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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.

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    $\begingroup$ Imagine that $\phi$ is periodic and therefore expandable as a discrete Fourier series. The high frequency plus low frequency decomposition of $\phi$ is a sum but the way to integrate over all choices of $\phi$ is to integrate over all the Fourier coefficients which is a product. $\endgroup$
    – Connor Behan
    Jun 19, 2022 at 1:00
  • $\begingroup$ I am still stupid. Suppose I write $\phi = \sum_{k=0}^{100} c_k e^{ikx} + \sum_{k=101}^{200} c_k e^{ikx} $. $d\phi = \sum_{k=0}^{100} c_k de^{ikx} + \sum_{k=101}^{200} c_k de^{ikx} $, or what else :( $\endgroup$
    – RandomUser
    Jun 19, 2022 at 1:05
  • $\begingroup$ This is a functional integral and functions are infinite dimensional vectors. When $\textbf{r} = x\hat{x} + y\hat{y} + z\hat{z}$, the volume element is $dx dy dz$ not $dr = dx \hat{x} + dy \hat{y} + dz \hat{z}$. $\endgroup$
    – Connor Behan
    Jun 19, 2022 at 1:26
  • $\begingroup$ I miserably don't get how the infinite dimensional aspect related to the question here :(. In QM, one can get the path integral from $<x't'|x,t> = \int dx'' <x't'|x''><x''| x',t>$ and inserting more x''', x''''... etc. I may write $x'' = x'' + \hat{x}''$, that applies to each variable. I can so far only relate the question to how changing of variables should work, e.g., via Jacobi in multivariable calculus. E.g., from (x,y) to (u,v) $|\partial x/\partial u \partial y/ \partial u, \partial x / \partial v \partial y / \partial v|$, here $\phi$ to $\phi, \hat{\phi}$, and not sure how to proceed $\endgroup$
    – RandomUser
    Jun 19, 2022 at 1:55
  • $\begingroup$ Seems to me the question is about how $d\phi$ to $d\phi$, $d\hat{\phi}$ $\endgroup$
    – RandomUser
    Jun 19, 2022 at 2:00

1 Answer 1

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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 $\mathrm{d}x\mathrm{d}y\mathrm{d}z=\mathrm{d}x\mathrm{d}y\mathrm{d}w$.

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  • $\begingroup$ (1) (Edited notations, sorry for the confusing) Thanks for your answer. I think, the range under the $\Pi$, i.e., $|k|<\Lambda$, is not to 'count' $k$, such that each $k$ has a different value and one can split into two $\Pi$ as the second line. From the QM path integral, it looks like $\int dx′′dx′′′⋯<x,t|x′′t′′><x′′t′′|x′′′t′′′><x′′′t′′′|⋯|x′t′>=\int \mathcal{D} x...$ (2.5.49) in Sakurai's modern quantum mechanics or Peskin (9.12). Each $x$ runs over $−\infty$ to $+\infty$. In QFT case, each $|k|$ (like $x''$, $x'''$, in QM) runs over all value below $\Lambda$. $\endgroup$
    – RandomUser
    Jun 19, 2022 at 5:24
  • $\begingroup$ (2) (Edited some wording, sorry for the confusing) The changing variable should include something like a Jacobi, here for $\phi \rightarrow \phi, \hat{\phi}$, the Jacobi may not 1. I am unable to find a 2*2 determinant, since from 1 to 2 variables, to get the value 1 as the result of changing variable. $\endgroup$
    – RandomUser
    Jun 19, 2022 at 5:26
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    $\begingroup$ @RandomUser for each fixed $k$, the integral over $\mathrm{d}\phi(k)$ runs from $-\infty$ to $+\infty$. It is as if the $\phi(k)$ are playing the role of $x'$ in the Sakurai expression you wrote. As for the Jacobian, it is $1$ because P&S are just renaming some of the integration variables (those with $|k| > b \Lambda$) without really making any other transformations. $\endgroup$
    – Níckolas Alves
    Jun 19, 2022 at 5:32
  • $\begingroup$ The trick in here is that the integration variables are the values of $\phi(k)$ themselves for each fixed $k$, just as in the Sakurai path integral it were the values $x(t)$ themselves for each fixed $t$. In particular, notice that the analogy between Sakurai and P&S is among $t$ and $k$ and among $x(t)$ and $\phi(k)$. $\endgroup$
    – Níckolas Alves
    Jun 19, 2022 at 5:33
  • $\begingroup$ So far, the only way I can justify P&S is to regard $\mathcal{D} \phi \mathcal{D} \hat{\phi}$ as schematic, to integrate the original variable but in a truncated range... Maybe there is a better way to demonstrate here. $\endgroup$
    – RandomUser
    Jun 19, 2022 at 5:56

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