Register to reply 
Regularization scheme independence? 
Share this thread: 
#1
Jan3113, 10:49 PM

P: 211

How can two different regularization schemes give the same physical results?
This seems unbelievable. If you impose the same renormalization conditions, then in all regularization schemes, the cutoff, or dimension, or the heavy masses get absorbed into constants in such a way that the leftover finite piece is the same? Is there a proof of this? 


#2
Feb113, 02:45 AM

Sci Advisor
Thanks
P: 2,329

This is a somewhat tricky issue. In general there can be a dependence on the regularization scheme, and one must be careful, how one defines the physical (finite) quantities, i.e., the renormalization scheme.
In a Dysonrenormalizable theory, the renormalization scheme for the Smatrix is fixed by a finite number of renormalization conditions, which can be formulated as conditions for a finite number of proper vertex functions. Then there is no need for any regularization scheme, but you can read the naive Feynman rules as rules to build the integrands of the loop integrals. The integrals themselves are usually divergent, but now you use the renormalization conditions to subtract these divergences. For diagrams with more than two loops this is complicated by the fact that a diagram in general has not only overall divergences but contains divergent subdiagrams. This is handled by the BPHZ scheme (Zimmermann's forrest formula). After you've done the appropriate subtractions from the integrand, you are left is a UV finite integral over the loop momenta. Using a regularization scheme as an intermediate step can simplify the task, because you can do the integrals first and then the subtractions. Here, you have to be careful to choose a renormalization scheme which does not violate the symmetries that may be important for the renormalizability of the theory. E.g., in QED the fourphoton vertex is superficially logarithmically divergent but in reality it's finite due to the gauge invariance, leading to WardTakahashi identities. If you violate gauge invariance (which is in fact very easy), such arguments won't apply. Here, you choose a renormalization scheme which keeps the regularized theory gauge invariant. The most elegant one is dimensional regularizatio, invented by 't Hooft in his proof of the renormalizability of nonabelian gauge theories. Another scheme is the PauliVillars regularization. Of course, you can also use the regularization free BPHZ renormalization, but this may become a bit cumbersome, because you cannot subtract the divergent vertex parts at 0 external momenta since the photon is massless, and you'll introduce artificial IR divergences which are not easily dealt with. So you have to subtract with the external momenta at some spacelike point, which introduces a renormalization scale. This is necessarily the case in any theory, containing massless particles. So here, dimensional regularization may be more convenient. Although many symmetries are robust under dimensional regularization, there is one exception, and that's for chiral models like the electroweak sector of the standard model. Here, the trouble is that there is no unique way to treat the matrix [itex]\gamma^5=\mathrm{i} \gamma^0 \gamma^1 \gamma^2 \gamma^3[/itex] in four dimensions or the LeviCivita tensor [itex]\epsilon^{\mu \nu \rho \sigma}[/itex]. The deeper reason for this issue, however, is that in general the renormalization conditions of the above kind are not complete. Here, you need additional constraints. E.g., there may anomalies occur, which means that not all symmetries of the classical action survive quantization. In the case of chiral couplings the [itex]\mathrm{U}(1)_L \times \mathrm{U}(1)_R[/itex] symmetry is anomalously broken and thus not all the currents are conserved. Here, in addition to the usual renormalization conditions one has to specify which linear combination of the Noether currents of the classical field theory has to stay conserved to get a unique specification of the renormalization scheme. In the case of the standard model the [itex]\mathrm{U}(1)_V[/itex] has to survive, because otherwise you run into trouble with electromagnetic gauge invariance. This also leads to a dimensional regularization scheme with special properties for the chiral [itex]\gamma^5[/itex] matrix in other than 4 spacetime dimensions (the socalled 't Hooft definition, according to which [itex]\gamma^5[/itex] anticommutes with the first 4 [itex]\gamma[/itex] matrices but commutes with all others). In any case, if the renormalization scheme is fixed completely, the Smatrix is independent of the used regularization scheme. This independence of the physical quantities (transitionmatrix elements, cross sections) leads to the socalled renormalization group equations, which describe, e.g., the running of the coupling with the renormalization scale, etc. A lot about this you find in my QFT manuscript (however, it still lacks a chapter on anomalies): http://fias.unifrankfurt.de/~hees/publ/lect.pdf 


#3
Feb113, 09:56 AM

P: 211

To me it looks like the renormalization conditions amount to a boundary condition at a single point. For example say for a 3point vertex V(k1,k2,k3) you want the renormalization condition V(0,0,0)=λ, where λ is experimentally measured at the kinematic point (0,0,0). In different regularization schemes, before enforcing the renormalization condition, you get different V_{i}'s that are different functions of k1,k2,k3. But once you enforce V_{i}(0,0,0)=λ on each of the V_{i}'s from different regularization schemes i, then all the V_{i}'s become the same? Getting all the V_{i}'s to agree at one kinetmatic point is like saying I can match Ax^6 and Bx by using the condition V(1)=1, thereby requiring A=B=1. But surely if the amplitude is x^6 that is different than the amplitude being x, even though both obey the renormalization condition? 


#4
Feb113, 05:26 PM

Sci Advisor
P: 1,190

Regularization scheme independence?



#5
Feb113, 07:10 PM

P: 211

Suppose you have an integral of the form I(p)=∫dx/(x+p) with renormalization condition I(k)=λ. If you take the derivative w.r.t. 'p', then the expression converges, so if you can expand about your renormalization point k: I(p)=I(k)+I'(k)*(pk)+... then regularization won't affect I'(k) since I'(k) is finite anyway, and different types of regularization correspond to different I(k) which is not yet equal to λ. Now in each regularization scheme I(k) is really something that's infinite, so in order to get I(k) in each scheme to equal λ you have to define counterterms or change your bare couplings such that I(k) becomes λ for each scheme? So each scheme has: I(p)=λ+I'(k)*(pk)+...? So is this the reason that vanhees71 said that you never have to consider divergent integrals or regularization? Just take the derivative of the integrand from the start! What's the point of things like dimensional reguarlization? Does taking derivatives violate some sort of symmetry? What's the name of just taking derivatives from the start instead of having a regularization scheme (edit: oh, I guess this is what vanhees71 calls subtraction scheme, and it's a renormalization scheme rather than a regularization scheme). Will integrals like I(p)=∫dx/(x+p) always be analytic in p so that the Taylor expansion is valid? addendum: Quick question. If you have: I(p)=I(k)+I'(k)*(pk)+... and choose your bare constants to cancel cutoffs in your regulator such that: I(k)=λ doesn't that affect I'(k), since I'(k) depends on those same bare constants? So you've made I(k) finite, but at the expense of I'(k)! 


#6
Feb213, 01:24 AM

Sci Advisor
Thanks
P: 2,329

That's indeed the way, BPHZ renormalization deals with the infinities in momentum space. See my qft manuscript,
http://fias.unifrankfurt.de/~hees/publ/lect.pdf (Sect. 5.9). 


Register to reply 
Related Discussions  
L1 Regularization Question  General Math  1  
Dimensional regularization  High Energy, Nuclear, Particle Physics  2  
Tikhonov regularization  General Math  3  
Dimensional regularization  Quantum Physics  1  
Dimensional regularization  General Math  3 