Question about Renormalization Group Equation

[Phileas.Fogg]

Hello,
in our lecture, we computed $$\beta^{\overline{MS}} , \gamma_m^{\overline{MS}}$$ for $$- \frac{\lambda}{4!} \phi^4$$ Theory.

These are:

$$\beta^{\overline{MS}} (\lambda_{\overline{MS}}) = b_1 \lambda_{\overline{MS}}^2 + O(\lambda_{\overline{MS}}^3)$$

$$\gamma_m^{\overline{MS}} (\lambda_{\overline{MS}}) = d_1 \lambda_{\overline{MS}} + O(\lambda_{\overline{MS}}^2)$$

Is there any change, if one goes to a $$- \frac{\lambda}{4!} \phi^2$$ Theory?

I think that the Lagrangian changes to:

$$L = \frac{1}{2} Z (\partial_{\mu} \phi_R)^2 - \frac{1}{2} m_0^2 Z \phi_R^2 - \frac{\lambda_0}{4!} Z \phi_R^2$$

But that has no effect on computing the beta- and gamma-function. Am I right?

Regards,
Mr. Fogg

 

[xepma]

A quadratic term like that is simply a mass term, with the mass played by the coefficient in front of the quadratic term. In the absence of interactions you would be dealing with a free theory, so all renormalization equations would turn out to be quite trivial.

The way to see it is to notice that the real mass term in the Lagrangian you wrote down is:

$$M^2\phi^2 = (\frac{1}{2}m_0^2Z-\frac{\lambda_0}{4!}Z)\phi^2$$

The coefficient for the $$\phi^4$$ is simply zero. But in the renormalization equations you wrote down there is no reference to the mass term, only to the interaction coupling constant. But that coefficient is zero...

Renormalization really only plays a role when you are dealing with interactions.

 

[Phileas.Fogg]

Hello,
thanks for your answer.

In the absence of interactions you would be dealing with a free theory, so all renormalization equations would turn out to be quite trivial.

So I compute the beta- and gamma-function in the same way, we did for phi^4 - Theory?

And the result is exactly the same like for phi^4-Theory?

Regards,
Mr. Fogg