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Question about epsilons that occurs in calculating beta functions in QFTs w dim reg

  1. Jul 7, 2005 #1
    Question about epsilons that occur in calculating beta functions in QFTs w dim reg

    In deriving the beta function of, say, QED using dimensional regularization we get the relation (up to 1 loop)

    [tex]\beta[e] = - \frac{\epsilon}{2} e - e \frac{d ln[Z_{e}]}{d ln[\mu]} \quad (1)[/tex]


    [tex]Z_{e} = 1 + \frac{e^{2} A}{\epsilon}[/tex]

    where e is the coupling, [itex]Z_{e}[/itex] is the renormalization of the coupling, [itex]\mu[/itex] is the arbitrary scale introduced to make the dimension of the coupling the same as if [itex]\epsilon[/itex] was zero, and A is some constant that does not depend on [itex]\mu[/itex] or [itex]\epsilon[/itex].

    Now if I do the following math

    [tex]\frac{d ln[Z_{e}]}{d ln[\mu]}[/tex]
    [tex]= \frac{1}{1 + \frac{e^{2} A}{\epsilon} } \frac{d}{d ln[\mu]} \left( 1 + \frac{e^{2} A}{\epsilon} \right)[/tex]
    [tex]= \frac{2 e A}{\epsilon + e^{2} A} \beta[e][/tex]

    Is this correct? It seems extremely elementary, but if I trust my results

    [tex]\beta[e] \left(1 + \frac{2 e^2 A}{\epsilon + e^{2} A} \right) = - \frac{\epsilon}{2}e[/tex]

    Taking the limit [itex]\epsilon[/itex] going to zero I get that the beta function for QED is zero up to one loop!

    What seems to be usually done is we taylor expand [itex] \frac{1}{1 + \frac{e^{2} A}{\epsilon} } \approx 1 - \frac{e^2 A}{\epsilon} [/itex] and so

    [tex]\frac{d ln[Z_{e}]}{d ln[\mu]}[/tex]
    [tex]= \frac{2 A e \beta[e]}{\epsilon} + \mathcal{O}[e^{4}] \quad (2)[/tex]

    and next we put (2) into (1) and iterate

    [tex]= - \frac{\epsilon}{2}e - \frac{2 A e^{2} \beta[e]}{\epsilon} + \mathcal{O}[e^{4}] [/tex]
    [tex]= - \frac{\epsilon}{2}e - 2 A e^{2} \left( - \frac{\epsilon}{2}e - \frac{2 A e^{2} \beta[e]}{\epsilon} + \mathcal{O}[e^{4}] \right) \frac{1}{\epsilon} + \mathcal{O}[e^{4}] [/tex]
    [tex]= - \frac{\epsilon}{2} e + A e^{3} + \frac{(2 A e^{2})^{2} \beta[e]}{\epsilon^{2}} + . . . [/tex]

    and we say as epsilon goes to zero it's approximately equal to

    [tex]A e^{3}[/tex]

    (For instance [itex]A = 1/12 \pi^{3}[/itex] for pure QED.)

    How can we taylor expand in powers of e when [itex]\epsilon[/itex] is supposed to be tiny in dimensional regularization? Even if we do taylor expand and iterate what about the [itex]1/\epsilon^{2}[/itex] and possibly even more infinite quantities?

    This issue really bothers me a lot because I'm sure there's something I'm not understanding here, since the QED coupling does indeed run as calculated, right? Any clarification would be deeply appreciated.
    Last edited: Jul 7, 2005
  2. jcsd
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