In the book An Introduction to Quantum Field Theory by Michael E. Peskin and Daniel V. Schroeder, they derive the Callan-Symanzik equation for the two-point function

\begin{equation} \left[M\frac{\partial}{\partial M}+\beta(\lambda)\frac{\partial}{\partial \lambda}+2\gamma(\lambda)\right]G^{(2)}(p)=0 \end{equation}

Then they want to change variable from $M$ to $p$ with space-like momentum $p=\sqrt{-p^2}$ and they find out (pg. 418) that

\begin{equation} \left[p\frac{\partial}{\partial p}-\beta(\lambda)\frac{\partial}{\partial \lambda}+2-2\gamma(\lambda)\right]G^{(2)}(p)=0 \end{equation}

I don't understand where the $+2$ came from. I did all the calculation, i found the overall minus sign, but i couldn't find out the origin of the factor $+2$.

Any Ideas?


I think it goes more or less as follows:

As written in the book the dependence of the two-point function on $p$ and $M$ reduces to $$G^{(2)}(p)=\frac{i}{p^2}g(-p^2/M^2).$$

Therefore one has $$p\frac{\partial G^{(2)}(p)}{\partial p}=-2G^{(2)}(p)-\frac{2i}{M^2}g'(-p^2/M^2)$$ Which follows just from the product rule of derivatives (where I have taken $i/p^2$ as first factor and $g(-p^2/M^2)$ as the second factor.) and the definition of $G^{(2)}(p)$ above. On the other hand $$M\frac{\partial G^{(2)}(p)}{\partial M}=\frac{2i}{M^2}g'(-p^2/M^2).$$ Now one can use the two equation to eliminate $g'$ and express derivatives w.r.t. $M$ in terms of derivatives w.r.t. $p$: $$M\frac{\partial G^{(2)}(p)}{\partial M}=-p\frac{\partial G^{(2)}(p)}{\partial p}-2G^{(2)}(p)$$ or $$M\frac{\partial }{\partial M}=-p\frac{\partial }{\partial p}-2$$


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