New Feynman Rules in QED from Counterterms

[blackie1008]

Hi,

I have been looking at the renormalisation of QED and been using Peskin & Schroeder. I understand (I think) what is going on, but I am slightly confused over 2 issues:

1. In the new feynman rules from the counterterms, the feynman diagrams all have a small circle with a cross in them...what do these represent? (P&S p332, fig 10.4)

2. why do we need the renormalization conditions? (P&S p331-331 eq 10.40)

Help would be much appreciated!

Thanks

 

[DarMM]

For question 1., the circles represent the counter-terms themselves.

For question 2., the renormalization conditions are required to fix the counterterms. For example the bare mass in the Lagrangian is $$m_{0}$$. We split this into the physical mass and the counterterm, $$m_{0} = m + \delta m$$. However you need the figure out what $$\delta m$$ should be so that the physical mass is $$m$$. The physical mass being $$m$$ means that $$\Sigma(m^{2}) = 0$$. This provides you with the equation you need to obtain $$\delta m$$.

 

[blackie1008]

Thanks for clearing up question 2, but as for question 1, by "representing the counterterms themselves" do you mean that it represents all the contributions that are taken into account?

 

[DarMM]

Thanks for clearing up question 2, but as for question 1, by "representing the counterterms themselves" do you mean that it represents all the contributions that are taken into account?

The black circle with for the interaction vertex represents $$\delta \lambda$$ to all orders. At a given order of perturbation theory it can only stand for $$\delta \lambda$$ to that order or less. For instance let's say you were computing at third order, you could have a diagram with one normal interaction vertex and one counter-term interaction vertex, the counter-term vertex will contain the counter-term to second order, so you will get $$\lambda$$ from the usual vertex and $$\lambda^{2}$$ from the counter-term vertex, giving you a third order contribution.

 

[blackie1008]

Ahhh ok think I have got it now!

Thanks for the help, its much appreciated!