# Dimensional regularization vs momentum cutoff

1. May 26, 2012

### nidnus

Hi everyone,

I have a 2D sigma model with supersymmetry on the worldsheet. It has both cubic and quartic interactions and I'm interested in the one loop correction to the worldsheet masses. When I calculate this with dimensional regularization I find that everything is zero as expected. In momentum cutoff $\Lambda$ however I find terms corresponding to infinite mass renormalization. That is

Dim reg:

$$\delta m^2 = 0$$

while in

momentum cutoff:

$$\delta m^2 = \alpha \Lambda^2 + \beta$$

where $\alpha,\beta$ are constants like $1/\pi$ etc.

My question is, what is the meaning of the terms popping up in the momentum cutoff? They shouldn't be physical but I feel I lack a proper argument for that statement.

Thanks a lot

2. May 27, 2012

### chrispb

People say that such terms are "pure counterterm", meaning adding a local counterterm of the form delta m^2*phi^2 term in your action is sufficient to cancel such divergences; there's no leftover bit that depends on p or mu, the renormalization scale.

3. May 27, 2012

### nidnus

Hi Chris,

Thanks for your answer. Indeed, the terms I find correspond to mass renormalization. However, I would expect $\delta m^2$ to be zero since its a susy model. What is more, the $\delta m^2$ in hard cutoff is different for boson and fermions implying that susy is broken which should not be the case. I want an argument for how to either get rid of the mass renormalization terms or why I can say they don't matter.

Thanks, Nid.

4. May 30, 2012

### chrispb

I would also expect it to be zero; I would expect that you'd find wavefunction renormalization that respects supersymmetry, provided you have no explicit SUSY-breaking terms in your Lagrangian in the first place. Have you verified your calculation?