space calculating the wave function via the 'ordinary' Schrödinger should end up with the same answer? However this does not happen because each observer is allowed/has a different value for the phase of the wave function based on their location. Therefore to rectify this, the Schrodinger equation has to be modified so as to cancel out this local location-dependent phase and once that is done everyone at every point using this 'modified' Schrödinger equation will end up with the same wave function?

I would appreciate any feedback as to whether I am on the right track here. Thanks in advance.

2. Jan 14, 2018 at 12:40 PM

Gauge symmetry is not a real symmetry. A symmetry acts on one state and takes it to another. A gauge transformation takes a state to the same state. The correct terminology is gauge invariance. The wave function is not a gauge invariant object. It doesn’t have to be because it is not an observable. However all observables like probability amplitudes, currents, etc. must be gauge invariant. Any gauge dependence must cancel out. That’s why you have both the canonical and kinematic momentum in the presence of the magnetic field, the canonical momentum is not gauge invariant.

One thing to note is that once you choose a gauge, the wavefunction must be continuous and single valued at every point in space. However, there are times when there are topological obstructions to this. Then you need to choose different gauge covering different areas which are related by enforcing these conditions on the wavefunction. For example, if you have a magnetic monopole, you must choose a different gauge to cover the north and south pole regions. However, given the singlevaluedness of the wavefunction, these gauge can only different by some quantized amount. For a magnetic monopole, this restricts the charge to be quantized in units of \frac{\hbar c}{2e}. This gives rise to various phenomena such as the quantization of the Hall conductance. The number of quanta n is a topological invariant, you cannot change it via smooth deformations.

3. Jan 14, 2018 at 9:29 PM

### arlesterc

Thanks for the distinction between gauge symmetry and gauge invariance. I think however based on the amount that I see gauge symmetry used it will be a long battle to knock it out of use. Or is there such a thing as gauge symmetry apart from gauge invariance?

4. Jan 15, 2018 at 8:14 PM