# Does the Ward identity rescue a zero photon mass?

1. Jun 11, 2013

### center o bass

In Peskin at page 248 he finds that if he calculates the vacuum polarization that

$$\Pi(q)^{\mu \nu} \propto g^{\mu \nu}\Lambda^2$$

a result which violates the Ward identity and would cause a non-zero photon mass $$M \propto \Lambda$$. But as Peskin states, the proof of the Ward identity is not valid when it requires a shift of integration variable and that is not a valid operation when the integral is divergent.

He then states that it is a way to rescue the Ward identity; namely trough dimensional regularization - a regularization method which respects the ward identity.

He then goes on and calculates the vacuum polarization and finds it respects the Ward identity.
I am however left a bit confused; it seems to be a choice which regulator to use and thus what mass the photon will get. Since the proof of the Ward identity does not hold for divergent integrals there seems to be no apriori reason to choose dimensional regularization other than to fit the theory to experiment.

So are we really left with a choice here?

2. Jun 11, 2013

### Physics Monkey

I would say the choice is binary. One can either use a regulator that automatically respects the U(1) symmetry or one that does not. A hard cutoff does not automatically respect the symmetry, e.g. the Ward identity can be violated unless one adds a certain naively gauge-non-invariant counterterm. On the other hand, a lattice regulator and dimensional regularization do automatically respect the symmetry.

This kind of thing is encountered in practical calculations in condensed matter physics all the time.

3. Jun 11, 2013

### center o bass

But then there is no theoretical reason why one should choose one which respects the ward identity? I.e. there is no theoretical reason why the photon should not gain a mass?

I've read a bit further, and it seems that this issue is also related with the 'axial anomaly'. From what I've understood if one also get a choice there; the axial current can be chosen to be conserved, but then the photon will get the wrong # of degrees of freedom.

4. Jun 11, 2013

### Avodyne

The theoretical reason is that we are trying to find a theory of photons and charged particles. In the real world, photons are massless and have spin one. It turns out that this is a natural consequence of a quantum field theory of a vector field with a gauge invariance, so that is the type of theory that we use.

Perturbative calculations require regularization at intermediate stages. It would make no sense to use a regularization scheme that violates gauge invariance, since that is one of the key input principles.

5. Jun 12, 2013

### Physics Monkey

Not sure exactly what you mean by "theoretical reason", but if you mean that, absent experiments, one isn't forced to choose, then I agree.

The basic picture I would advocate is that to define a quantum field theory, you must include a regulator. The choice of the regulator is part of the definition of the theory. Experimentally, the photon is very close to being exactly massless, so a simple solution is to posit that it is exactly massless and to use a regulator which preserves this choice.

Of course, one can always imagine that the mass just happens to be very small, but if you use one of these "bad" regulators, then you find that typically the mass is of the order of the cutoff. So there are some theoretical considerations that suggest that if the mass is very small (we can never be sure it is zero), then the most natural thing is to use a regulator which explicitly keeps the mass at zero. Then if it later turns out that there is a small mass, one can introduce it using some low energy physics.

However, I should also emphasize that even if you use a "bad" regulator, you can still describe a massless state by adding some bare A^2 term to cancel the unwanted mass. It's really a question of how you regulate things and the physics you want to describe.