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Why does dimensional regularization respect the Ward identity?

  1. Jun 11, 2013 #1
    It is often stated that this is the case, but I have often wondered if it is a general statement or just something that we observe to be the case when calculating the relevant loop corrections. Can it be proven generally? Is it somehow easy to see?
     
  2. jcsd
  3. Jun 11, 2013 #2
    Dimensional regularization is a fairly general mathematical technique. You can always use it for calculations in QFT. As far as I know, in QFT, the technique basically just amounts to introducing extra parameters (regulators) to change an integral originally in some weird coordinate system into a simple euclidean integral [basically the infinitesimal elements and the integration bounds get retrofitted by the introduction of the regulator].

    The Ward identity, on the other hand, is a result of QFT. So it only holds in special circumstances.

    You might want to look in Peskin and Schroeder, they have a nice little discussion about dimensional regularization which uses the Gamma function very sneakily.

    Edit: Maybe you can't always use it: if you're being very creative, you may have to check whether the regulator you introduce is going to cause something fishy to happen. But in Peskin and Schroeder, you'll see they're doing a very innocuous mathematical manipulation.
     
    Last edited: Jun 11, 2013
  4. Jun 11, 2013 #3
    I agree, but the ward identity holds when one uses dimensional regularization, while it does not when using for example a simple cutoff regulator.

    When searching the net I found that this was due to the fact that integrals in a general dimensional respects a translational substitution; i.e that

    $$\int d^{d} p f(p+q) = \int d^{d} p f(p).$$

    This is not true for a cut-off regulator and since the proof of ward identity involves such a substitution, dimensional regularization respects it, while a cutoff regulator does not.
     
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