Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The ln(m^2) term in dimensional regularization

  1. Jan 11, 2012 #1
    Using dimensional regularization we frequently end up with a term 2/(4-d)+ ln(somthing with mass dimension), c.f. Peskin page 250~251, and Peskin said the scale of the logarithm is hidden in the 2/(4-d) term. How is this so? No matter how hard I try to look at 2/(4-d), I see a purely dimensionless number.
     
  2. jcsd
  3. Jan 11, 2012 #2

    Avodyne

    User Avatar
    Science Advisor

    Try reading a better book such as Srednicki.
     
  4. Jan 12, 2012 #3

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    I couldn't agree more! Peskin-Schroeder is sloppy at too many places. Particularly to have logarithms with dimensionful arguments in the section about renormalization and the renormalization group is quite ironic. The whole physicss point of renormalization is the introduction of an energy-momentum scale.

    Admittedly this is not so easy to comprehend when one starts with dimensional regularization right away. This regularization is not more than a quite clever mathematical trick to make sense of divergent integrals in intermediate steps of the perturbative evaluation of vertex/Green's functions. Here the scale only enters quite indirectly by assuming that all coupling paramaters in the Lagrangian keep their energy-momentum dimension as in four-dimensional spacetime. In order to keep the action dimensionless (since [itex]\hbar[/itex] is set to 1, actions are measured as dimensionless quantities), one has to put a energy-momentum scale [itex]\mu[/itex] in front of the interaction terms in the Lagrangian. If you do this everywhere correctly all your logarithms appear with strictly dimensionless arguments as it MUST be!

    The physics of this scale is somewhat hidden. The physically most appropriate way to renormalize is the BPHZ formalism, where you keep the model at four dimensions. You simply read the Feynman rules as rules for the integrands and not as rules for the whole integral, which often are divergent and are thus not properly defined. Then you provide renormalization conditions for your vertex functions (depicted by one-particle irreducible amputated diagrams). As can be shown by formal perturbation theory, in the case of a renormalizable theory you need a finite set of renormalization conditions. To fulfill these conditions, you make subtractions from the integrand as explained by Zimmermann's "forest formula". The physical meaning is to subtract appropriate counter terms for the wave-function normalization constants, the masses, and couplings of the model allready in the Lagrangian. Then everything is expressed in a set of finite parameters since after these subtractions are done all the integrals become finite and your vertex function are well defined.

    Now, you are not totally free to choose your renormalization conditions. The reason is that you must make sure that your Hamiltonian must remain Hermitean and thus the counter terms to the parameters of the theory must be real (I suppose you have a model with only real paramaters for sake of simplicit). E.g., if you calculate the self-energy of a particle, this usually leads to the necessity to subtract a wave-function normalization and a mass counterterm, and these should be real. Thus you are not allowed to subtract the self-energy at a point, where it becomes complex. E.g., take the photon self-energy in QED. This quantity becomes complex for [itex]p^2=s>4m^2[/itex]. thus, you have to subtract at a four momentum with [itex]s<4 m^2[/itex]. As long as the electron mass is finite, you can subtract at [itex]s=0[/itex] (which is the original choice in the BPHZ formalism). For the electron self-energy the same holds true since, there is a threshold at [itex]s=m^2[/itex]. In this choice of the renormalization, your scale is the electron mass, and all your results for the renormalized vertex functions have dimensionless arguments in the logarithms, because there is the electron mass which cancels the dimensions of the momentum dependent parts of the logarithm's argument.

    This changes when you take the limit of vanishing electron mass. Then you are not allowed to subtract the vertex functions at vanishing external momenta since the thresholds of the loop integrals is already at this point. Then you must subtract at a point where all internal momenta are space like, and thus you must specify a energy-momentum scale [itex]\Lambda[/itex] in the spacelike regime of the vertex functions. This is then your renormalization scale, which you must introduce precisely to make your renormalized vertex functions well defined, and this scale then leads to proper dimensionless arguments for the logarithms in your vertex functions.

    You find some more details about this approach in my lecture notes on QFT at my home page:

    http://fias.uni-frankfurt.de/~hees/publ/lect.pdf
     
  5. Jan 13, 2012 #4
    Thanks, I get the part for dimensional regularization, and I'll save the rest when I get there.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The ln(m^2) term in dimensional regularization
Loading...