When P&S explain Wilson's method of integrating out high momenta they start from the Euclidean path integral of the $\phi^4$-theory (eq. (12.3)) and then define in the following:

$$\hat{\phi}(k):= \begin{cases} \phi(k)\hspace{1cm}\text{ for } b\Lambda\leq \lvert k\rvert < \Lambda,\\ 0\hspace{1.55cm}\text{ otherwise,} \end{cases} $$ where $b<1$ is some fraction and furthermore introduce some new $\phi$ like $$\phi(k):= \begin{cases} 0\hspace{1cm}\text{ for } b\Lambda\leq \lvert k\rvert\\ \phi(k)\hspace{0.5cm}\text{for } \lvert k\rvert< b\Lambda. \end{cases} $$ Then the (Euclidean) Lagrangian of the $\phi^4$-theory is expanded in the newly defined quantities (which is equivalent with a replacement of the old $\phi$ by new $\phi + \hat{\phi}$):

$$\begin{align}Z =&\int \mathcal{D}\phi e^{-\int {\cal L}(\phi)}\cr &\times\int\mathcal{D}\hat{\phi}\exp\left(-\int d^dx\left[\frac{1}{2}(\partial\hat{\phi})^2 + \frac{m^2}{2}\hat{\phi}^2 + \lambda (\ldots \frac{1}{4}\phi^2\hat{\phi}^2 + \ldots)\right]\right).\end{align} \tag{12.5}$$

On top of the quartic terms in $Z$ also the term $\frac{1}{2}m^2\hat{\phi}^2$ will be treated as perturbation. Therefore as unperturbed Lagrangian ${\cal L}_0$ one has

$$\int {\cal L}_0 = \frac{1}{2}\int_{b\lambda\leq |k|<\Lambda} \frac{d^dk}{(2\pi)^d} \hat{\phi}^\ast(k)k^2 \hat{\phi}(k)\tag{12.7}.$$

In the next step P&S introduce a "problem-adapted" propagator (actually I don't know how to write a Wick contraction which is used by P&S, instead I use an "overline"):

$$\overline{\hat{\phi}(k)\hat{\phi}(p)}:= \frac{\int\mathcal{D}\hat{\phi} e^{-\int {\cal L}_0}\hat{\phi}(k)\hat{\phi}(p)}{\int\mathcal{D}\hat{\phi} e^{-\int {\cal L}_0}}=\frac{1}{k^2}(2\pi)^d\delta^{(d)}(k+p)\Theta(k) \tag{12.8}$$ where $$\Theta(k):= \begin{cases} 1\hspace{1cm}\text{ for } b\Lambda\leq \lvert k\rvert < \Lambda,\\ 0\hspace{1.55cm}\text{ otherwise.} \end{cases} \tag{12.9} $$

Here is my 1. question: How is the (very) RHS $\sim \Theta(k)$ of the propagator obtained?

P&S then proceed with the "integration out" of the term $\sim \phi^2\overline{\hat{\phi}\hat{\phi}}$, i.e. the evaluation of the term:

$$-\int d^dx\frac{\lambda}{4}\phi^2 \overline{\hat{\phi}\hat{\phi}}.\tag{12.10}$$

Here comes my 2. question: Apparently this term is "kind of" part of the exponentiated Lagrangian in the path integral (12.5), however, I have no clue how to get such a term for the following reasons: According to the definition of the propagator the contraction happens with Fourier components of $\hat{\phi}$ whereas in the exponentiated Lagrangian $\hat{\phi}:=\hat{\phi}(x)$. Actually, I have no problem of using Fourier components (instead of $\hat{\phi}(x)$) as long as the concerned term is quadratic since the Plancherel theorem can be used to transform the position space integral in a momentum space integral. However, in case of a quartic term like $\phi^2\overline{\hat{\phi}\hat{\phi}}$ I have no idea. In top of it how to get/generate the term

$$\int\mathcal{D}\hat{\phi}e^{-\int {\cal L}_0} \text{?}$$

which has to be in the denominator according to the definition of the propagator.

Actually my doubts already start at the generating functional of the $\phi^4$-theory when it is expressed in Fourier functional space (this is the P&S formula (12.1)). In order to make any sense of such an integral the exponentiated action should also be an integral over Fourier space with a Lagrangian composed of Fourier components of $\phi$. Again, as long as it is the free K-G-theory, I have no problem with it, however, how to change the $\phi^4$ term in the interacting theory into the $\phi^4$ term composed of Fourier components, I have no idea.


2 Answers 2


Here are a few hints...

  1. The factor $\Theta(k)$ is zero whenever $k$ is in the range where $\hat\phi(k)=0$. This enforces the fact that if $\hat\phi(k)=0$, then the integrand is zero, and therefore the integral must also be zero.

  2. The term being questioned comes from a term $\int d^dx\ \phi^2 \hat\phi^2$ in the integrand, which in turn comes from expanding the exponent to first order in $\lambda$. Since we're only doing the functional integral over $\hat\phi$, this is the same as having a term $\hat\phi^2(x)$, integrated over $x$ with an $x$-dependent coefficient. Thus we are essentially evaluating $\langle 0|\hat\phi^2(x)|0\rangle$ and then integrating the result over $x$ with an $x$-dependent coefficient. To define $\hat\phi(x)$, use linearity: $\hat\phi(x)$ is the Fourier transform of $\hat\phi(k)$, with the understanding that the latter is zero outside the allowed range of $k$. Thus we're evaluating a suitable integral of $\langle 0|\hat\phi(k)\hat\phi(p)|0\rangle$, which is zero when $|k|\neq |p|$.

The denominator $\int \mathcal{D}\hat\phi\ \exp(-\int L_0)$ is just a convenient normalization factor. The starting equation (12.3) in P&S might be misleading in this respect, because they "set $J=0$ for simplicity," which reduces the whole original partition function to a useless normalization factor. What they should have done is set $J(k)=0$ for $b\Lambda\leq |k|<\Lambda$ and retained $J(k)$ for $k<b\Lambda$. Then the partition function would be a functional of the remaining part of $J$, whereas $\int \mathcal{D}\hat\phi\ \exp(-\int L_0)$ is still just a normalization factor. Normalization factors may be chosen for convenience in intermediate steps, as long as the final result is properly normalized overall. (P&S don't keep track of the overall final normalization because it's beside the point of what they're trying to demonstrate.)

To write the $\phi^4(x)$ term in Fourier components, just write $\phi(x)$ in Fourier components and raise it to the fourth power. The integral over $x$ gives a delta-function of the sum of the momenta.

  • $\begingroup$ Thank you very much for the prompt answer. I still have to digest what you have explained. It is possible that I'll ask you a bit later on some details. But already a more obvious question: When you say that P&S don't keep track of the overall final normalization, does it mean that they just "forgot" to put the normalization factor in the formula (12.5) ? And my 1. question actually concerns the whole expression of the evaluated propagator, not only the $\Theta(k)$-dependence. $\endgroup$ Oct 28, 2019 at 14:47

Chiral Anomaly has already given a good answer. Here is another answer using slightly different words.

  1. The $\Theta$-function on the RHS of eq. (12.8) reflects where in momentum space the heavy modes $\hat{\phi}$ live.

  2. It is still possible to Fourier transform action terms from position space to momentum space. The $x$-integration becomes part of a Dirac delta distribution that imposes total momentum conservation. When performing the Fourier transformation, it is important to keep track of where the light modes $\phi$ and where the heavy modes $\hat{\phi}$ live in momentum space.

    Eq. (12.10) becomes one correction term in the effective action $\int {\cal L}_{\rm eff}(\phi)$.

    Concerning the normalization of the expectation value (12.8) and in other places, this is part of the systematics of perturbation theory in QFT. In practice, this is easiest to calculate by introducing sources $\hat{J}$ for the heavy modes, which at the end of the calculation is turned off. (One aspect is e.g. that it is enough to consider connected Feynman diagrams, cf. my Phys.SE answer here.)


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