# Why is it important that the vector current should be conserved in QED?

In Quantum Field Theory and the Standard Model by MD Schwartz in the chapter about the anomalies, he derives from the equation of motions and the Noether currents of a effective massless QED Lagrangian that the vector current is exactly conserved, while the axial current has an anomaly and remarks:

Thus, classically the vector symmetry is exactly conserved, which is important since it is the one that couples to QED, while the chiral symmetry is only conserved in the massless limit.

Why is it so important that the current that couples to QED should be conserved, I suppose it is for unitarity reasons, but I would like to see an explicit argument.

• Since QED is a gauge theory, It would be simply inconsistent a non conserved current. For example, the low-energy theorem by Weinberg based on Lorentz symmetry and gauge invariance dictates that the scattering amplitude can be non trivial only if charges are conserved. – TwoBs Dec 23 '14 at 8:35

There is following statement: massless particles with both of helicities $\pm 1$ can't be represented by 4-vector field $A_{\mu}$. The only field (up to equivalence) which represents corresponding particles is $F_{\mu \nu}$. If you decide to represent these particles by $A_{\mu}$, then it won't be 4-vector: $$A_{\mu}(x) \to \Lambda_{\mu}^{\ \nu}A_{\nu}(\Lambda x) + \partial_{\mu}\psi (x) ,$$ or, equivalently, $$\tag 1 \epsilon_{\mu}(p) \to \Lambda_{\mu}^{\ \nu}\epsilon_{\nu}(p) + p_{\mu}\psi(p^{2}).$$ So if we build theory of interaction of some matter field with $A$-field (we need it because it represents the inverse square law, while $F_{\mu \nu}$-interaction doesn't), we need to verify that interaction processes are lorentz-invariant, i.e., second summand in $(1)$ doesn't affect on physical amplitude. It can be shown in the soft-photons limit that it's really true only if total charge in process is conserved. But conservation of charge is nothing but 4-vector current conservation in integral form.

So you see that 4-current conservation is necessary for Lorentz-invariance of QED (as 4-momentum conservation and the equivalence principle is necessary for Lorentz-invariance of gravitation theory).

There's a glib argument to see why conservedness of the current is needed for gauge invariance. The "coupling" between the photon and the current is given by

$$L \supset A_\mu J^\mu.$$

Under a gauge transformation, $A_\mu \to A_\mu + \partial_\mu \Lambda(x)$ so

$$L \to L + J^\mu \partial_\mu \Lambda.$$

After integration by parts and throwing away a boundary term, the change in the action is

$$\delta S = \int d^4x \, \Lambda(x) \, \partial_\mu J_\mu.$$

Since this must vanish for all choices of $\Lambda(x)$, $\partial_\mu J_\mu = 0.$

• The problem with your answer is that the conservation of current is valid only on shell, for fields that satisfy the eq. of motions. – TwoBs Dec 23 '14 at 15:23
• I don't really agree - conserved currents don't renormalize to all orders in perturbation theory. This is a honest theorem in QFT, see the Peskin-Schroeder chapter on RG or Collins' book for a proof. – Guest201444 Dec 23 '14 at 15:51
• I think you misunderstood my criticism. of course they don't get renormalized (meaning the anomalous dimension is zero). But reamains the fact that the conservation of the current is for fields that satisfy the equations of motion whereas the fields in the action do not necessarily obey the eq. of motion. I think there is a sense in which your answer is morally correct that goes under the name of Ward identity (one can show that any insertion of $\partial_\mu J^\mu$ on correlations will vanish etc, but this is not what you have written and I encourage you to improve your answer. – TwoBs Dec 23 '14 at 16:18
• The non-renormalization statement means more than "anomalous dimension is zero" - it means that the Ward identities remain true to all orders in PT, regardless of any equations of motion. So $\partial_\mu J_\mu = 0$ as an operator identity (up to contact terms). This is all textbook stuff, not something that is 'morally correct in some sense', and I don't plan on spending any more time on this question. – Guest201444 Dec 23 '14 at 17:13
• If vector current were anomalous then Ward identity would have been violated. That in turn implies that the unitarity of the theory will be violated. The point that Ward identity is necessary for preserving unitarity is explained in the QFT book by Lewis Ryder. – SRS Jul 20 '15 at 8:00