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I have been attempting to go through Chapter 12 of Peskin & Schroeder, but I have been having a very tough time. In particular I have been having trouble following this chapter much beyond page 397 beginning with integrating out high momentum degrees of freedom.

In Section 12.1, they start with the following generating functional for the $\phi^4$ theory: $$Z = \int[\mathcal{D}\phi]_\Lambda \exp\Big(-\int d^dx \Big[\frac{1}{2}(\partial_\mu\phi)^2 + \frac{1}{2}m^2\phi^2 + \frac{\lambda}{4!}\phi^4\Big]\Big) \tag{12.3}.$$

Letting $b < 1$ and $\Lambda$ some UV cutoff, they define
$$\hat{\phi} = \begin{cases} \phi(k) , \quad b\Lambda \leq |k| < \Lambda\\ 0 \end{cases}$$ and redefine $\phi(k)$ as $$\phi = \begin{cases} \phi(k) , \quad |k| < b\Lambda\\ 0 \end{cases}$$ so that we replace the original $\phi$ by $\phi + \hat{\phi}$. This leads to the functional integral: $$Z = \int \mathcal{D}\phi e^{-\int \mathcal{L}(\phi)} \int \mathcal{D} \hat{\phi} \exp\Big(-\int d^dx\big[\frac{1}{2}(\partial_\mu \hat{\phi})^2 + \frac{1}{2} m^2\hat{\phi}^2 + \lambda(\frac{1}{6} \phi^3 \hat{\phi} + \frac{1}{4}\phi^2 \hat{\phi}^2 + \frac{1}{6}\phi \hat{\phi}^3 + \frac{1}{4!}\hat{\phi}^4)\big]\Big). \tag{12.5}$$

Their goal is to perform the integral over $\hat{\phi}$ to transform (12.5) into an expression of the form: $$Z = \int[\mathcal{D}\phi]_{b\Lambda} \exp\Big(-\int d^dx \mathcal{L}_\text{eff}\Big)$$ where $\mathcal{L}_\text{eff}$ contains only the Fourier components $\phi(k)$ with $|k| < b\Lambda$. To do this, they treat all quartic terms in (12.5) as perturbations including the mass term. They state the leading-order term in the Lagrangian that involves $\hat{\phi}$ is $$\int\mathcal{L} = \frac{1}{2} \int_{b\Lambda \leq |k| < \lambda} \frac{d^dk}{(2\pi)^d} \hat{\phi}^*(k)k^2\hat{\phi}(k). \tag{12.7}$$

This is where I begin to become confused. They state that (12.7) leads to a propagator, but how? Furthermore, by keeping the remaining $\hat{\phi}$ terms in (12.5) as perturbations, they claim that one can use this propagator along with Wick's theorem to determine their contributions. They somehow get some Feynman diagrams out of this approach. What exactly is going on here?

I'm hoping that I can follow the rest of the chapter once I understand what they are doing in these early steps. Also, are there another resources available to learn Wilsonian renormalization and the renormalization group from that covers similar material?

More specifically, these are my questions:

  1. How does (12.7) define a propagator?
  2. How does one construct Feynman rules from this procedure? I cannot see how they obtained the Feynman diagrams that they did (or why we would even want them in the first place).
  3. How does one use Wick's theorem to integrate out the high momentum modes?
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    $\begingroup$ Maybe this can be of use physics.stackexchange.com/q/695742 $\endgroup$
    – schris38
    Commented Apr 11, 2023 at 8:26
  • $\begingroup$ @schris38 Thank you for the link, I saw that post and though it helped I am still for the most part lost. Especially when it comes to involving diagrams. $\endgroup$
    – CBBAM
    Commented Apr 11, 2023 at 8:33
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    $\begingroup$ This link by Tong will be helpful (sect. 3.4). The important point is to perform path integrals only for the high energy modes. $\endgroup$
    – Siam
    Commented Apr 11, 2023 at 9:27

1 Answer 1

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The answer to this question can be found in the note on statistical field theory by Tong that I gave in my comment above, but in case you are interested, I will put it on this site as an answer as well. For details, please refer to the link for specific examples of calculations.

Let us start from eq.(12.5): $$Z = \int \mathcal{D}\phi e^{-\int \mathcal{L}(\phi)} \int \mathcal{D} \hat{\phi} e^{-\int \mathcal{L}_0(\hat\phi)-\int \mathcal{L}_{\mathrm{int}}(\phi,\hat\phi)}. $$ $$\mathcal{L}_0(\hat\phi)=\frac{1}{2}(\partial_\mu \hat{\phi})^2 + \frac{1}{2} m^2\hat{\phi}^2 $$ $$\mathcal{L}_{\mathrm{int}}(\phi,\hat\phi)=\lambda(\frac{1}{6} \phi^3 \hat{\phi} + \frac{1}{4}\phi^2 \hat{\phi}^2 + \frac{1}{6}\phi \hat{\phi}^3 + \frac{1}{4!}\hat{\phi}^4)\big] $$

Here, we want to know the effective action which is obtained by integrating out the high-energy mode $\hat\phi$. So, what we have to calculate is the following $W[\phi]$: $$e^{-\Gamma[\phi]}=\int \mathcal{D} \hat{\phi} e^{-\int \mathcal{L}_0(\hat\phi)-\int \mathcal{L}_{\mathrm{int}}(\phi,\hat\phi)}.$$

According to the textbook knowledge about 1PI effective action, $\Gamma[\phi]$ consists of one-particle irreducible Feynman diagrams. In this case, Feynman rules can be read by $$\int \mathcal{D} \hat{\phi} e^{-\int \mathcal{L}_0(\hat\phi)-\int \mathcal{L}_{\mathrm{int}}(\phi,\hat\phi)}.$$ Here we should note that the integral region about momentum is limited to $b\Lambda\leq |k|\leq\Lambda$.

Once we calculate the $\Gamma[\phi]$, the effective action for this system is given by $$Z = \int \mathcal{D}\phi e^{-S(\phi)} e^{-\Gamma[\phi]} ,$$ $$S_{\mathrm{eff}}[\phi]=S[\phi]+\Gamma[\phi]$$ This is what we want to know.

Summary:

  1. What we want to know is a low energy effective action which is obtained by integrating out the high energy modes on the partition function.
  2. Except for some specific models, we can’t do the integral about high-energy modes exactly, so we must rely on perturbative calculations: $$e^{-\Gamma[\phi]}=\int \mathcal{D} \hat{\phi} e^{-\int\mathcal{L}_0(\hat\phi)-\int \mathcal{L}_{\mathrm{int}}(\phi,\hat\phi)}.$$ Since the perturbative treatment is not different from the usual method of computing the partition function, we can follow the previous chapter on Peskin about propagators, Feynman rules, and Wick's theorem. In this case, it is better to take log on both sides of the above equation and then expand about log. (see the link above for details)
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  • $\begingroup$ Thank you! When you say we can follow the previous chapters of Peskin, do you know which chapters these are? In particular, about obtaining Feynman rules from a Lagrangian. $\endgroup$
    – CBBAM
    Commented Apr 12, 2023 at 1:52
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    $\begingroup$ For example, sect 9.2 of Peskin, or ch.9 and ch.10 of Srednicki. $\endgroup$
    – Siam
    Commented Apr 12, 2023 at 2:20
  • $\begingroup$ I will have a look, thanks again. $\endgroup$
    – CBBAM
    Commented Apr 12, 2023 at 3:11

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