# What is the meaning of the local gauge transformation exactly?

1. Jul 6, 2011

### timewalker

What is the meaning of the local gauge transformation exactly??

These days I'm studying. [D.J. Griffiths, Introduction to Elementary Particles 2nd Edition, Chapter 10. Gauge Theories] Here the Section 3. Local Gauge Invariance, the author gives the Dirac Lagrangian, $\mathcal{L}=i \hbar c \bar{\psi} \gamma^{\mu} \partial_{\mu} \psi - mc^{2} \bar{\psi} \psi$ and show this is invariant under the global phase(gauge) transformation which is defined by $\psi \rightarrow e^{i \theta} \psi$. As auther mentioned, I already awared about this from quantum mechanics. After that, the auther introduce a local gauge transformation. At first I just passed this. I just regarded the auther asking to me just like "then, what about $\theta$ is not a constant??". Then, he abruptly put $\lambda (x) = -\frac{\hbar c}{q} \theta(x)$.

Now I confused. What the hell this 'charge' q come from? How can someone arrive that conclusion that he should introduce 'charge' here? So I visited my professor and he answered just "this is the 'photon'."

I confused AGAIN. What the?? Where is the right position should I begin my study step by step?

I thought this $\lambda(x)$ is just a gauge function that makes the Lagrangian invariant under the given local gauge transformation. In fact, after few lines the author indeed mention about a new field that transform under the local gauge transformation as $A_{\mu} \rightarrow A_{\mu} + \partial_{\mu} \lambda$ But he said this contents talking about a 'particle'. How can I understand this?

Last edited: Jul 6, 2011
2. Jul 6, 2011

### bapowell

Re: What is the meaning of the local gauge transformation exactly??

I see how you are confused. The charge $q$ is the generator of the U(1) (phase) transformation. Every group has generators -- members out of which the full group can be constructed. In the case of U(1), it's just a number; for the rotation group, for example, the generators are matrices.

The $\lambda(x)$ does not by itself make anything invariant. What happens is as follows: you perform the local (gauge) U(1) transformation on the Lagrangian. You will find that, as it is written, your Lagrangian will not be invariant under this transformation. In order to make it invariant, one needs to change the ordinary derivative, $\partial_\mu$, with something called a covariant derivative: $D_\mu = \partial_\mu + iqA_\mu$, where $A_\mu$ is the gauge field. In the case of electrodynamics, this is nothing more than the vector potential with well-known gauge transformation law $A_\mu \rightarrow A_\mu + \partial_\mu \lambda$.

So, if you replace the ordinary derivatives in your Lagrangian by these covariant derivatives, you will arrive at a gauge invariant theory. What you have succeeded in doing is incredibly profound -- you have introduced the photon, $A_\mu$, into the free electron theory!! By imposing gauge invariance -- a local dependence on the electron phase -- you have introduced the electrodynamic interaction.

And yes, I did associate the photon -- a particle -- with the vector potential -- a field. This is quantum field theory; if I remember correctly, Griffiths does not discuss fields -- so that might be new to you.

With regards to how to continue your studies, I would suggest familiarizing yourself with group theory, in particular, Lie groups as applied to particle physics. There are several good texts: Kaku's Quantum Field Theory, Cornwell's Group Theory in Physics, Georgi's Lie Algebras in Particle Physics, or Ryder's Quantum Field Theory should all be helpful.

3. Jul 6, 2011

### Bill_K

Re: What is the meaning of the local gauge transformation exactly??

I'd say that by introducing local gauge invariance you only have some interesting mathematics: a gauge field Aμ which might someday be light, and a constant q which could be the electric charge but is not yet. At this stage, Aμ carries neither energy or momentum, Gauss' Law does not hold, and q's for different particles do not have to be integer.

What really makes it a photon is when you add the FμνFμν term. The FμνFμν term is what says, Let there be light. Local gauge invariance only says, Let light be possible.

4. Jul 6, 2011

### timewalker

Re: What is the meaning of the local gauge transformation exactly??

At first I thank you so much for your kindness. Everyone who I asked didn't understood my situation. TT But I think all that is because of my little effort.

Anyway, now, I understand what you mean. In my point of view, you mentioned U(1) group structure. Actually, I took a course of abstract algebra when I was junior, for my second degree in mathematics(at that time, in fact, I thought it's not so helpful for physics :( ....), at that class I've learned that U(1) group is one parameter group among the general unitary group U(n) whose linear(matrix) representation is given by unitary matrix. In other words, U(1) and a circle group on the complex plane $T={z \in \mathbb{C} | |z|=1 }$are homomorphic. In that sens, I understood your explain that the generator of U(1) is just a number because any complex number whose norm is 1 has its polar form $e^{i \theta}$. So that our phase transformation have this freedom to choose an element(certain theta) among this group, therefore there are U(1) symmetry. And it's generator can be given by consider infinitesimal one. Right?

And the $\lambda$ was my miss. The book modified the lagrangian by introducing a new field $A_{\mu}$ so that making the new lagrangian become invariant under the local gauge transformation. Not by just the lambda. Modification of lagrangian is necessary I understand that. You are absolutely right. Thanks for pointing out that.

Here I wonder that the reason why you mentioned 'in the case of electrodynamics'. It sounds like unless we are talking about such a specific case(like electrodynamics), we are considering more abstract concept which covers any interactions?(it's our hope? To find governing systematic way to describe all interactions?)

Finally, so, all of this works succeeded by forcing there be the gauge invariance to a certain theory, and we found there should be a field like $A_{\mu}$ for that invariance. And its locality of gauge invariance explains its particle nature. Is my understanding right?

Last edited: Jul 6, 2011
5. Jul 6, 2011

### timewalker

Re: What is the meaning of the local gauge transformation exactly??

6. Jul 6, 2011

### bapowell

Re: What is the meaning of the local gauge transformation exactly??

Yes, indeed. My description gives you the equations of motion of the electron with the electromagnetic coupling; you must add the $F^{\mu \nu}F_{\mu \nu}$ term to complete the picture -- this is the equation of motion of the photon as Bill pointed out. Schematically, one wants:

free electrons + free photons + electron-photon interaction

7. Jul 6, 2011

### bapowell

Re: What is the meaning of the local gauge transformation exactly??

No problem timewalker, happy to help.
Gauge invariance is indeed a general concept, and applies to other interactions beyond electrodynamics. I think I referred specifically to ED because I was calling $A_\mu$ the vector potential which is familiar from ED, but is by no means unique to it. Apologies. In fact, all of the Standard Model of particle physics is based on these kinds of gauge interactions; the weak and strong forces are based on higher rank gauge groups (SU(2) and SU(3), specifically).

Not quite. In quantum field theory, one starts with quantum fields and interprets the quantized excitations of these fields as particles. This interpretation is true for free fields as well as those participating in gauge interactions. What the imposition of local gauge invariance does for you is that it creates an interacting theory out of a free theory.

8. Jul 6, 2011

### timewalker

Re: What is the meaning of the local gauge transformation exactly??

Then what about the 'photon' we've mentioned? So you mean that here we are just talking about 'gauge bosons' that intermediating certain interactions? What you mean is, thus, this story can not be applied to another particles i.e. fermions of SM like quarks and electron, muon, neutrino, etc.?

9. Jul 6, 2011

### bapowell

Re: What is the meaning of the local gauge transformation exactly??

The photon is the particle associated with excitations in the field $A_\mu$, just as electrons are the particles associated with the fermionic field $\psi$. But the imposition of gauge invariance does not imply this interpretation. What the imposition of gauge invariance did was it brought $A_\mu$, and hence, the photon, into the picture. So, yes, we talking about gauge fields facilitating interactions, and the excitations of these fields would indeed be the gauge bosons that we observe in colliders.

Not sure what you mean about applying the story to other particles. If you are referring to the possibility of fermions filling the role of gauge fields, this is not forbidden. However, if you start off with fermionic matter fields, the gauge fields will be vectors. This has to do with the fact that while fermionic matter fields happen to transform under the so-called fundamental representation of the gauge group, the gauge fields transform under the vector rep.

10. Jul 6, 2011

### timewalker

Re: What is the meaning of the local gauge transformation exactly??

I thank you so much :D All of these are very helpful explanations to me. Again, thank you so much! Now I noticed what I should do now and what is beyond this study for my further work. Thank you so much again. :D

11. Jul 11, 2011

### Lapidus

Re: What is the meaning of the local gauge transformation exactly??

The U(1) symmetry applies for the matter fields in the Lagrangian. If you only got a EM Lagrangian, there is no U(1) symmetry, only this redundant description, called gauge symmetry.

But then when dealing with the strong force, we have quarks and gluons, both have SU(3) symmetry, one in the fundamental the other in the adjoint representation of SU(3).

Are the photons also in some U(1) representation, which I miss?

Or what goes on here? Can someone unconfuse me here? thanks