why does dimensional regularization need counterterms ??


by zetafunction
Tags: counterterms, dimensional, regularization
zetafunction
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#1
Mar24-10, 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/(d-4) [/tex] which is ONLY a divergent quantity that could be 'absobed' or re-parametered as a divergent universal constant 'a'
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tom.stoer
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#2
Mar24-10, 05:48 PM
Sci Advisor
P: 5,307
Several integrals like self-energy 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/(d-4).

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.
zetafunction
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#3
Mar25-10, 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/(d-4) [/tex] when d-->4 perhaps it would suffice to define the divergent quantity

[tex] a=1/(d-4) [/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

tom.stoer
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#4
Mar27-10, 03:07 AM
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P: 5,307

why does dimensional regularization need counterterms ??


Quote Quote by zetafunction View Post
however the pole at d=4 is SINGLE , it will give only a divergent quantity in every integral proportional to

[tex] 1/(d-4) [/tex] when d-->4 perhaps it would suffice to define the divergent quantity

[tex] a=1/(d-4) [/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
You should have a look at the MS or MS-bar subtraction scheme (MS = minimal subtraction). Indeed only the infinities are removed, but nevertheless counter terms are required.

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.
zetafunction
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#5
Mar27-10, 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]
tom.stoer
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#6
Mar27-10, 04:47 AM
Sci Advisor
P: 5,307
Quote Quote by zetafunction View Post
... anyway zeta function regularization has the problem of a pole at s=1
That's no problem. You introduce s just to regularize a kind of spectral- or Dirichlet series. The physical value is s=0, the pole at s=1 is unphysical. Then you continue the zeta function to s=0 avoiding s=1.


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