# On-shell renormalization (Srednicki ch31)

1. Mar 12, 2010

### LAHLH

Hi,

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

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 $$\Pi(-m^2)=0$$ and $$\Pi'(-m^2)=0$$ (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) :

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

Thus

$$\Pi(k^2)=-\frac{\lambda}{16\pi^2}\kappa_Bm^2-Ak^2$$

But now, imposing $$\Pi(-m^2)=0$$ leads to:

$$A=\frac{\lambda}{16\pi^2}\kappa_B$$

So,

$$\Pi(k^2)=-\frac{\lambda}{16\pi^2}\kappa_B(m^2+k^2)$$

Finally, imposing $$\Pi'(-m^2)=0$$ leads to:

$$-\frac{\lambda}{16\pi^2}\kappa_B=0$$

and thus $$\Pi(k^2)=0$$

I really don't know what I'm doing that could be possibly wrong. Appreciate any help whatsoever, thanks.

2. Mar 13, 2010

### Avodyne

There's nothing wrong. Pi(k^2)=0 at one loop in the OS scheme.

3. Mar 14, 2010

### LAHLH

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