On-shell renormalization (Srednicki ch31)

In summary, the conversation discusses the imposition of the usual on-shell renormalization scheme in \phi^4 theory. The self energy (equation 31.5) is deduced by Srednicki and is represented by \Pi(k^2). Imposing \Pi(-m^2)=0 and \Pi'(-m^2)=0 leads to the conclusion that \Pi(k^2)=0 at one loop in the OS scheme. There were no mistakes made in the process.
  • #1
LAHLH
409
1
Hi,

In [tex] \phi^4[/tex] theory, Srednicki deduces the self energy (equation 31.5), as:
[tex] \Pi(k^2)=\frac{\lambda}{2(4\pi)^2)} \left[ \frac{2}{\epsilon}+1+ln\left(\frac{\mu}{m^2}\right)\right]m^2-Ak^2-Bm^2 [/tex]

I understand how he gets to this just fine, but now I'm trying to impose the usual on-shel (OS) renormalization scheme, which I believe means [tex] \Pi(-m^2)=0 [/tex] and [tex] \Pi'(-m^2)=0 [/tex] (Which are due to needing the exact prop to have poles and residues in correspondance with the Lehman-Callen form of it).

I'm having some issues doing this and I really don't understand why. Firstly I set (to remove the mu depedence and the 1/epsilon infinity) :

[tex] B=\frac{\lambda}{16\pi^2}\left[ \frac{1}{\epsilon}+\frac{1}{2}+\kappa_B+ln\left(\frac{\mu}{m^2}\right)\right] [/tex].

Thus

[tex] \Pi(k^2)=-\frac{\lambda}{16\pi^2}\kappa_Bm^2-Ak^2 [/tex]

But now, imposing [tex] \Pi(-m^2)=0 [/tex] leads to:

[tex] A=\frac{\lambda}{16\pi^2}\kappa_B [/tex]

So,

[tex] \Pi(k^2)=-\frac{\lambda}{16\pi^2}\kappa_B(m^2+k^2) [/tex]

Finally, imposing [tex] \Pi'(-m^2)=0 [/tex] leads to:

[tex] -\frac{\lambda}{16\pi^2}\kappa_B=0 [/tex]

and thus [tex] \Pi(k^2)=0 [/tex]

I really don't know what I'm doing that could be possibly wrong. Appreciate any help whatsoever, thanks.
 
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  • #2
There's nothing wrong. Pi(k^2)=0 at one loop in the OS scheme.
 
  • #3
Avodyne said:
There's nothing wrong. Pi(k^2)=0 at one loop in the OS scheme.

Oh, I see, I thought that must have meant I had made a mistake. Thanks very much Avodyne
 

1. What is on-shell renormalization?

On-shell renormalization is a process in quantum field theory that involves removing divergences from a theory by redefining the parameters and fields of the theory in terms of their physical, or on-shell, values. This is done by imposing certain conditions, such as the renormalization conditions, on the theory.

2. Why is on-shell renormalization important?

On-shell renormalization is important because it allows us to make meaningful predictions and calculations in quantum field theory. Without renormalization, the theory would be plagued by divergences, making it impossible to extract any meaningful physical information.

3. What is the difference between on-shell renormalization and off-shell renormalization?

The main difference between on-shell and off-shell renormalization lies in the conditions that are imposed on the theory. On-shell renormalization involves imposing conditions on the physical values of the parameters and fields, while off-shell renormalization involves imposing conditions on arbitrary values of these quantities.

4. How is on-shell renormalization carried out?

The process of on-shell renormalization involves calculating the divergent terms in the perturbative expansion of a theory and then redefining the parameters and fields in terms of their physical values. This is typically done using the renormalization group equations and the renormalization conditions, which are determined by physical observables.

5. What are the challenges of on-shell renormalization?

One of the main challenges of on-shell renormalization is dealing with the large number of divergences that arise in quantum field theory calculations. These divergences need to be properly identified and removed in order to obtain meaningful results. Another challenge is choosing appropriate renormalization conditions, which can vary depending on the specific theory being studied.

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