Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Why zeta regularization only valid at one-loop?

I mean there are zeta regularizations for multiple zeta sums. Also we could use the zeta regularization iteratively on each variable to obtain finite corrections to multiple loop diagrams.

share|improve this question
1  
Possibly related: physics.stackexchange.com/q/12235/2451 –  Qmechanic Jul 20 '11 at 18:50
add comment

1 Answer 1

The standard use of zeta-regularization in quantum field theory is to replace the ill defined one-loop functional determinant with something finite:

$$\begin{align} \det(H) &= \exp(\mathrm{tr}\log H) := \exp(-\zeta'_H(0))\,, \\ \zeta_H(s) &= \mathrm{tr} H^{-s} = \frac1{\Gamma(s)}\int_0^\infty\mathrm{d}t\, t^{s-1}\mathrm{tr}(\exp(-H t)) \end{align}$$

This is a generalization of the Riemann zeta-function $\zeta(s)=\zeta(s,q)$ and Hurwitz zeta-function $\zeta(s,q)=\sum_{n=1}^\infty (q+n)^{-s}$ since $$\zeta_H(s) = \mathrm{tr} H^{-s} = \sum_{n} \lambda_n^{-s}\,.$$ For example, for the harmonic oscillator, $\lambda_n=n+\frac12$ so $\zeta_H(s)=\zeta(s,\frac12)$.

This obviously only works at one-loop because the quantum corrections only look like a functional determinant at one-loop.

But related ideas can be extended to multiple loops. (How does the following compare to your scheme? It's hard to tell exactly what you're saying in your paper...)

In your previous question I mentioned operator regularization (e.g. PhysRevD.35.3854) as an extension of zeta-regularization. Here, the functional determinant is written using $$\begin{align} \det(H) &= \exp(\mathrm{tr}\log H) \\ \log H &:= \lim_{s\to0}\Big[-\frac{d^m}{d s^m}\Big[\frac{s^{m-1}}{m!}H^{-s}\Big]\Big]\,, \end{align}$$ for some large enough (only actually need $m=1$) integer $m$. With $m=1$ and writing $H^{-s}=\frac1{\Gamma(s)}\int_0^\infty \mathrm{d}t\, t^{s-1}\exp(-H t)$ you recover the zeta-regularized form given above.

Similarly you can also regularize the propagators $H^{-1}$ as $$ H^{-1} = \frac{d}{d H}\log H := \lim_{s\to0}\Big[\frac{d^m}{d s^m}\Big[\frac{s^{m}}{m!}H^{-s-1}\Big]\Big]\,. $$ This is the underlying basics of operator regularization.

In arXiv:1006.1806, this was generalized to $$ H^{-n} := \lim_{s\to0}\Big[\frac{d^m}{d s^m}\Big[ \big(1+\alpha_1s+\dots+\alpha_ms^m\big)\frac{s^{m}}{m!}H^{-s-n}\Big]\Big]\,. $$ where the $\alpha_i$ are initially arbitrary finite constants and $m$ is the loop order (or higher). Comparisons were made with some dimensionally regularized Feynman integrals and it was found that for a choice of $\alpha_i$ they matched. These constants are fixed according the renormalization scheme that is used. I'm not sure if a "minimal subtraction" scheme exists for operator regularization (unless you do a comparison with dimensional regularization for each diagram!) or if a physical/on-shell renormalization scheme has to be used to ensure consistent results.

A much more solid paper is Operator regularization and multiloop Green's functions. There they define the subtraction operator (with out the $\alpha_i$ constants) $$ \mathcal S^{(m)}\big(\prod_i A_i^{-1}\big) = \lim_{s\to0}\frac{d^m}{d s^m}\Big[\frac{s^{m}}{m!}\Big(\prod_i A_i^{-s-1}\Big)\Big] $$ and show that its naive application to obtain finite amplitudes breaks unitarity. (The results of arXiv:1006.1806 suggest that if you include the $\alpha_i$'s and choose them correctly, you should be able to get correct results...) Then they actually use Bogolibov's recursion formula to show how to construct a consistent $\mathcal R$-operator from $\mathcal S$.

Detail computations are given for both the naive and sophisticated use of operator regularization in massive $\phi^4_4$ theory.

Note that using a BPHZ-like scheme essentially reintroduces counterterms (hidden in the subtractions in each diagram) to the method and definitely breaks the simplicity of the original $\zeta$-regularization.

share|improve this answer
1  
Note that all of this opens up the question of whether it's possible to use the approach of arXiv:1006.1806 consistently fix the constants using a MS-like scheme. I.e. If it's possible to get away with not applying physical renormalization conditions or BPHZ-like subtractions at each order... –  Simon Aug 1 '11 at 8:16
    
However, why can not simply set the divergent terms to 0 , i mean if you have the sum of divergent quantities $ \sum a_{n} e^{-n} $ then this is divergent in the limit e-->0 then you simply 'ignore' this quantities and keep always the FINITE PART in the sense of zeta regularization $ \zeta (1)= \gamma$ and $ 1+2^{m}+3^{m}+.......................= \zeta (-m) $ so no counterterms are needed –  Jose Javier Garcia Aug 1 '11 at 10:37
    
@Jose: As I've been saying for days (!) in order to have unitary scattering amplitudes you need to be careful and consistent about how you renormalize the naively divergent Feynman integrals. The final paper I linked to has an example of what goes wrong if you're not careful. –  Simon Aug 1 '11 at 11:31
    
@Jose: Read about BPHZ renormalization and the classic papers on how analytic, dimensional and differential renormalization are additive renormalizations and equivalent to BPHZ up to finite counterterms. Also interesting is the Hopf algebra underlying renormalization. –  Simon Aug 1 '11 at 11:38
1  
ok i get it my paper is wrong :) however i still hoped to define it or give it a definition to give a finite meaning for divergent integrals in mathematics in the same fashion you give finite results for divergent series in Zeta regularization :) , i hoped to get a grant for my ideas at my university but failed , i thought that once you got the correct maths results the physics results would be obvious :) –  Jose Javier Garcia Aug 1 '11 at 13:29
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.