On the derivation of Eq. (12.5) in Peskin and Shroeder's QFT

Question

I stuck at the derivation of Eq. (12.5) in Peskin and Schroeder's QFT. The authors tried to from (12.3)

$$ Z = \int [\mathcal{D}\phi]_{\Lambda} \exp\left(-\int d^dx \left[\frac{1}{2} (\partial_{\mu} \phi)^2 + \frac{1}{2} m^2 \phi^2 + \frac{\lambda}{4!} \phi^4 \right]\right) \tag{12.3} $$ where $$ [\mathcal{D}\phi]_{\Lambda} =\prod_{|k|<\Lambda} d\phi(k) \tag{12.4} $$

The authors then introduced a new set of variables as $\phi = \phi + \hat{\phi}$, where new $\phi$ is the old one for $|k|< b \Lambda$, otherwise zero; and $\hat{\phi}$ equals the old one for $b\Lambda \leq |k| < \Lambda $, otherwise zero.

Eq. (12.3) was replaced as $$ Z = \int \mathcal{D}\phi \int \mathcal{D}\hat{\phi}\exp\left(-\int d^dx \left[\frac{1}{2} (\partial_{\mu} \phi +\partial_{\mu} \hat{\phi} )^2 + \frac{1}{2} m^2 (\phi + \hat{\phi})^2 + \frac{\lambda}{4!} (\phi+\hat{\phi})^4 \right]\right) \\ \cdots \tag{12.5} $$

I naively expect $\mathcal{D} \phi = \mathcal{D}\phi + \mathcal{D} \hat{\phi} $, as what happened in the Lagrangian. But, (12.5) looks like a multiplication, as $\int \mathcal{D} \phi \int \mathcal{D} \hat{\phi}$. My stupid question is, why here is the multiplication in the integral variables?

I may view as changing variables in multivariable calculus. Nevertheless, the difference in the old and new variables, as a Jacobi, may not be 1. I am unable to work out the Jacobi in the transformation.

Answer 1

It is given on Eq. (12.4). One has \begin{align} [\mathcal{D}\phi]_{\Lambda} &= \prod_{|k|<\Lambda} \mathrm{d}\phi(k), \\ &= \prod_{|k|<b\Lambda} \mathrm{d}\phi(k) \prod_{b\Lambda<|k|<\Lambda} \mathrm{d}\phi(k), \\ &= \prod_{|k|<b\Lambda} \mathrm{d}\phi(k) \prod_{b\Lambda<|k|<\Lambda} \mathrm{d}\hat{\phi}(k), \\ &= \mathcal{D}\phi \mathcal{D}\hat{\phi}. \end{align}

As mentioned in the comments, the matter is that it is an integral in many dimensions. What P&S are doing is pretty much making a change of variables such as $x \to x$, $y \to y$, $z \to w$, so the volume element transforms as $\mathrm{d}x\mathrm{d}y\mathrm{d}z=\mathrm{d}x\mathrm{d}y\mathrm{d}w$.