# Gauge invariance and conserved current in SU(N)

1. Nov 13, 2013

### michael879

Hi, so I'm trying to derive the charge conservation law for a general SU(N) gauge field theory by using gauge invariance. For U(1) this is trivial, but for the more general SU(N) I seem to be stuck... So if anyone sees any flaws in my logic below please help!

Starting with the Lagrangian density:
$L=-\dfrac{1}{2}Tr(F_{\mu\nu}F^{\mu\nu}) + 2Tr(J_\mu A^\mu)$
where:
$F_{\mu\nu} = \dfrac{i}{g}[D_\mu, D_\nu]$
$D_\mu = \partial_\mu - igA_\mu$

The gauge symmetry transformation is:
$A_\mu \rightarrow UA_\mu U^\dagger + \dfrac{i}{g}U\partial_\mu U^\dagger$
giving:
$D_\mu \rightarrow UD_\mu U^\dagger$ *note this is still a derivative operator
$F_{\mu\nu} \rightarrow UF_{\mu\nu}U^\dagger$
so that trivially, the first term in the Lagrangian is unchanged. My problem is with the second term. I know from previous research I've done that the answer is:
$[D_\mu,J^\mu] = 0$
and that J transforms as:
$J_\mu \rightarrow UJ_\mu U^\dagger$

HOWEVER, if I plug this gauge transformation into the Lagrangian I arrive at:
$\delta{L} = 2\dfrac{i}{g}Tr(J^\mu\partial_\mu{U^\dagger}U)$
and I see absolutely no way to derive the continuity equation from this, considering it doesn't even depend on A!
The reason I know something is fishy (and why I've finally come here after hours of staring at this problem) is that if I make J transform like:
$J_\mu \rightarrow J_\mu U^\dagger$
I get:
$\delta{L} = 2\dfrac{i}{g}Tr([D_\mu,J^\mu]U)$
which is exactly what I was looking for! The only problem is, if J transforms like this then the continuity equation doesn't hold up under gauge transformations!! So there is a clear contradiction here, I just don't see where it is. Any help would be greatly appreciated!

2. Nov 14, 2013

### michael879

ok, I've done a little more work on this and found the following:
1) The contradiction is my fault, I assumed J did not transform and then proceeded to assume it did
2) Even though the continuity equation is trivial to derive from the equations of motion, I believe it is NOT a consequence of gauge invariance. See below:

First, I looked at the equations of motion and verified that in fact the following two statements are true:
1) The continuity equation holds:
$[D^\mu,J_\mu] = 0$
2) To preserve gauge invariance J must transform as:
$J_\mu \rightarrow UJ_\mu U^\dagger$

Instead of looking at arbitrary gauge transformations, I looked at infinitesimal ones:
$U = 1 + i\epsilon + \bigcirc(\epsilon^2)$
so that the new transformations are:
$\delta{A_\mu} = [D_\mu,\epsilon] + \bigcirc(\epsilon^2)$
$\delta{J_\mu} = -ig[J_\mu,\epsilon] + \bigcirc(\epsilon^2)$
$\delta{L} = Tr(J_\mu[D^\mu,\epsilon] - ig[J_\mu,\epsilon]A^\mu) + \bigcirc(\epsilon^2)$
from which it is very straight forward to show that:
$\delta{L} = J^a_\mu\partial^\mu\epsilon_a + \bigcirc(\epsilon^2)$
and therefore gauge invariance implies that J is a conserved current in ALL Yang-Mills theories!! This is astonishing and I'm not sure I believe it... If it is true then there are actually TWO conserved currents in SU(N) gauge theories: a matter charge current and a field charge current. What symmetry provides this second conservation law??

*Combining the two continuity equations gives you:
$[A_\mu,J^\mu] = 0$
which means that A and J are "orthogonal" to each other...
and combining them with the equations of motion you get the second continuity equation for the fields:
$\partial_\mu J^\mu_{field} = ig\partial_\mu([A_\nu,F^{\mu\nu}]) = 0$
[STRIKE]You can then substitute back in the equations of motion and you get:
$\tiny{-\dfrac{i}{g}\partial_\mu J^\mu_{field} = \dfrac{1}{2}F_{\mu\nu}F^{\mu\nu} - A_\mu J^\mu}$
Which means that $\tiny{A_\mu J^\mu = \dfrac{1}{2}F_{\mu\nu}F^{\mu\nu}}$ and an SU(N) Yang-Mills theory with sources is IDENTICAL to a free field theory (with an overall factor of -1 in the lagrangian)!!!

none of this sounds even close to right, what am I doing wrong??[/STRIKE]
*edit* I actually forgot the commutator in the field current definition. The final result ends up being the tautology:
$[F_{\mu\nu},F^{\mu\nu}] = 0$
I still do find it bizarre that there are two separable currents that are both conserved, but only 1 continuous gauge symmetry...

Last edited: Nov 14, 2013
3. Nov 14, 2013

### Chopin

I'm still learning quite a bit about gauge theories myself, so it's entirely possible that I'm incorrect here. But my understanding was that charge conservation is a consequence of the global symmetry only, not the local gauge symmetry. Entirely by coincidence, I happened to be reading over Tong's lecture notes on QFT today (http://www.damtp.cam.ac.uk/user/dt281/qft/six.pdf), and on page 138 he talks about gauge invariance and charge conservation, and mentions that charge conservation in QED comes out of the gauge variation only because the gauge group also contains the global symmetry as a subgroup.

So I'm not sure if this is relevant at all, but is it possible that the reason you're having difficulty here is that examining local gauge variations isn't a valid path to charge conservation at all?

4. Nov 15, 2013

### Avodyne

In Yang-Mills theory, the coupling to an external current is not gauge invariant. See p.200 of Pierre Ramond's "Field Theory: A Modern Primer" (revised version).

5. Nov 15, 2013

### michael879

1) I can't give you a reference so I could be wrong, but I'm pretty sure local gauge invariance is required to give charge conservation w/o the Euler-Lagrange equations
2) local invariance is a generalization of global invariance, so the global transformation should be included in my analysis

6. Nov 15, 2013

### michael879

I don't have access to this book but surely you're mistaken... The standard model respects 3 gauge transformations and included external currents coupled to the gauge fields. And the math above proves that as long as the current is transformed as well, gauge invariance is preserved

7. Nov 15, 2013

### Chopin

I'm not sure I see how this statement could be correct. It's perfectly possible to have charge conservation without gauge invariance--even a plain old N-multiplet of complex scalar fields has N conserved currents, and there's no gauge symmetry at all there. Noether's theorem says you get a conserved current out of any 1-parameter symmetry group. So a local gauge symmetry doesn't even qualify for that--it's an infinite-parameter symmetry group, so it's no more applicable for Noether's theorem than a discrete symmetry like C or T.

I thought that the way this worked was that you started out with local gauge invariance being a property of the free photon field only. Then you try to work out how photons could interact with a matter field, and you discover that you can preserve local gauge invariance if you couple them to an existing conserved current in the matter field, turning the field's global symmetry into a local one in the process. If that's the case, then charge conservation is a necessary prerequisite for an interacting theory with local gauge invariance, not a consequence of it. Perhaps others with more experience on the subject can confirm or refute this.

Fair enough, but if it is the case that only the global symmetry is required to prove charge conservation, then it would seem to me that including the full local symmetry is at best overcomplicating things.

Last edited: Nov 15, 2013
8. Nov 15, 2013

### michael879

complex scalar fields with conserved currents DO obey gauge symmetries, and that's exactly where the conservation laws come from. For example, the free massive complex scalar field obeys a U(1) gauge symmetry.

As for Noether's theorem, you are talking about her first theorem. Her second theorem is specifically for infinite-parameter symmetries and is perfectly applicable to local gauge symmetries. In fact, you CAN'T derive the charge conservation law by only using the global gauge symmetry!

Thats just a chicken vs. egg thing. Its a matter of interpretation whether symmetries produce conservation laws or conservation laws produce symmetries.

The point I'm trying to make here, is that I have found it is very straightforward to show the following (and I have done so in my previous posts):

1) You have a free SU(N) gauge field that respects the local symmetry
2) You have an arbitrary current (details irrelevant) that you wish to couple to the vector gauge field
3) You want the resulting theory to retain the local gauge symmetry

=>

From the Euler-Lagrange equations:
1) J must transform properly under gauge transformations
2) The sum of J and the field's charge current must be conserved
3) J commutes with the covariant derivative (a modified continuity equation)

From enforcing gauge symmetry:
4) J must be a conserved current
5) The field's charge current must be conserved
6) J must be "orthogonal" to A

1,2, and 3 are very straightforward and I haven't derived them here. 4,5, and 6 I have derived, but I am very skeptical of them.

9. Nov 15, 2013

### michael879

*edit* sorry, you can derive A conservation law using the global symmetry AND the Euler-Lagrange equations. However this is a less powerful law than the one I derived using only the local symmetry

10. Nov 15, 2013

### michael879

ok so I finally figured out (mostly) what I was doing wrong. Only when I assume J transforms under gauge transformations do I find that it is conserved! If I leave J alone, and apply the infinitesimal gauge transformation I arrive at:
$\delta{L}=-\epsilon[D_\mu,J^\mu] = 0$
which is exactly what the EL equations give! This makes MUCH more sense, but now I'm a little confused about the transformation properties of the EL equations. As shown above, under the gauge transformation L transforms as:
$L \rightarrow L + \partial_\mu X^\mu \equiv L$
However the EL equation
$\left[D_\nu,F^{\mu\nu}\right] = J^\mu$
transforms like
$\left[D_\nu,F^{\mu\nu}\right] \rightarrow U[D_\nu,F^{\mu\nu}]U^\dagger = J^\mu$
which is NOT invariant under the gauge symmetry! (in general U and J wouldn't commute).

What is going on here???

11. Nov 15, 2013

### samalkhaiat

12. Nov 16, 2013

### michael879

Wow thank you! I will go through this, and hopefully it will answer my questions. I have an 8 hour flight ahead of me so expect another post in ~10 hours