Integrating high momentum modes using Wilson's approach to renormalization

Question

In Section 12.1 of Peskin & Schroeder, they introduce the renormalization group for $\phi^4$ theory.

Let $b < 1$ and $\Lambda$ some UV cutoff. 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: $$Z = \int \mathcal{D}\phi e^{-\int \mathcal{L}(\phi)} \int \mathcal{D} \hat{\phi} \exp\Big(-\int d^dx\big[\frac{1}{2}(\partial_\mu \hat{\phi})^2 + \frac{1}{2} m^2\hat{\phi}^2 + \lambda(\frac{1}{6} \phi^3 \hat{\phi} + \frac{1}{4}\phi^2 \hat{\phi}^2 + \frac{1}{6}\phi \hat{\phi}^3 + \frac{1}{4!}\hat{\phi}^4)\big]\Big). \tag{12.5}$$

They introduce a term $\mathcal{L}_\text{eff}$ and note that it involves only the Fourier components $\phi(k)$ with $|k| < b\Lambda$. To carry out the integrals over the high momentum modes $\hat{\phi}$ they say on p. 396-397

...we will see below that the new terms in $\mathcal{L}_\text{eff}$ can be written in diagrammatic form. In this analysis, we treat the quartic terms in (12.5), all proportional to $\lambda$, as perturbations. Since we are mainly interested in the situation $m^2 \ll \Lambda^2$, we will also treat the mass term $\frac{1}{2}m^2\hat{\phi}^2$ as a perturbation. Then the leading-order term in the portion of the Lagrangian involving $\hat{\phi}$ is $$\int \mathcal{L}_0 = \frac{1}{2} \int_{b\Lambda \leq |k| < \Lambda} \frac{d^d k}{(2\pi)^d} \hat{\phi}^*(k) k^2 \hat{\phi}(k). \tag{12.7}$$

I am having a hard time following their work.

1. Where did (12.7) come from?

2. By quartic terms do they mean all terms of the form $\phi^n \hat{\phi}^m$ such that $n + m = 4$?

Answer 1

Assuming that you are comfortable with eq. (12.5) of Peskin-Schroeder, $$\begin{align}Z &= \ldots \\ &= \int \! \mathcal{D} \phi \, e^{-\int \mathcal{L}(\phi)} \! \! \int \! \mathcal{D}\hat{\phi} \\ &{} \exp \left(\!- \!\int \! \! d^d x \left[\frac{1}{2}\left(\partial_\mu \hat{\phi}\right)^2 \!+\frac{m^2}{2}\hat{\phi}^2 +\lambda \left( \frac{1}{6}\phi^3 \hat{\phi}+\frac{1}{4}\phi^2 \hat{\phi}^2 + \frac{1}{6} \phi \hat{\phi}^3 +\frac{1}{4!} \hat{\phi}^4 \right) \right] \right), \end{align} $$ they define $$\mathcal{L}_0= \frac{1}{2}\left(\partial_\mu \hat{\phi} \right)^2.$$ Using the Fourier decomposition $$\hat{\phi}(x) = \int \! \frac{d^d k}{(2\pi)^d} \, e^{i k x}\, \hat{\phi}(k)= \! \! \!\int \limits_{b \Lambda \le |k| \lt \Lambda} \! \! \frac{d^d k}{(2 \pi)^d} \, e^{ikx}\, \hat{\phi}(k), $$ where $\hat{\phi}^\ast \!(x)=\hat{\phi}(x)$ implies $\hat{\phi}^\ast\!(k) = \hat{\phi}(-k)$, they compute $$\begin{align}\int \! d^d x \, \mathcal{L}_0 &= \frac{1}{2} \int \! d^d x \, \left(\partial_\mu \hat{\phi}(x)\right)^2 \\ &= \frac{1}{2} \int \! d^d x \int\! \frac{d^d \ell}{(2 \pi)^d} \, i \ell_\mu \, e^{i \ell x} \, \hat{\phi}(\ell) \int \! \frac{d^d k}{(2 \pi)^d}\, i k_\mu \, e^{ikx} \, \hat{\phi}(k) \\ &= - \frac{1}{2} \int \! \frac{d^d \ell}{(2 \pi)^d}\int \frac{d^d k}{(2 \pi)^d} \, \ell \cdot k \, \, \hat{\phi}(\ell) \, \hat{\phi}(k) \, (2\pi)^d \delta^{(d)}(\ell +k) \\ &= \frac{1}{2} \int \frac{d^dk}{(2 \pi)^d}\, k^2 \, \hat{\phi}(-k) \, \hat{\phi}(k) \\&=\frac{1}{2} \! \! \int\limits_{b \Lambda \le |k| \lt \Lambda} \! \! \frac{d^d k}{(2 \pi)^d} \, \hat{\phi}^\ast \!(k) \, k^2 \, \hat{\phi}(k), \end{align} $$ being just the formula given in eq. (12.7) of the book.

Answer 2

1. Eq. (12.7) is the kinetic term for the high/heavy modes $\hat{\phi}$. Since the mass term is treated as a perturbation term, eq. (12.7) is the free part of the perturbative action for the path integral over high/heavy modes $\hat{\phi}$, i.e. the inner path integral in eq. (12.5). Eq. (12.7) therefore leads to the free high/heavy 2-point function/propagator (12.8).

2. Yes.