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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?

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1 Answer 1

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.

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Thanks for the answer. I'll try to read it carefully soon. –  joshphysics Jul 11 '13 at 6:43
    
No problem. Let me know if you have any question or if something is not sufficiently clear.@joshphysics –  drake Jul 11 '13 at 6:49
    
+1: I finally got around to reading this. This confirms everything that I would have hoped to be true. There's one follow-up I'd like to ask. It seems to me that if one is willing to choose something super pathological, then one might find a "regulator" that does not lead to the same results when physical quantities are written in terms of physical quantities. My point (which is why I put quotes around "regulator") is; "what exactly is the definition of the term "regulator;" how are we restricted in the way we decide to parameterize divergences? –  joshphysics Dec 13 '13 at 21:54
    
@joshphysics It's hard to answer if you don't specify what a "super pathological" regulator means to you. The point is that the "pathology" in the regulator can be removed or compensated by the equivalent pathology in the bare Lagrangian. I'd say that the regulator in the last paragraph of my answer is pathological enough. –  drake Dec 15 '13 at 21:58
    
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 $3\epsilon$"? –  joshphysics Dec 15 '13 at 22:13

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