11
$\begingroup$

I am reading Peskin and Schroeder, chapter ten, and my Lagrangian is $$ \mathcal{L}=\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2-\frac{\lambda}{4!}z^2\phi^4+\frac{1}{2}\delta_Z(\partial_\mu\phi_r)^2-\frac{1}{2}\delta_m\phi^2_r-\frac{\delta_\lambda}{4!}z^2\phi^4. $$

It was my understanding that terms of power 2 in the fields always give Feynman rules that are propagators. However, it appears that the counterterms with power 2 in the fields give a Feynman rule looking like $$i(p^2\delta_z-\delta_m),$$ instead of something with a denominator that would be more familiar. Like $\frac{i}{p^2-m^2+i\epsilon}$, from the first terms. Why is this the case? Is the idea that any term with power 2 in the fields gives a propagator wrong?

$\endgroup$
7
+50
$\begingroup$

Consider $\phi^4$ theory: $$ \mathcal L=\frac12 Z_1(\partial\phi)^2-\frac12 Z_m m^2\phi^2-\frac{1}{4!}\lambda_0\phi^4 $$

There are two approaches to perturbation theory:

First

The propagator is given by $$ \Delta=\frac{1}{Z_1p^2-Z_m m^2} $$ and there is one type of vertex, with value $$ -i\lambda_0 $$

Second

The propagator is given by $$ \Delta=\frac{1}{p^2-m^2} $$ and there are two types of vertices, with value $$ -i((Z_1-1)p^2-(Z_m-1)m^2),\qquad -i\lambda_0 $$

The two approaches are completely equivalent, and they give rise to the exact same expression for a given scattering process.

Note that the coefficients $Z_1,Z_m$ depend on the expansion parameter $\lambda$. This means that the first approach is more cumbersome because it is in general not clear which diagrams contribute to a given order in perturbation theory, inasmuch as both the vertices and the propagators contain powers of $\lambda$. On the other hand, the second approach leads to more diagrams (because there is one more vertex) but it is more convenient (because the propagators are independent of $\lambda$).

$\endgroup$
4
$\begingroup$

I want to add to AccidentalFourierTransform's answer:

Assuming the $\delta$'s are small, then we can expand the renormalized term in powers of $(\delta_2p^2-\delta_m)$: $$\frac{i}{Z_2p^2-Z_mm^2}=\frac{i}{p^2-m^2 }\left(1+\frac{\delta_2p^2-\delta_m}{p^2-m^2}\right)^{-1}=\frac{i}{p^2-m^2 }\left(1-\frac{\delta_2p^2-\delta_m}{p^2-m^2}+\dots\right)=\frac{i}{p^2-m^2 }+\frac{i}{p^2-m^2 }\left(i\delta_2p^2-i\delta_m\right)\frac{i}{p^2-m^2 }+\dots$$

Which is the sum of all the diagrams consisting of the original term + the counter-term, so by identifying $\frac{i}{p^2-m^2}$ as the momentum term, we identify $i(\delta_2p^2-\delta_m)$ as the momentum counter-term.

$\endgroup$
  • $\begingroup$ but some of the $\delta$ terms are also divergent, so how do we make sense out of the expansion? $\endgroup$ – M. Zeng Jan 1 '18 at 2:31
  • $\begingroup$ @M.Zeng The assumption that the $\delta$'s are small is used many times in RG computations. e.g: To one loop order $\frac{\partial\log{(Z)}}{\partial \mu}=\frac{\partial\log{(1+\delta)}}{\partial \mu} \approx \frac{\partial\delta}{\partial \mu}$. However, to be frank, I cannot argue to the validity of this assumption. $\endgroup$ – EZLearner Feb 5 '18 at 20:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.