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Dimension of interaction in a QFT theory

  1. May 27, 2015 #1
    The problem statement.

    When an exercises say " the interaction in a QFT has dimensions Δ" , what does it mean?, it means the field or the Lagrangian has this mass dimension?

    In this exercise I'm trying to find the classical beta function (β-function) for the assciated couling.
  2. jcsd
  3. May 27, 2015 #2


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    It sounds like they mean the power-counting dimension of the operator corresponding to the interaction in question, where the dimension of the field is determined from the kinetic term. So, in 4d for a scalar ##\phi##, the dimension of ##\phi^4## would be 4, while the dimension of ##\phi^2 (\partial \phi)^2## would be 6, etc.
  4. Jun 2, 2015 #3
    Oh I see , I think i got it , but still I dont see how to compute the beta function to get to the form
    I really appreciate the help :)

    The full exercise is :

    Show that if the interaction in a QFT is has dimension Δ then there is a classical β function for the associated coupling given by ,
    Discuss what happens to interactions where Δ > d or Δ < d.

    Any hits or insight will do , thank you very much :D
  5. Jun 2, 2015 #4


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    I'll outline this, because once the concepts are put together, there really isn't a lot of work left for you to do.I will consider the case of a single field, but you should generalize the argument for multiple fields + allow for derivatives in the coupling.

    Consider an interaction of the form

    $$S_\text{int} = \int d^dx g_0 \Phi^r$$

    and do the rescaling ##x\rightarrow \lambda x##. Under this transformation, the fields transform according to their scaling dimension, ##\delta##, namely ##\Phi\rightarrow \lambda^{-\delta}\Phi##. In fact, we will find that

    $$S_\text{int} \rightarrow \int d^dx g_0 \lambda^\kappa \Phi^r,$$

    for some exponent ##\kappa## that you should relate to ##d## and ##\Delta##. The theory is obviously only invariant under the scale transformation if the exponent ##\kappa## is zero for all of the terms in the action, but we can still define an effective coupling ##g(\lambda) = g_0 \lambda^\kappa##. The corresponding beta function, defined as

    $$\beta_g(\lambda) = \lambda \frac{dg(\lambda)}{d\lambda},$$

    can then be computed.
  6. Jun 2, 2015 #5
    Ahaa I solved it . It was a lot easier that I thought !

    thank for your time :)
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