# Gauge Symmetry in QM

1. Nov 17, 2012

### the_kid

Hi all,

I'm taking graduate level QM I and trying to wrap my head around the notion of gauge symmetry. For some reason I've struggled with this concept more than others. I don't really have a specific question; I'm more looking to see if someone has a succinct explanation of the relevant details - or perhaps a good source to read.

Thanks!

-TK

2. Nov 17, 2012

### the_kid

OK, I've done some more reading on my own and now have a specific question. Hopefully this will get this thread moving.

It is my understanding that for a quantum system to be locally gauge invariant, it must be in the presence of an EM field. I understand mathematically why this is true (at least roughly), but I'm wondering if there is a physical explanation for it. Any input?

3. Nov 18, 2012

### dextercioby

Actually the gauge symmetry is classical in origin, and most of the quantum systems also admit a classical picture (especially in field theory). So that you can write down a fully classical lagrangian (density if needed) as a sum between the purely gauge field and the "matter field", i.e. the Schrödinger, Dirac fields. Then apply specific methods of quantization (i.e. converting clasical Poisson/Dirac brackets to (anti)commutators) to get a meaningful quantum theory.

Yes, the EM field in vacuum is the most proeminent example of gauge field, but the linearized (Fierz-Pauli) gravity field is also a gauge field. Likewise for a collection of covector fields (so-called Yang-Mills fields).

4. Nov 18, 2012

### andrien

EM field is not any necessity,however the idea of local gauge invariance came from electromagnetism which is an abelian gauge field i.e. the commutator [Aμ,Av] vanishes which is not the case with non abelian gauge fields.It is used in the work of yang and mills.

5. Nov 18, 2012

### Jano L.

I do not think this is a good way to describe what gauge invariance is. It is not the system that is invariant, but the predicted quantities, like probability or average values, are invariant. In non-relativistic theory, the situation is as follows:

-there is Schroedinger's equation for $\psi$, which can be formulated in the presence of external electromagnetic field; this is accomplished by use of electromagnetic potentials $\varphi, \mathbf A$;

-$|\psi(r)|^2$ is interpreted as probability density that certain configuration $r$ occurs; and $\frac{1}{m}\mathrm{Re} (\psi^* (\mathbf p - \frac{q}{c}\mathbf A)~ \psi)$ is intepreted as probability current density in space of these configurations;

- the Schroedinger equation is such that the change of electromagnetic potentials which leaves fields E,B the same (change of gauge) does not change the above kind of quantities.

Thus the gauge invariance is the invariance of probabilities calculated from Born's rule. Or it can be invariance of all other quantities calculated from these, such as average expected position or momentum.