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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

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

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  • $\begingroup$ $Z^{-n/2}\partial_M <\phi(x_1)...\phi(x_n)> \neq \partial_M (Z^{-n/2} <\phi(x_1)...\phi(x_n)>)$. 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
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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.

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