Question about the epsilon in using dim reg to derive the beta function of QFTs
lost.and.lon...@gmail.com
regularization we get the relation (up to 1 loop)
beta[e] = - (epsilon/2) e - e dlog[Z_e]/dln[mu] (1)
and
Z_e = 1 + e^2 A / epsilon
where e is the coupling, Z_e is the renormalization of the coupling, A
is some constant that does not depend on mu or epsilon and mu is the
arbitrary scale introduced to make the dimension of the coupling the
same as if epsilon was zero.
Now if I do the following math
dlog[Z_e]/dln[mu]
= (1/(1 + e^2 A / epsilon))(d/dln[mu])( 1 + e^2 A / epsilon )
= (2 e A/(epsilon + e^2 A)) beta[e]
Is this correct? It seems extremely elementary, but if I trust my
results
beta[e](1 + 2 e^2 A/(epsilon + e^2 A) ) = - (epsilon/2)e
Taking the limit epsilon 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 the denominator of
1/(1 + e^2 A / epsilon) \approx 1 - e^2 A / epsilon and so
dlog[Z_e]/dln[mu]
= 2 A e beta[e] / epsilon + Order[e^4] (2)
and next we iterate it, putting (1) for beta into (2)
beta[e]
= - (epsilon/2) e - 2 A e^2 beta[e] / epsilon + Order[e^4]
= - (epsilon/2) e - 2 A e^2 (- (epsilon/2) e - 2 A e^2 beta[e] /
epsilon + Order[e^4]) / epsilon + Order[e^4]
= - (epsilon/2) e + A e^3 + (2 A e^2)^2 beta[e] / epsilon^2 + . . .
and we say it's approximately equal to
= A e^3
(For instance A = 1/12pi^3 for pure QED.)
How can we ignore the epsilon? How can we taylor expand in powers of e
when epsilon is supposed to be tiny in dimensional regularization? Even
if we do taylor expand and iterate what about the 1/epsilon^2 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.