why does dimensional regularization need counterterms ??by zetafunction Tags: counterterms, dimensional, regularization 

#1
Mar2410, 04:26 PM

P: 399

if all the integrals in 'dimensional regularization' are FINITE why do we need counterterms ?? in fact all the poles of the Gamma function are simple hence the only divergent quantity is the limit as d tends to 4 of
[tex] 1/(d4) [/tex] which is ONLY a divergent quantity that could be 'absobed' or reparametered as a divergent universal constant 'a' 



#2
Mar2410, 05:48 PM

Sci Advisor
P: 5,307

Several integrals like selfenergy vacuum polarization produce infinities which can be written as functions of a cutoff. The only special thing about dim. reg. is that you do not use a cutoff in terms of mass or energy but that you get a pole in 1/(d4).
If you have an infinite quantity you can write it as (finite quantity + infinite quantity), so that means that the calculation is ambiguous. The counter term ensures that the subtraction of all infinite quantities in all Feynman integrals of the theory is done consistently. 



#3
Mar2510, 05:51 PM

P: 399

however the pole at d=4 is SINGLE , it will give only a divergent quantity in every integral proportional to
[tex] 1/(d4) [/tex] when d>4 perhaps it would suffice to define the divergent quantity [tex] a=1/(d4) [/tex] to be 'fixed' by experiments another question if dimension of our space was for example d=3.956778899797696969695.. instead of d=4 , since there would be no poles (gamma function is perfectly defined for negative numbers except when they are negative integers) would dimensional regularization give the CORRECT finite answer, that's it if the dimension was different from an integer , there would be no poles inside the Gamma function and everything would be OK 



#4
Mar2710, 03:07 AM

Sci Advisor
P: 5,307

why does dimensional regularization need counterterms ??I would not take the approach with varying dimension too seriously; it is simply a method to parameterize the infinities w/o breaking several invariances. If you use Hamiltonian methods with mode expansion other methods e.g. heat kernel or zeta function regularization are more appropriate. In that case you do not see any deviation from D=4. 



#5
Mar2710, 04:12 AM

P: 399

thanks Tom i wil give these methods (Zeta regularization and Heat Kernel) a look, anyway zeta function regularization has the problem of a pole at s=1 [tex] \zeta(1)= \infty [/tex]




#6
Mar2710, 04:47 AM

Sci Advisor
P: 5,307




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