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Question about epsilons that occurs in calculating beta functions in QFTs w dim reg 
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#1
Jul705, 08:55 PM

P: 32

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] and [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]\beta[e][/tex] [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. 


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