# Bob makes a local gauge trans., can Alice undo with some fields?

1. Mar 26, 2013

### Spinnor

Say Alice gives Bob the wave-function, a momentum eigenstate, of a charged particle. Bob then makes a local gauge transformation on the wave-function,

ψ --> exp[iqθ(X,t)]ψ.

Can Alice now undo the local gauge transformation with the right addition (or subtraction?) of electromagnetic fields?

Hope that was worded right. Thanks for any help!

2. Mar 26, 2013

### fzero

You have a (probably common) misconception of what gauge invariance means. A gauge invariance is a redundancy in the degrees of freedom that we use to describe a system. This means that the degrees of freedom controlled by a gauge transformation are not observables. So it does not make sense to say that an observer will perform a gauge transformation.

If you study the details of how to quantize a system with gauge invariance, either canonically or via a path integral, you will see the terms "gauge choice" or "gauge fixing." These are formal methods introduced to account for the fact that the extra gauge degrees of freedom need to be consistently removed to have a sensible quantum theory. It is important to show that the quantization procedure leaves the observables independent of the choice of gauge. This is done in any decent textbook on quantum field theory.

You might ask, why were the redundant degrees of freedom introduced in the first place? The short answer is that naive quantization of a field that is a Lorentz vector results in the situation that some of the components of the field seem to have a negative norm, $|\chi|^2 < 0$.. Since the norm of a state must be positive in order to have a sensible description of probabilities in quantum mechanics, we must find some way to get rid of these states. It turns out that gauge invariance provides such a mechanism. Removing the redundant gauge degrees of freedom also removes the negative norm states, leaving a sensible quantum theory.

In ordinary quantum mechanics, it should already be clear that absolute phases of wavefunctions are not observables. Observables themselves can be expressed as expectation values of Hermitian operators computed in the particular state

$$\langle A \rangle = \int dV ~ \psi^* \hat{A} \psi,$$

so it's clear that the phase of $\psi$ cancels against the phase of $\psi^*$. Certain measurements can be used to measure a relative phase between two states, such as are responsible for interference patterns, but the common part of the phase shared by two states is similarly not an observable.

Finally, the electromagnetic field is defined in terms of the electromagnetic potentials in precisely such a way that the fields are gauge-invariant, while the potentials transform. Consistent with the unobservable nature of gauge choice, changing the electromagnetic field in a lab cannot lead to a gauge transformation on some state.

3. Mar 27, 2013

### Spinnor

It is good to know what you don't know. There is a nice presentation of gauge invariance of Maxwell's equations in "Gauge Theories in Particle Physics" Aitchision and Hey. I will reread that section with your post in mind.

Thank you!