Perturbative Renormalization in Phi 4 Theory

In summary, the conversation discusses the renormalization of the Lagrangian at 1-loop order and finding the amplitude of a specific term to order λ. The use of dimensional regularization and the counter-term at this order are also mentioned, along with a mistake in the counter-term. A related thread on the same issue is also referenced.
  • #1
TL;DR Summary
I seem to have a misunderstanding as to how counterterms actually get rid of the divergences in amplitudes.
For example, after the Lagrangian is renormalized at 1-loop order, it is of the form
$$\mathcal{L}=\frac{1}{2}\partial^{\mu}\Phi\partial_{\mu}\Phi-\frac{1}{2}m^2\Phi^2-\frac{\lambda\Phi^4}{4!}-\frac{1}{2}\delta_m^2\Phi^2-\frac{\delta_{\lambda}\Phi^4}{4!}$$.

So if I were to attempt to find the amplitude of $$\bra{p'}(\Phi(x_1)\Phi(x_2))\ket{p}$$ to order $\lambda$, I would get

$$\bra{p'}\Phi(x_1)\Phi(x_2)\ket{p}=\bra{\Omega}a_{p'}\Phi(x)\Phi(x) a_{p}^{\dagger}e^{i\int d^4y (\lambda+\delta_{\lambda})}\ket{\Omega}\\=(e^{i(p'\cdot x_1-p\cdot x_2)}+e^{i(p'\cdot x_2-p\cdot x_1)}-i(\lambda+\delta_{\lambda})\int e^{i(p'-p)\cdot x}\int\frac{d^4k}{(2\pi)^4}\frac{ie^{-ik\cdot (x_1-x)}}{k^2-m^2+i\epsilon}\int\frac{d^4q}{(2\pi)^4}\frac{ie^{-iq \cdot (x_2-x)}}{q^2-m^2+i\epsilon}d^4x)$$
From here, I would dimensionally regularize
and use $$\delta_{\lambda}=\frac{3\lambda^2}{32\pi^2}(\frac{2}{\epsilon}-\gamma+\log{4\pi})$$, which is at order $\lamba^2$, so it dosent cancel out the divergence of this integral. What is it about renormalization that I'm misunderstanding?
 
Physics news on Phys.org

Suggested for: Perturbative Renormalization in Phi 4 Theory

Replies
5
Views
720
Replies
10
Views
885
Replies
6
Views
763
Replies
1
Views
696
Replies
0
Views
534
Replies
1
Views
2K
Replies
1
Views
1K
Replies
2
Views
748
Replies
4
Views
2K
Replies
6
Views
798
Back
Top