Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Regulator-scheme-independence in QFT

+ 7 like - 0 dislike
3003 views

Are there general conditions (preservation of symmetries for example) under which after regularization and renormalization in a given renormalizable QFT, results obtained for physical quantities are regulator-scheme-independent?

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user joshphysics
asked Mar 25, 2013 in Theoretical Physics by joshphysics (835 points) [ no revision ]
retagged Mar 31, 2014
Most voted comments show all comments
(2) I am not sure if things give the same results independent of scheme, but perhaps in the IR they will. The fact that when you look in the PDG they tell you 'we used MS bar here' indicates it matters what scheme you use. But I have never properly understood this.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user DJBunk
I thought it should only depend on the initial conditions to which IR fixed point the QFT considered tends to flow (if there is one) in the IR limit and the number and kinds of fixed points depends only on the RNG equations and nothing else?

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user Dilaton
@Qmechanic Thanks!

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user joshphysics
The reason why I find the question problematic is the problematic ill-defined concept "physical quantities". The key problem with the renormalization scheme dependence is that "some particular physical quantities" get mixed with others. As long as such mixing is possible, it essentially means that there's some freedom to redefine what we mean by "the" physical quantities, and this ambiguity of the mixing may be counted as a part of the renormalization scheme. Some extra symmetry constraints may forbid some/most/all mixings, but that would need a more refined discussion.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user Luboš Motl
@LubošMotl Your point is well-taken. I deliberately left the wording vague because I need to think more about how to phrase my question more sharply. I was hoping that the result of my asking this (admittedly vague) question would have been commentary that would lead to a sharper version that captures a concept on which I'm unclear, but for whom I'm as of yet unable to form a precise question.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user joshphysics
Most recent comments show all comments
Assuming your regulator preserved the symmetries of the model, isn't this sort of what the Callan Symanzik equation does for us? It essentially says no physical quantity can't depend on how we did our regulation. Although I might be misreading your question.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user DJBunk
@DJBunk Hmm perhaps that's right; are you saying that given any regulator the RG flow equations show that once we've applied renormalization conditions, all physical quantities will agree in the IR for different regulatuon schemes? It seems to me that this is probably true for all regulators that lead to expressions for regulated integrals that agree up to divergent terms.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user joshphysics

1 Answer

+ 7 like - 0 dislike

By definition, a renormalizable quantum field theory (RQFT) has the following two properties (only the first one matters in regard to this question):

i) Existence of a formal continuum limit: The ultraviolet cut-off may be taken to infinite, the physical quantities are independent of the regularization procedure (and of the renormalization subtraction point, if it applies).

ii) There are no Landau-like poles: All the (adimensionalized) couplings are asymptotically safe (roughly, their value remain finite for all values — including arbitrarily high values — of the cut-off.) (Footnote: Here one has to notice that there are Gaussian and non Gaussian fixed points.)

Thus, the answer to this question is: "The only condition is the renormalizability of the theory." The fact that in renormalizable theories some results seem to depend on the regularization procedure (dimensional regularization, Pauli-Villars, sharp cut-off in momentum space, lattice, covariant and non-covariant higher derivatives,etc.) and on the renormalization subtraction point (for example, minimal subtraction MS or renormalization at a given momentum) is due to the fact that what we call 'results' in QFT are expressions that relate a measurable magnitude, such as a cross section, to non-measurable magnitudes, such as coupling constants, which depend on the regularization or renormalization prescription. If we could express the measurable magnitudes in terms of other measurable magnitudes, then these relations would not depend on the regularization or renormalization prescription. That is, in QFT results usually have the form:

$$P_i=P_i \, (c_1, …, c_n)$$

where $P_i$ are physical (directly measurable) magnitudes, such as cross-sections at different values of the incoming momenta, and $c_i$ are renormalized, but not physical, parameters, such as renormalized coupling constants. The $c_i$'s are finite and regularization/renormalization dependent. The $P_i$'s are finite and renormalization/regularization independent. Therefore the equations above are regularization/renormalization dependent. However, if we could obtain an expression that involved only physical magnitudes $P_i$,

$$P_i=f_i\, (P_1,…, P_{i-1}, P_{i+1},… ,P_m)\,,$$

then the relation would be regularization/renormalization independent.

Example: Considerer the following regularized (à la Pauli-Villars) matrix element (it is not a cross-section, but it is directly related) before renormalization (up to pure numbers everywhere)

$$A(s,t,u)=g_B+g_B^2\,(\ln\Lambda^2/s+\ln\Lambda^2/t+\ln\Lambda^2/u)$$

where $g_B$ is the bare coupling constant, $\Lambda$ is the cut-off, and $s, t, u$ are the Mandelstan variable. At a different energy, one obviously has

$$A(s',t',u')=g_B+g_B^2(\ln\Lambda^2/s'+\ln\Lambda^2/t'+\ln\Lambda^2/u')$$

And then

$$A(s,t,u)=A(s',t',u')+A^2(s',t',u')\,(\ln s'/s+\ln t'/t+\ln u'/u)$$

This equation relates physical magnitudes and is regularization/renormalization independent. If we had chosen dimensional regularization, we would have obtained (up to pure numbers):

$$A(s,t,u)=g_B+g_B^2\,(1/\epsilon +\ln\mu^2/s+\ln\mu^2/t+\ln\mu^2/u)$$

$$A(s',t',u')=g_B+g_B^2\,(1/\epsilon +\ln\mu^2/s'+\ln\mu^2/t'+\ln\mu^2/u')$$

And again

$$A(s,t,u)=A(s',t',u')+A^2(s',t',u')\,(\ln s'/s+\ln t'/t+\ln u'/u)$$

is regularization/renormalization independent. The amplitudes $A$ are the previous $P_i$. The problem is that matrix elements aren't usually that simple and, in general, it is not possible to get rid of non measurable parameters. But the reason is technical rather than fundamental. The best we can usually do is to choose some $s',t',u'$ that not correspond to any physical configuration so that the "coupling" is an element matrix at a non-physical point of momentum space. This is called momentum-dependent subtraction. But even this is often problematic for technical reasons so that we have to use minimal subtraction, where the renormalized coupling does not correspond to any amplitude. These couplings are the previous $c$'s.

Symmetries and regulators

Let's assume that a classical theory has some given symmetries. Then there are two alternatives:

i) There is not any regularization that respects all the symmetries. Then, there is an anomaly. If this anomaly does not destroy essential properties of the quantum theory, such as unitarity or existence of a vacuum, then the quantum theory has fewer symmetries than the classical one, but the quantum theory is consistent. These are anomalies related to global (non-gauge) symmetries.

ii) There exists at least one regularization that respects all the symmetries of the theory. Nevertheless, we are not forced to use one of these regularizations. We can use one regularization that doesn't respect the symmetries of the classical theory, provided that we add all the (counter)terms to the action (in the path integral) compatible with the symmetries preserved by both the classical theory and the regularization. For example, in QED one can use a gauge-violating regularization, then the only thing one has to do is to add a term $\sim A^2$ to the action. Therefore, the fact that a regularization respects a symmetry has nothing to do with the dependence of results on the regularization. One can use the regularization one likes the best as long as one is consistent. Of course, in most cases, regularizations that respect the symmetries are technically more convenient.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user drake
answered Jul 11, 2013 by drake (885 points) [ no revision ]
Most voted comments show all comments
@drake: I admit I don't have a solid argument in mind, roughly speaking, I feel insecure if the form of lagrangian is changed, that is, I'm no longer confident to claim the renormalized lagrangian still describes the same physics as the original one. One confusion, for example, with a term like $\beta A^2$ in QED, what would be the difference if we just take a Proca lagrangian and try to renormalize it? Besides, another worry is the breaking of manifest gauge invariance, could it damage results like Ward identity so badly beyond remedy? I'm not certain if it is just a matter of convenience.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user Jia Yiyang
@JiaYiyang If one regularizes QED with a sharp cut-off in momentum space, one needs a mass counter-term to cancel an effective mass term coming from the cut-off. You ensure gauge invariance through renormalization conditions.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user drake
@drake: Is it done explicitly anywhere? It would be reassuring to see some details.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user Jia Yiyang
@JiaYiyang I don't know or remember any reference. It is easy to check it by oneself.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user drake
@drake: ok then I'll try when I get some time. Thanks.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user Jia Yiyang
Most recent comments show all comments
Hi drake, I have doubts about your last paragraph on the claim "the fact that a regularization respects a symmetry has nothing to do with the dependence of results on the regularization". Say in QED we have two measurable physical quantities $P_1$ and $P_2$, if we have two different regularization schemes that respect symmetry so that the renormalization only gives a multiplicative change of coupling constant $\alpha$, then it is quite evident that $P_1$ will always be related to $P_2$ in the same way, because, the functional forms of $P_1(\alpha)$ and $P_2(\alpha)$ will be independent of

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user Jia Yiyang
regularization scheme, the functional form is fixed by the form of the lagrangian, different regularization schemes can only result in different relations between renormalized coupling and bare coupling constants. Now if you choose a bad regularization so that you need to add a new term, say $\beta A^2$ in the lagragian, it is not transparent at all to me that $P_1(\alpha,\beta)$ and $P_2(\alpha,\beta)$ will be related in the same way as in the previous case. What would you say?

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user Jia Yiyang

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOver$\varnothing$low
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...