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The renormalization group equations for the $n$-point Green’s function $$\Gamma(n) = \langle \psi_{x_1} \dots \psi_{x_n}\rangle $$ in a four-dimensional massless field theory are $$\mu \frac{d}{d \mu} \tilde{\Gamma}(n) (g) = 0$$ where the coupling g is defined at mass scale $\mu$.

This is easily seen to be equivalent to $$ (\beta \frac{\partial}{\partial g} + n )\tilde{\Gamma}(n) = 0,\tag{1} $$ where $\beta(g) = \mu \frac{d g}{d \mu} $ and where the field $\psi$ has mass dimension one and the Green’s function is a homogeneous function of degree $n$ in the field.

This is a statement from my reading but I am just trying to verify equation $(1)$. In renormalisation, $\psi \rightarrow Z_{\psi} \psi$ and given that the Green's function is a homogenous function of degree $n$, in the renormalised Green's function, we now have a factor of $(Z_{\psi})^n$ in each term. So, $$\frac{d}{d \mu} \tilde \Gamma = \frac{\partial \tilde \Gamma}{\partial \mu} + \frac{\partial \tilde \Gamma}{\partial Z_{\psi}} \frac{\partial Z_{\psi}}{\partial \mu}$$ I would say that $$\frac{\partial \tilde \Gamma}{\partial Z_{\psi}} = n (Z_{\psi})^{-1}\tilde \Gamma$$ but this does not seem to match with equation (1).

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  • $\begingroup$ The reason for your confusion is that $\psi$ has mass dimension one only in free theory. In an interacting theory it will have anomalous dimension, and the anomalous dimension is determined precisely by the derivative of $Z_\psi$. $\endgroup$ – Peter Kravchuk Feb 24 '16 at 5:00
  • $\begingroup$ Ah I see, so only in an interacting theory are we renormalizing the parameters in the lagrangian. I am a little confused though - where does the factor of $n$ come from in the equation $(1)$ then? Thanks! $\endgroup$ – CAF Feb 24 '16 at 7:58
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I think I have managed to find the answer so I'll post what I have and any other comments are appreciated:

For the Green's function to be homogeneous of degree $n$ and the fields in four space time dimensions to be of mass dimension $1$ (which can be argued on dimensional arguments from the action) then this means $\tilde{{\Gamma}}^{(n)} = \mu^n f(g)$. So, $$\frac{d}{d\mu} \tilde{{\Gamma}}^{(n)} = \frac{\partial}{\partial \mu} \tilde{{\Gamma}}^{(n)} + \frac{\partial g}{\partial \mu} \frac{\partial}{\partial g} \tilde{{\Gamma}}^{(n)}$$ If $$\mu \frac{d}{d\mu} \tilde{{\Gamma}}^{(n)} = 0$$ then performing the derivatives above and inputting yields the result.

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