# Gauge invariance of QED if the photon has a mass

1. Oct 6, 2011

### shakespeare86

Hi, excuse the funny title :).
In his book on quantum field theory Zee says (pag 245, fouth line) that QED gauge simmetry follows from the conservation of the current j=ψ γ^μ ψ (with the bar on the first spinor).
I'm confused because that current is the noether current resulting from the invariance of QED lagrangian under a global phase trasformation, in which the spinor trasforms under U(1) and the vector potential A do not transform, and this global symmetry wouldn't be spoilt if the photon had a mass!
So what I get from this observation is that whatever mass m the photon has, j is a conserved quantity. On the other hand, gauge invariance (that is a local transformation, under which A transforms too) holds only if m=0.
So is Zee right or wrong?

2. Oct 6, 2011

### tom.stoer

It's confusing.

Of course gauge symmetry does not follow from current conservation, but current conservation follows (by Noether's theorme) from a (global or local) symmetry.

If you gauge a global symmetry you can again derive a current - and it agrees with the current derived from the global symmetry. For m>0 the local gauge symmetry does no longer hold and the current is no longer conserved; the latter can be derived from the field equations w/o, too.

3. Oct 6, 2011

### shakespeare86

So is Zee wrong?
I think if j is conserved, you just can't argue that you have gauge invariance, because j would be conserved also if the photon has a mass.
Do you agree?

4. Oct 6, 2011

### shakespeare86

if you give the photon a mass, QED lagrangian is still invariant under U(1) global trasformation, because the mass term does not change!
Under a global transformation A -> A!
So whatever mass we give to the photon, it seems to me that j is a conserved quantity.

5. Oct 6, 2011

### Bill_K

What Zee says is rather long and convoluted! But I think what it boils down to is yes, he actually does say that gauge invariance follows from current conservation.

His argument extends over II.6 and II.7. He starts with a photon propagator with mass μ, takes any diagram with external electron lines, and proves the Ward-Takahaski identity (which is a generalization of current conservation). He then uses this to show that the amplitude for the diagram is independent of μ, and hence one may set μ = 0. But the Lagrangian with μ = 0 is gauge invariant.

6. Oct 6, 2011

### shakespeare86

yes i agree.
The amplitude for the diagrams is independent of m just thanks to current conservation, that makes the k^mu k^nu / m^2 in the propagator irrilevant.
So j is a conserved quantity whatever m>0 we take and it sounds wrong that from this he can deduce gauge invariance that holds just if m=0.

So said that, what he wrote few lines below is an argument to say that from gauge invariance follows current conservation.
Do you agree?
So
gauge invariance -> current conservation is a good implication
the opposite implication is not.

Last edited: Oct 6, 2011
7. Oct 7, 2011

### tom.stoer

In principle "symmetry => current conservation" is the classical Noether theorem which can be translated to QFT (Ward-Takahashi).
Here local gauge symmety is a rather special topic, but the Noether theorem remains unchanged. There are many differences between global and local symmetries, but regarding the application of the Noether theorem they are identical.
In practice there are two types of conserved currents. For most of them I would say that "current conservation => symmetry" is correct - but you have to check this for each indivisual case, i.e. you have to identify the symmetry and then work backwards, i.e. recunstruct the current from the symmetry.

I give you simple example for a current which is not defined via a symmetry: take an arbitrary scalar field in 1+1 dim. spacetime and construct

$$j^\mu = \epsilon^{\mu\nu}\partial_\nu\phi$$

This current is trivially conserved due to the antisymmetry of the epsilon, but of course this is not related to any global or local symmetry.

8. Oct 7, 2011

### dextercioby

In the context of field theory, gauge invariance doesn't follow from anything. Instead, gauge invariance generates the interactions between massive or massless fields.

9. Oct 7, 2011

### Bill_K

I think we're quite familiar with the usual relationship between Noether's Theorem and current conservation. But has anyone read Zee's argument, and can comment on it? Because he arrives at an unconventional result.

To restate, he starts with a fermion minimally coupled to a massive spin one boson having mass μ. Now a massive spin one boson is clearly not gauge invariant. And we all know it is not renormalizable. Nevertheless he derives, or appears to derive, a Ward-Takahashi identity for the system. His conclusion is that although the longitudinal mode is present, it decouples from all physical processes, and that therefore "we can set the photon mass μ equal to zero with impunity." The gauge-dependent terms involving the mass μ fail to contribute to the Feynman amplitude, and in that sense he claims to have "proved gauge invariance."

10. Oct 7, 2011

### shakespeare86

Yes you are right: he says that.
Infact in this way he is not proving gauge invariance and most of all he can't say that from d j =0 follows gauge invariance, as now i see we all agree :)!

He has just proven that if we start with a massive spin one boson, then we can take the limit m->0 safely because the k^mu k^nu / m^2 term in the propagator do not contribute to the feynman diagrams.
In this way, we can arrive at a theory where m=0 and that is gauge invariant.
This is what I think he should say.

The reason why this argoment works is that giving the photon a mass m, we do not spoil GLOBAL U(1) symmetry of QED and thus j is still a conserved corrent!
Thus, we can neglet the k^mu k^nu / m^2 term!
I don't think that massive QED is not renormalizable, because the bad terms k^nu / m^2 just goes away thanks to current conservation.

A non abelian theory is qualitatively different because when we give a mass to the gauge boson, we spoil SU(2) GLOBAL simmetry, besides SU(2) local symmetry.
Thus, we haven't a conserved current and the k^mu k^nu /m^2 terms stays there and seems to contribuite to the diagrams unless you use 't Hooft gauge.

So, summing up:
-in an abelian theory, whatever mass m we give to the gauge boson we have a conserved current associated with U(1) GLOBAL simmetry. j is of course a current associated with the matter field. This makes this theory renormalizable at first sight whatever m. Anyway we have, of course, gauge invariance just at m=0.

-in a non abelian theory, global symmetry holds just when the gauge boson do not have a mass. In this case we have global and local SU(2) symmetry. When we give them a mass, we have not a conserved (matter) current, we have just a (gauge bosons + matter) conserved current, that is not what couples to the gauge boson propagator, and the k^mu k^nu term do not goes away, thus making renormalization not easy to be proven.
If you agree I'd like to open other related discussions :).
Thanks for the interesting conversation.

11. Oct 9, 2011

### Bill_K

Zee is not the only one who claims to deduce local gauge invariance from current conservation and not the other way around. Weinberg makes a similar deduction.

On p.342 he says, "In the above discussion we have started with the existence of massless spin one particles [coupling to a conserved current] and have been led to infer the invariance of the matter action under a local gauge transformation (8.1.12), (8.1.13). As usually presented, the derivation runs in the opposite direction. That is, one starts with a global internal symmetry and asks what must be done to promote this to a local symmetry.

12. Oct 9, 2011

### vanhees71

Zee is, as usual in his book "QFT in a nutshell" not quite right.

First of all, current conservation is weaker than gauge invariance since it follows already from a global symmetry. E.g., take the Dirac-Yukawa Lagrangian

$$\mathcal{L}_{\text{Dirac}} = \overline{\psi}(\mathrm{i} \gamma^{\mu} \partial_{\mu} - m \psi + \frac{1}{2} (\partial_{\mu} \phi) (\partial^{\mu} \phi) - g \overline{\psi} \psi \phi.$$

Since this Lagrangian is invariant under the global transformation

$$\psi \rightarrow \exp(\mathrm{i} \alpha) \psi, \quad \overline{\psi} \rightarrow \exp(-\mathrm{i} \alpha) \overline{\psi}, \quad \phi \rightarrow \phi$$

with $\alpha=\text{const}$, the usual current $j^{\mu} = \overline{\psi} \gamma^{\mu} \psi$ is conserved, but for sure this Lagrangian is not invariant under local gauge transformations, i.e., you cannot make $\alpha$ a function of the space-time variables $x$ since the term with the derivative of the Dirac field leads to an additional term which needs to be compensated by the introduction of a gauge field, which is a vector field, with help of which you can define a covariant derivative via $\partial_{\mu} + \mathrm{i} q A_{\mu}$.

In the most simple form, you keep the vector field massless, and this is natural in the sense that a massless vector field with a finite number of spin-like degrees of freedom (for usual massless vector particles these are the two polarization states, e.g., given by the states of helicity 1 and -1), from Poincare invariance the massless vector field must necessarily be introduced as a gauge field.

Local gauge invariance has important consequences for the Green's and the 1PI vertex functions, the already mentioned Ward-Takahashi identities (or generalizations for non-Abelian gauge theories, called Slavnov-Taylor identities). These WTIs are also important for renormalizability: If a gauge theory is superficially renormalizable, i.e., contains only the allowed terms with not too many derivatives and/or fields, it is really renormalizable, although there are vertices that are not finite by power counting. E.g., in QED the four-photon vertex is of superficial degree of divergence 0 and thus naively logarithmically divergent. If it were really divergent, one would be in trouble since there is no gauge invariant local four-photon term for the Lagrangian which at the same time is renormalizable. Fortunately, the WTIs make the potentially divergent parts exactly 0 due to gauge symmetry, and QED is renormalizable. The same mechanism holds for spontaneously broken gauge symmetries ("Higgs mechanism") for both abelian and non-abelian gauge theories, leading to massive vector bosons without breaking gauge invariance.

In the abelian case, there's however another possibility, i.e., one can have a massive gauge boson, coupled to a conserved current (conserved because of a gauged global symmetry) without violating the gauge invariance due to the vector-boson mass term. This is known as the Stückelberg mechanism. That goes as follows: You start with the free Lagrangian for a massive vector field, i.e.,

$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{m^2}{2} A_{\mu} A^{\mu}$$.

This Lagrangian is not invariant under gauge transformations,

$$A_{\mu} \rightarrow A_{\mu}-\partial_{\mu} \chi$$

because of the mass term. This can be repaired by introducing a real scalar field $\phi$ in the following way

$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{m}{2} A_{\mu} A^{\mu} + \frac{1}{2} (\partial_{\mu} \phi) (\partial^{\mu} \phi)-m \phi \partial_{\mu} A^{\mu}.$$

Then the Lagrangian is invariant under gauge transformations (up to a total divergence, i.e., the action is invariant)

$$A_{\mu} \rightarrow A_{\mu}-\partial_{\mu} \chi, \quad \phi \rightarrow \phi-m \chi.$$

Of course, this finally leads to a field theory of free massive vector bosons and a free massless scalar boson and is not very exciting.

However, of course, now you are free to introduce other particles. One application is the Kroll-Lee-Zumino model for $\rho$ mesons, pions, and photons leading to the famous (renormalizable) vector-meson dominance model which gives good effective descriptions for the elastic pion-scattering phase shift in the 11-spin-isospin channel and the electromagnetic form factor of the pion. It's also easy to add $\omega$ mesons.

Quantizing this kind of models with the Faddeev-Popov method allows a gauge fixing similar to the $R_{\xi}$ gauges in electroweak theory, showing that indeed the interacting-particle spectrum consists of a massive vector meson and the matter fields. The field, $\phi$ (in this context called Stückelberg ghost), decouples as the Faddeev-Popov ghosts do. The difference is that Faddev-Popov ghosts are scalar Grassmann fields while the Stückelberg ghost is a usual scalar field. Thus, the massive vector field together with the Stückelberg and Faddeev-Popov ghosts make 3 effective physical field degrees of freedom as it should be for a massive vector field. At the same time the model is gauge invariant very similar to QED and thus manifestly renormalizable in the $R_{\xi}$ gauge.

13. Oct 9, 2011

### Bill_K

I agree, or at least he is often imprecise. Which is why I thought it worth mentioning that Weinberg also presents a similar argument. But after tracing Weinberg's spiderweb of assumptions back through the various earlier chapters, I think he has overstated the case also.

14. Oct 10, 2011

### DrDu

Hendrik,

isn't the Stueckelberg method just another example of a broken gauge symmetry for the phi field?

15. Oct 10, 2011

### genneth

For the experts (I'm not sure how to state this in a way that's beginner friendly):

The issue is whether you assume that current conservation is "on-shell" or "off-shell" as well. Classically, this is the question of whether the current conservation is a result of the equations of motion, or if it is in fact completely independent. Assuming the latter gives a local gauge theory.

From the point of view of quantum theory, one should keep in mind that choice of Lagrangian is only (less than?) half the story --- one also needs to pick the kinematic space. Local gauge symmetry is really about the latter.

16. Oct 11, 2011

### Avodyne

But the only way to get the latter is to explicitly add a gauge field to the theory. Or, as Weinberg does, explicitly postulate a massless particle that couples directly to the current (which then ends up being the quantum excitation of a gauge field). If you don't explicitly add a gauge field (in one way or another), then you cannot have a current that is conserved off shell.

17. Oct 11, 2011

### tom.stoer

I think this is not correct.

There is a global axial UA(1) symmetry in QED w/o any gauge field. The axial current is conserved classically; the current is not conserved after quantization b/c of the so-called axial anomaly. If you look at the same situation in QCD (w/o QED and w/o the usual U(1) symmetry of QED, i.e. w/o photons!) then this UA(1) is conserved, i.e. there is no axial anomaly in QCD, the gluon contribution to the triangle anomaly cancels exactly.

So I can't really see where the gauge field enteres the stage in this case

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