Register to reply

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

by lonelyphysicist
Tags: beta, epsilons, functions, occurs, qfts
Share this thread:
Jul7-05, 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]


[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.
Phys.Org News Partner Physics news on
Engineers develop new sensor to detect tiny individual nanoparticles
Tiny particles have big potential in debate over nuclear proliferation
Ray tracing and beyond

Register to reply

Related Discussions
Need help calculating wavelengths of L(alpha) and L(beta Advanced Physics Homework 5
Need help calculating wavelengths of L(alpha) and L(beta Introductory Physics Homework 0
Software for calculating beta decay High Energy, Nuclear, Particle Physics 4
Calculating spherical Bessel functions General Math 0
Calculating values of trig functions General Math 4