In chapter 12.2 p. 410 of Peskin and Schroeder the Callan-Symanzik equation is derived. I understand the relation between (connected) renormalized and non-renormalized Green's functions given by $$ G^{(n)}(x_1,\ldots,x_n)=Z^{-n/2}G_0^{(n)}(x_1,\ldots,x_n) $$ where $Z$ is given by the field strength renormalization $$ \phi=Z^{-1/2}\phi_0 $$ and then with a shift in our renormalization scale $M\rightarrow M+\delta M$ we have a rescaling of $\lambda\rightarrow \lambda + \delta\lambda$ and $\phi\rightarrow (1+\delta\eta)\phi$ where we notice that this definition gives $$ Z^{1/2}=(1-\delta\eta) $$ What I don't understand is how this gives a shift in $G^{(n)}$ which is given by the equation below 12.37 of the book, namely $$ G^{(n)}\rightarrow G^{(n)}(1+n\delta\eta) $$ Notice that the coefficient of $\delta\eta$ is $n$ (power of fields) and not $\eta$.
1 Answer
It is assumed that the bare quantities do not depend on the renormalization scale, i.e. if $M\rightarrow M + \delta M$, $\lambda \rightarrow \lambda + \delta\lambda$ and $\phi \rightarrow (1+\delta\eta)\phi$ they do not change. In particular the bare Greensfunction $G_0$ and $\phi_0$ do not change.
We observe that
$\phi \sim \frac{1}{Z^{1/2}}$
therefore if $\phi \rightarrow (1+\delta\eta)\phi$ then we get:
$$Z^{1/2} \rightarrow Z^{1/2} (1- \delta\eta) \quad\text{but also} \quad Z^{-1/2} \rightarrow Z^{-1/2} (1+ \delta\eta)$$
as mentioned in the post.
However, $G^{(n)}(x_1,\ldots,x_n) \sim (Z^{-1/2})^n$, remember the bare Greensfunction does not depend on the renormalization scaling. Therefore
$$G^{(n)}(x_1,\ldots,x_n) \rightarrow G^{(n)}(x_1,\ldots,x_n)(1+ \delta\eta)^n \approx G^{(n)}(x_1,\ldots,x_n)(1+n\delta\eta + O(\delta\eta^2) )$$
or shortly:
$G^{(n)}(x_1,\ldots,x_n) \rightarrow G^{(n)}(x_1,\ldots,x_n)(1+n\delta\eta)$
Actually $G^{(n)}$ is the $n$-point function, i.e. is the vacuum expectance value of $n$ field operators. If 1 field operator scales like $\phi \rightarrow (1+\delta\eta)\phi$, $n$ of those scale like
$\phi^n \rightarrow (1+n\delta\eta)\phi^n$.
