Consider the standard $\phi^4$ theory:

$\mathcal{L}=\frac{1}{2}(\partial_\mu \phi)^2 - \frac{1}{2} m^2_0 \phi^2 - \frac{\lambda_0}{4!}\phi^4$

We define the renormalized field $\phi = Z^{1/2} \phi_r$ and set some renormalization conditions at some scale $M$. The physical observables are now correlation functions of the form $<\phi_r(x_1)...\phi_r(x_n)>~=~Z^{-n/2}<\phi(x_1)...\phi(x_n)>$.

The physical intuition is that the scale $M$ doesn't impact the observables of the theory, we can define the theory at any other scale and obtain the same answer. So $\frac{d}{dM} <\phi_r(x_1)...\phi_r(x_n)>=0$.

Carrying out this computation and multiplying by $M$ we obtain the following:

$\big[ -\frac{n}{2} M \partial_M \log(Z) + M \partial_M + \beta(\lambda)\partial_{\lambda} \big] <\phi(x_1)...\phi(x_n)> = 0$

This is almost the Callan-Symanzik equation derived in Peskin (equation 12.41), but instead of the above differential operator acting on $<\phi(x_1)...\phi(x_n)>$ it acts on $<\phi_r(x_1)...\phi_r(x_n)>$.

Where is the mix up occurring here? These notes seem to agree with me http://www.damtp.cam.ac.uk/user/dbs26/AQFT/Wilsonchap.pdf

  • $\begingroup$ This question is still open. The below answer doesn't obtain the result from Peskin. $\endgroup$ – ranques Jun 21 '18 at 16:30

When you take the derivative of the equation $<\phi_r(x_1)...\phi_r(x_n)>=Z^{-n/2}<\phi(x_1)...\phi(x_n)>$ you will get what you wrote times $Z^{-n/2}$, because $\frac{\partial }{\partial M}Z^{-n/2}=-\frac{n}{2}Z^{-n/2}\frac{\partial }{\partial M}logZ$, and the other terms just carry this factor as well. Take this factor into the correlation function, and it becomes the renormalized one.

  • $\begingroup$ $Z^{-n/2}\partial_M <\phi(x_1)...\phi(x_n)> \neq \partial_M (Z^{-n/2} <\phi(x_1)...\phi(x_n)>)$. If you carry out the integration by parts you get $(M \partial_M + \beta(\lambda) \partial_{\lambda}) <\phi_r(x_1)...\phi_r(x_n)>=0$, which is yet another form of Callan-Symanzik that shows up in various sources, but not what Peskin has. $\endgroup$ – ranques Jun 19 '18 at 17:11

If you take a look at Eq. (12.38), you will find that $$\frac{d}{dM} \langle\phi_r(x_1)...\phi_r(x_n)\rangle=0$$ is not the case. Instead, on the right-hand side, there should be $$\frac{n\delta\eta G^{(n)}}{\delta M}.$$ Or, you can write $\frac{d}{dM} \langle\phi_0(x_1)...\phi_0(x_n)\rangle=0$ which is equivalent. In short, the renormalized $G$ is changing with $M$, as like there is a flow. Only at a fixed point is the theory not changing anymore.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.