Self energy logarithmic divergence due to chiral symmetry.

[center o bass]

In peskin at page 319 right above equation (10.6) he writes

"If the constant term in a taylor expansion of the self energy were proportional to the cutoff $\Lambda$, the electron mass shift would also have a term proportional to $\Lambda$. But the electron mass shift must actually be proportional to $m$ since chiral symmetry would forbid a mass shift if $m$ were zero."

So chiral symmetry is a symmetry between right and left handed fields. If the mass is zero the Lagrangian has this symmetry and the axial current is conserved classically. But how does this symmetry restrict the particle from getting a mass?

Is there an restriction within QFT which prevents a massless particle from gaining mass in mass renormalization?

 

[andrien]

If there will be a mass term then γ₅ will not commute with the Hamiltonian.

 

[center o bass]

If there will be a mass term then γ₅ will not commute with the Hamiltonian.

And if $\gamma_5$ commutes with the Hamiltonian for the bare theory; it must also do so for the renormalized theory?

 

[andrien]

why not?

 

[center o bass]

why not?

I would say that the properties action can't change since the bare and the renormalized theory are essentially the same theory, but perturbing about different coupling constants?

However I have read that it's not always true that if a theory is massless, one always obtain obtain a zero mass shift.

In these notes for example:

http://www.google.no/url?sa=t&rct=j...=-mHf2NPyluthnweklmsZzA&bvm=bv.47810305,d.bGE

at page 83 footnote 26.

 

[andrien]

Not all regularization schemes are very sensible,but it is expected that it should be zero.

 

[center o bass]

Not all regularization schemes are very sensible,but it is expected that it should be zero.

Alright, but would you say that my argument above holds? I.e that the action of the renormalized theory must have the same properties as the bare theory and thus a symmetry of the Lagrangian in the bare theory is a symmetry in the renormalized Lagrangian?

 

[Avodyne]

Yes, it is easy to show (via formal manipulations of the path integral) that the effective action (which is finite) has the same symmetries as the classical action, provided that the functional measure in the path integral is invariant (this is not true when the symmetry is anomalous).