20
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Hint: the propagator is an all order resummation of terms related to the terms you write. $\endgroup$
    – innisfree
    May 26, 2017 at 7:50
  • $\begingroup$ related: Counterterm Lagrangian and Renormalisation?. $\endgroup$ May 26, 2017 at 10:07
  • $\begingroup$ Try this related exercise: In free field theory, treat $m^2$ like an interaction term and determine the full propagator by summing over all Feynman diagrams. Do you get the same answer as the massive propagator? $\endgroup$
    – Prahar
    Jul 13, 2020 at 20:11

4 Answers 4

10
+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 $\lambda$).

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ but some of the $\delta$ terms are also divergent, so how do we make sense out of the expansion? $\endgroup$
    – M. Zeng
    Jan 1, 2018 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, 2018 at 20:40
4
$\begingroup$

This comes from treating $$\frac{1}{2}(\partial_\mu\phi_r)^2-\frac{1}{2}m^2\phi_r^2$$ as the free Lagrangian, and treating $$-\frac{\lambda}{4!}\phi_r^4+\frac{1}{2}\delta_Z(\partial_\mu\phi_r)^2-\frac{1}{2}\delta_m\phi^2_r-\frac{\delta_\lambda}{4!}\phi_r^4$$ as the perturbation. The propagators are then defined as the time-ordered two-point correlation functions of the "free" field theories, just as before. In this case, the propagator of the scalar field is $$D_F(x-y)\equiv \langle0|T\{\phi(x)\phi(y)\}|0\rangle=\int\frac{d^4p}{(2\pi)^4}\frac{i}{p^2-m^2+i\epsilon}e^{-ip\cdot(x-y)}.$$ The only difference is that we have changed what we are viewing as the "free" theory. This is basically just changing the center point of the perturbation expansion.

The vertices come from the perturbation terms. For example, to the lowest order (one vertex), we expand $$\exp\left[i\int\mathcal{L}\right]\approx\exp\left[i\int\mathcal{L}_0\right]\Big[1+i\int dx^4 \Big(-\frac{\lambda}{4!}\phi_r^4$$ $$+\frac{1}{2}\delta_Z(\partial_\mu\phi_r)^2-\frac{1}{2}\delta_m\phi^2_r-\frac{\delta_\lambda}{4!}\phi_r^4\Big) + ...\Big].$$ The terms $-(\lambda/4!)\phi_r^4$ and $-(\delta_\lambda/4!)\phi_r^4$ both yield vertices that connect four propagators just as in normal perturbation theory. The term $(\delta_Z/2)(\partial_u\phi_r)^2-(\delta_m/2)\phi_r^2$ yields a vertex that connects two propagators. The derivative terms makes it slightly more complicated to see what the vertex will look like, but by looking at the formula for $D_F(x-y)$ above, we can see that $\partial_\mu$ will just pull down an extra factor of $p_\mu$ from the connected propagators (both propagators will be constrained to have the same momentum by four-momentum conservation).

$\endgroup$
1
  • $\begingroup$ This is the most helpful answer. Could you elaborate on why $\partial_\mu \phi_r$ will be equal to $p_\mu \phi_r$? I'm having trouble seeing how the derivative pulls down this extra factor $p_\mu$. $\endgroup$
    – CBBAM
    Mar 22 at 4:23
0
$\begingroup$

Just to add to EZLearner's answer:

As M. Zeng has pointed out that the geometric series could not be rigorously used I would like to expand on the answer. The propagator of a free field has the form

$$\frac{i}{Zp^2-Zm_0^2}$$

We want the two-point function of this theory to give the following $$\frac{i}{p^2-m^2}$$ When a particle is on-shell. This is because we measure $m$ to be it's mass

If we set our sights on the free field propagator, we can produce the 2 point function through the sum of all 1 particle irreducible combinations. The sum of all 1 particle irreducible diagrams is given by $-i\Sigma(p)$. We can now write the two-point function as

$$\Gamma^{(2)} = \frac{i}{Zp^2-Zm_0^2} + \frac{i}{Zp^2-Zm_0^2}(-i\Sigma(p))\frac{i}{Zp^2-Zm_0^2}+\frac{i}{Zp^2-Zm_0^2}(-i\Sigma(p))\frac{i}{Zp^2-Zm_0^2}(-i\Sigma(p))\frac{i}{Zp^2-Zm_0^2} +\cdots$$

We can use the geometric sum here because the magnitude of $\Sigma(p)\frac{1}{Zp^2-Zm_0^2}$ depends on our choice for $Z$ so we can always make this smaller than 1. If we apply the geometric sum we get the following

$$\Gamma^{(2)} = \frac{i}{Zp^2-Zm_0^2} \frac{1}{1 - \Sigma(p)\frac{1}{Zp^2-Zm_0^2}}$$

$$\Gamma^{(2)} = \frac{i}{Zp^2-Zm_0^2 - \Sigma(p)}$$

If we apply the definitions $Z = 1 + \delta_\phi$, $Zm_0^2 = m^2 + \delta_m$

$$\Gamma^{(2)} = \frac{i}{p^2 - m^2 + \delta_\phi p^2-\delta_m - \Sigma(p)}$$

If we want this to reduce to $\frac{i}{p^2-m^2}$ when $p = m$ we can impose that $\delta_\phi m^2-\delta_m = \Sigma(m)$

We can also do the following

$$\Gamma^{(2)} = \frac{i}{p^2 - m^2} \frac{1}{1 - \frac{\Sigma(p) - (\delta_\phi p^2-\delta_m)}{p^2 - m^2}}$$

$$\Gamma^{(2)} = \frac{i}{p^2 - m^2} \frac{1}{1 - (-i\Sigma(p) + i( \delta_\phi p^2-\delta_m))\frac{i}{p^2 - m^2}}$$

$$\Gamma^{(2)} = \frac{i}{p^2 - m^2} + \frac{i}{p^2 - m^2}(-i\Sigma(p) + i( \delta_\phi p^2-\delta_m)) \frac{i}{p^2 - m^2} + \frac{i}{p^2 - m^2}(-i\Sigma(p) + i( \delta_\phi p^2-\delta_m)) \frac{i}{p^2 - m^2}(-i\Sigma(p) + i( \delta_\phi p^2-\delta_m)) \frac{i}{p^2 - m^2}$$

We can expand this out because $\delta_\phi p^2-\delta_m - \Sigma(p) $ is a finite quantity we can define to be less than $1$. We can see here that for a set of infinite 1 particle irreducible diagrams there is a counter term of the form $i( \delta_\phi p^2-\delta_m)$.

Hope this is helpful.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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