Is the Renormalization Group Equation for the n-Point Green's Function Correct?

  • Thread starter Thread starter CAF123
  • Start date Start date
  • Tags Tags
    Group
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
CAF123
Gold Member
Messages
2,918
Reaction score
87

Homework Statement


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 is $$\mu \frac{d}{d \mu} \tilde{\Gamma}(n) (g) = 0$$ where the coupling g is defined at mass scale ##\mu##.

Show that this is equivalent to $$ (\beta \frac{\partial}{\partial g} + n )\tilde{\Gamma}(n) = 0, $$ where ##\beta(g) = \mu \frac{d g}{d \mu}. ##The field ##\psi## has mass dimension one and the Green’s function is a homogeneous function of degree n in the field.

Homework Equations


[/B]
function of homogenous degree n is one in which the exponents of each term all add up to n.

Renormalisation of fields

The Attempt at a Solution


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 $$\frac{\partial \tilde \Gamma}{\partial Z_{\psi}} = n (Z_{\psi})^{n-1}\tilde \Gamma$$ but this does not seem to give me correct result.

Did I assume something incorrect? Thanks!
 
Physics news on Phys.org
CAF123 said:

Homework Statement


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 is $$\mu \frac{d}{d \mu} \tilde{\Gamma}(n) (g) = 0$$ where the coupling g is defined at mass scale ##\mu##.

Show that this is equivalent to $$ (\beta \frac{\partial}{\partial g} + n )\tilde{\Gamma}(n) = 0, $$

You are missing a ##\gamma## in that equation, next to the factor of n , did you realize this?