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Dimensional Regularisation - Contracting/Commuting Gamma Matrices

  1. Aug 26, 2011 #1

    After having solved some problems I encountered by using Google and often being linked to threads here, I finally decided to register, especially because I sometimes have problems for which I don't find solutions here and now want to ask them by myself :)

    Like the following: I am currently doing my first calculations on loop diagrams working with Peskin/Schroeder and I try to regularize the QED Electron-Vertex-function by dimensional regularisation (I want to solve Problem 7.2: Calculating Z1 and Z2 by dimensional regularisation and manual cutoff and seeing whether Z1=Z2 is fulfilled or not).

    I did something, and it seems to be wrong, and I don't understand why.

    Looking at the numerator:
    [itex]\bar{u}(p^\prime)({\not} k \gamma^\mu {\not} k^\prime + m^2 \gamma^\mu - 2 m (k+k^\prime)^\mu)u(p)[/itex]

    I used the usual 4-dimensional contraction and anticommutation rules + Dirac equation to get to the final formula where you have only [itex]\gamma^\mu[/itex] and [itex]\sigma^{\mu \nu}q_\nu[/itex] with certain coefficients. Afterwards I rewrote everything in d dimensions and used the general formulae for dimensional regularisation in order to calculate the integrals. I crosschecked my solution with one I found on the internet (http://www-personal.umich.edu/~jbourj/peskin/homework 6.pdf) and I basically get the same result, except for an additional 2! He gets it because he uses the d-dimensional contraction rules, ending up with an additional factor of [itex](2-\epsilon)^2[/itex], whose mixed term cancels a [itex]\frac{1}{2} \frac{1}{\epsilon}[/itex] to get a finite value of 2.

    Now I am a bit confused: Peskin Schroeder also gives the d-dimensional contraction formulae so I understand how I am supposed to get the same result. And it seems to be neccessary to contract in d dimensions if I want to regularize it in that way. But I don't understand why it is forbidden to start doing the Dirac algebra in 4 dimensions and then, if I end up with an divergent integral, doing dimensional regularisation, put everything in d dimensions and calculate it and put d=4 at the end. [Doing this, I won't get this peculiar 2 I talked about earlier]

    I mean: I see that the two ways give different results, so they can't be equivalent, but it confuses me because the step going into d dimensions is just a temporary aid to see where the divergence occurs and how to renormalize it, but at the end I will set d=4. So intuitively it should not matter at which point I start to generalize and at which point I will specify again, at least for the finite part, shouldn't it?

    What I am asking for is maybe an explanation / argument why it matters if I contract in d or in 4 dimensions. And do I really have to calculate/contract+commute everything in d dimensions if I already know that in the end I want to do dimensional regularisation? Even though I just do it to extract the divergent part [itex]1/\epsilon[/itex]?
  2. jcsd
  3. Aug 26, 2011 #2


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    Yep, it matters, and you should contract in d dimensions. The reason is that gauge invariance, from which the Ward identity you like to prove follows, is closely related to the space-time structure. E.g., you have to use [tex]g_{\mu \nu} g^{\mu \nu}=d[/tex] at the intermediate steps of the calculation.

    This issue becomes not really involved in QED, but as soon as you consider chiral theories with [itex]\gamma_5[/itex] and [itex]\epsilon_{\mu \nu \rho \sigma}[/itex] which have specific properties in 4 dimensions, and it's not a priori clear, how to treat these quantities in d dimensions. The answer is subtle and has to do with anomalies. To get the right answer you must introduce these quantities that are special in 4 dimensions in a specific way dependent on the physics. This problem occurs in any renormalization scheme since you can shuffle between the vector and the axialvector anomaly at will, and the physics decides which regularization is the one that respects the correct symmetries.
  4. Aug 27, 2011 #3
    Ok, thank you. I am not quite sure yet how going from 4 to d dimension affects things like gauge invariance, but I guess it will be clearer after learning more about the subject of renormalization.

    At least I now know for sure that I really do things wrong if I don't work with the proper d-dimensional contraction scheme.

  5. Aug 28, 2011 #4


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    The main merit of dim. reg. is that it doesn't destroy gauge symmetries. That's why it's so widely used in perturbative calculations in gauge theories. It has been invented by 't Hooft during the work on his PhD thesis to have a convenient regularization precisely for this reason! This makes the organization of the renormalization business of gauge theories much more convenient than other regularizations which may destroy some symmetries.
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