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Regulator-scheme-independence in QFT

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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 (810 points) [ no revision ]
retagged Mar 31, 2014
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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
This is murky water for me for indeed, so take everything I say with a grain of salt. I think there are 2 separate issues going on. (1) Whether the physics quantities are regulator independent. Like I said before, to my knowledge this is the content of the CS equation - we literally a physical quantity that we can measure and take a derivative with respect to mu and say this has got to be zero.

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

1 Answer

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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 (880 points) [ no revision ]
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Thanks for the answer. I'll try to read it carefully soon.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user joshphysics
That's fair. Here's more specifically what I mean. In my mind, a regulator is a scheme by which otherwise ill-defined (divergent) expressions are made well-defined by making them depend on some parameter $\epsilon$, say, in such a way that they diverge when $\epsilon$ limits to some value. Could my scheme be something like "replace every divergent integral one encounters with the expression $3\epsilon$"?

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user joshphysics
@joshphysics With $3\epsilon$ ? What do you mean? The regulated theory must reduce to the original one in some (formal) way.

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user drake
Right; that's my point. In my comment above I asked "what exactly is the definition of the term "regulator;" how are we restricted in the way we decide to parameterize divergences?" If the statement "the physical quantities are independent of the regularization procedure" is to have precise meaning, then the term "regularization procedure" must as well. When people prove that theories are renormalizable, I would hope that they do so for any "sufficiently sensible" procedure. What exactly do you mean by "The regulated theory must reduce to the original one in some (formal) way."

This post imported from StackExchange Physics at 2014-03-31 22:27 (UCT), posted by SE-user joshphysics
Hi @joshphysics Ok, I don't have a complete, rigorous definition of the term regulator, just an intuitive one. Two properties the regularized integral must fulfill and which are not by your scheme "replace every divergent integral one encounters with the expression 3ϵ" is that there must be a one to one correspondence between the regularized and unregularized integrals and that you must recover the unregularized integrals from the regularized ones in some given way.

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

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