A Does The Use Of The Zeta Function Bypass Renormalization

bhobba

Mentor
I am trying to figure out if the use of the Zeta function allows renormalization to be bypassed. I have formed a preliminary view but would like to hear what others think:

Thanks
Bill

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A. Neumaier

I am trying to figure out if the use of the Zeta function allows renormalization to be bypassed.
No, it cannot be bypassed in this way. Ones still needs the subtractions and determine their value such that the divergences cancel. Zeta-regularization is only a way to do the bookkeeping.

vanhees71

Gold Member
You have to disinguish between regularization and renormalization: Regularization is some way to make sense of the divergent integrals you get from evaluating the Feynman rules for diagrams with loops. There are many possibilities around in the literature. The most simple is a cut-off, but that's uncomfortable, because it destroys Lorentz and gauge invariance. The most used regularization nowadays is dimensional regularization, where you analytically continue the space-time integrals over 4-dimensional spacetime momenta to an arbitrary d-dimensional momenta with $d$ not necessarily integer. Another one is $\zeta$-function regularization. These have the advantage of making the calculations Lorentz and gauge invariant at any step.

At the end you want the result for the physical values of the parameters introduced in the regularization prescription (i.e., cut-off to $\infty$ or dimensions to $4$). The divergent integrals are of course divergent when taking this limit. So you have to subtract the divergent part, and for renormalizable theories this can be done in a way that is equivalent to adding contributions to the Lagrangian which are of the same form as the terms already there, which lumps the infinities into the unobservable "bare normalization and coupling constants as well as masses", and everything is expressed in terms of the finite coupling constants that are observed.

A way to renormalize without an explicit regularization step is to read the Feynman rules as Feynman rules for the integrands of the loop integrals and you subtract the counterterms from them before you do the then finite integrals. This is known as BPHZ renormalization (named after Boguliubov and Parasiuk, Hepp, and Zimmermann). For details, see my QFT script:

king vitamin

Gold Member
Some - but not all! - divergences are "automatically" renormalized by zeta/dimensional regularization (both methods lie within the general purview of analytic continuation methods of regularization). For example, consider the integral
$$I_{d,s} = \int \frac{d^d p}{(2 \pi)^d} \frac{1}{(p^2 + m^2)^s}.$$
This is clearly divergent for any $d>2s$, but analytic continuation gives you the answer
$$I_{d,s} = \frac{\Gamma(s - d/2)}{\Gamma(s) (4 \pi)^{d/2}}m^{d - 2s}.$$
This expression has poles if $s - d/2$ is a negative integer, but is perfectly finite everywhere else. So then loop integrals of this form are finite for all odd $d$ provided $s$ is an integer.

But certain properties of renormalization need to agree between all regularization schemes. For example, the flow of the beta functions, the analytic properties of Callan-Symanzik equations, anomalous dimensions in scale invariant theories, etc. So clearly zeta reg can't get rid of all divergences, since it also needs to be able to contain this important physical information.