Relationship between Wilson's RG and the Callan-Symanzik Equation's normalization scale

Question

I have taken a Quantum Field Theory course recently in which we first derived the Callan-Symanzik equation and then discussed Wilson's Renormalization. However, I don't think I have a clear understanding of the procedures and how they relate to each other. For the sake of this question, let's restrict ourselves to massless theories. Let's also say we have the following normalization scheme:

$$ \Gamma^{(2)}(p^2)|_{p^2=0} = 0 $$ $$\Gamma^{(4)}(p_1,p_2,p_3,p_4)|_{p_i = \mu e_i} = \lambda_{(\mu)}$$

where ${e_i}$ are a set of four vectors such that $e_1+e_2+e_3+e_4=0$. The normalization scheme depends on the scale normalization $\mu$ and the associated equation for the correlation functions when the normalization scheme is changed is the Callan-Symanzik equation.

$$(\mu \frac{\partial}{\partial \mu} + \beta(\lambda) \frac{\partial}{\partial \lambda} + n \gamma (\lambda)) \langle \phi(x_1)...\phi(x_2)\rangle= 0$$

On the other hand, we have Wilson's approach to renormalization. We start of with an action and a cut-off, integrate out to get the transformed action. This changes the cut-off of the new action so we perform a variable change to return the cut-off to its original value. So if $L$ is the parameter of the RG flow, we have the following relationship between correlation functions:

$$ \langle\phi(x_1)...\phi(x_2)\rangle_{A'} = Z^{\frac{n}{2}}(L) \langle \phi(Lx_1)... \phi(Lx_n) \rangle_{A} $$

where we have $RG_L[A] = A'$. We obtain the same relationship in the case of a normalization scheme. I am pretty sure the parameter $L$ given by the ratio of cut-offs in Wilsons RG somehow is the same as choosing a normalizing a scheme $\mu$ and both these methods are equivalent. However, I am not sure how to rigorously justify it.

I've also heard that $\gamma(\lambda)$ is the anomalous dimension of Wilsons RG but don't know why. So any help is appreciated.

Answer 1

Clearly, Wilson's approach and perturbative RG based on dimensional regularization are NOT the same thing. In Wilson's approach, you use a UV cut off and your flow equations explicitly depend on it. In dimensional regularization, the coefficients in flow equations are all finite. So mathematically, they are indeed different.

Nevertheless, they both give the same scaling behavior and the set of critical exponents for the same theory. To see how it happens, it is useful to have a clear definition of a RG transformation. For more information, I refer you to Zinn-Justin's book on the subject, which has inspired what comes in the following:

Assume that you have a field theory formally defined by a set of parameters $\{ \alpha_i\}$ and its n-point functions $W^{(n)}(x_1,...,x_n)$. A renormalization group transformation with a parameter $b$ is defined by s mapping from $\{ \alpha_i \} \to \{ \alpha_i(b) \}$ if $$\alpha_i(1)=\alpha_i$$ and for every n-point function we have: $$W^{(n)}_{b=1}(bx_1,...,bx_n)-Z(b)^{-n/2}W^{(n)}_{b}(x_1,...,x_n)=R^{(n)}_b(x_1,...,x_n)$$ where $W_b^{(n)}$ means the correlation function is calculated using an action with parameters $\alpha_i(b)$ and $R^{(n)}_b$ is an arbitrary function with only one condition of decaying faster than any power of $b$ for $b\to \infty$. One can easily show that we will get similar transformation rules for the inverse of the correlators.

Now, different $Z(b)$ and $R^{(n)}$ result in different RG transformations:

For perturbative RG, we assume that $R^{(n)}=0$ and $W^{(n)}$ depends on $b$ only through $\{ \alpha_i(b) \}$ such that: $$W^{(n)}_b(x_1,...,x_n)=W^{(n)}(x_1,...,x_n;\{\alpha_i(b)\})$$

and that the map $\{ \alpha_i\} \to \{ \alpha_i(b) \}$ is invertible such that: $$Z(b)^{n/2}W^{(n)}(bx_1,...,bx_n;\{\alpha^0_i(b)\})=W^{(n)}(x_1,...,x_n;\{ \alpha_i\})$$ Going to Fourier space and defining $b \sim p_i/\mu$ and the putting the derivative of R.H.S with respect to $\mu$ equal to zero will give the CS equations.

In Wilson's approach, there is an extra parameter which is the UV cut off. So the RG transformation will be: $$W^{(n)}(bx_1,...,bx_n;\Lambda_0;\{\alpha^0_i\})=Z(b)^{-n/2}W^{(n)}(x_1,...,x_n;b\Lambda_0;\{ \alpha_i(b)\})$$ or: $$Z(b)^{n/2}W^{(n)}(bx_1,...,bx_n;\frac{\Lambda_0}{b};\{\alpha^0_i(b)\})=W^{(n)}(x_1,...,x_n;\Lambda_0;\{ \alpha_i\})$$ Take $b=\Lambda_0/\Lambda$ and the L.H.S becomes a function of $\Lambda$. Since the R.H.S doesn't depend on $\Lambda$ we put the derivative of R.H.S with respect to $\Lambda$ equal to zero and we will get Wilsonian RG.

These approaches can give totally different looking equations, what they share in common is that they have the same scaling behavior with the same relevant and irrelevant operators.

There are other RG schemes as well:

You can add a soft UV cut off and then study the flow of parameters in terms of the momentum of the soft cut off, this will result in Polchinski's functional RG approach.

In another approach, we assign an IR cut off to the theory and see how the parameters change as we change the IR cut off. This gives Wetterich's functional RG.

The point is that, any RG scheme should satisfy the RG transformation property mentioned at the beginning.

Answer 2

First, have a look at this wiki article. Then rename $M$ your $L$, $g$ your $A$ (and $\lambda$) and $G^{(n)}$ your correlation function, as customary. The shift $M\rightarrow M+\delta M$ induces the shift $$g\rightarrow g+\delta g $$ and the rescaling $$\phi \rightarrow(1+\delta\eta)\phi,$$ where we defined $\delta\eta = \frac{\delta\phi}{\phi}$. Now, differentiating in $M$ your relationship between correlation functions, the left-hand side is zero, since it does not depend on $M$ (we may take it as unrenormalized). Hence, taking into account that $Z^{\frac{1}{2}}=1-\delta\eta$, we get $$ 0=\frac{d}{dM}Z^{\frac{n}{2}}G^{(n)} = \frac{\partial G^{(n)}}{\partial M}+ \frac{\partial G^{(n)}}{\partial g}\frac{\partial g}{\partial M} - n \frac{\partial\eta}{\partial M} G^{(n)}.$$ Multiply by $M$ and you get your CS equation.

PS $\gamma=0$ is what you get by naive dimensional analysis, hence the "anomalous" $\gamma\neq 0$.