- #1
nikol
- 13
- 0
I would like to ask if anyone can give me a hand with the understanding of the counterterms. I am reading by myself Chapter 10 of Peskin & Shroeder and got stuck in the middle of their example of how to renormalize [itex]\phi^{4}[/itex] theory. What is puzzling me is how to obtain from the new Lagrangian (equation 10.18):
[itex]\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}_{r}-\frac{1}{2}m^{2}\phi^{2}_{r}-\frac{\lambda}{4!}\phi^{4}_{r}+\frac{1}{2}\delta_{Z}(\partial_{\mu}\phi)^{2}_{r}-\frac{1}{2}\delta_{m}\phi^{2}_{r}-\frac{\delta_{\lambda}}{4!}\phi^{4}_{r}[/itex]
the first new vertex (shown on fig.10.3, the single line with the[itex] \otimes[/itex]), equal to:
[itex]i(p^{2}\delta_{Z}-\delta_{m})[/itex]
If you ask me, looking at terms (#4 + #5) in the above Lagrangian, this is a kinetic energy term and I imagine should bring a propagator into the Feynman rules (not a vertex!), that should look something like this:
[itex]\frac{i}{\delta_{Z}p^{2}-\delta_{m}}[/itex])
Obviously I am not right and I would like to clarify what I am missing. Thanks in advance to anyone that tries to help.
[itex]\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}_{r}-\frac{1}{2}m^{2}\phi^{2}_{r}-\frac{\lambda}{4!}\phi^{4}_{r}+\frac{1}{2}\delta_{Z}(\partial_{\mu}\phi)^{2}_{r}-\frac{1}{2}\delta_{m}\phi^{2}_{r}-\frac{\delta_{\lambda}}{4!}\phi^{4}_{r}[/itex]
the first new vertex (shown on fig.10.3, the single line with the[itex] \otimes[/itex]), equal to:
[itex]i(p^{2}\delta_{Z}-\delta_{m})[/itex]
If you ask me, looking at terms (#4 + #5) in the above Lagrangian, this is a kinetic energy term and I imagine should bring a propagator into the Feynman rules (not a vertex!), that should look something like this:
[itex]\frac{i}{\delta_{Z}p^{2}-\delta_{m}}[/itex])
Obviously I am not right and I would like to clarify what I am missing. Thanks in advance to anyone that tries to help.