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Dimensional regularization vs momentum cutoff

  1. May 26, 2012 #1
    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. jcsd
  3. May 27, 2012 #2
    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.
  4. May 27, 2012 #3
    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.
  5. May 30, 2012 #4
    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?
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