# Photon, spontaneous mass

1. Jun 17, 2014

### geoduck

I read somewhere that gauge symmetry prevents the photon from acquiring a mass. The argument seems to go that the 1-loop correction to the photon won't contain a term independent of the external momentum due to gauge invariance, so there is no need for a bare mass counter-term.

So should that statement be modified to gauge symmetry prevents the photon from acquiring a bare mass?

Can't you always set the renormalized mass equal to zero, even if gauge symmetry is lacking? Like a $\phi^4$ theory?

Also, shouldn't the relationship between bare mass and renormalized mass be that they will always be proportional to each other, because there are no other parameters in the theory with dimensions of mass? Then it should follow that the bare mass can always be set to zero?

2. Jun 17, 2014

### atyy

3. Jun 17, 2014

### andrien

There is quite a subtlety with putting bare mass equal to zero for photon. In fact, if you will put bare mass of photon equal to zero, you will find that with a convergence factor included physical mass of photon goes quadratic with the cut-off !

This is as bad as it sounds, the 1 loop correction to photon propagator will include a mass term in zero momentum limit coming from the polarization tensor and it is not zero. Gauge invariance and lorentz invariance can not alone make it zero because this tensor can still have a pole at k2=0.

4. Jun 21, 2014

### geoduck

It seems with dimensional regularization you can put the bare mass to zero because the 1-loop won't contribute a momentum-independent term.

But with cut-off you'll get a term that goes quadratic with cut-off. But then can't you set the bare mass equal to opposite of this cut-off, so that the renormalized mass is zero?

It can have a pole at k2=0 if you adjust the bare mass to cancel the quadratic cutoff term?