Renormalization-Scheme-Independent Perturbation Theory by Resumming Logarithms

0.21 MB·English

by  Chris Dams

Renormalization-Scheme-Independent
Perturbation Theory by Resumming Logarithms
Chris Dams∗, Ronald Kleiss†
Institutefor Theoretical Physics
Radboud University
5 Toernooiveld 1
0 6525 ED Nijmegen
0 The Netherlands
2
n December 17, 2004
a
J

6
Abstract
2
Results of perturbation theory in quantum field theory generally de-

4 pend on the renormalization scheme that is in use. In particular, they
v depend on the scale. We try to make perturbation theory scheme invari-
4 antbyre-expandingwithrespecttoaschemeinvariantquantity. Further-
5 more, we investigate whether the potentially large logarithms in such an
2
expansion cause inaccuracy and how this can be improved.
2
1
4 1 Introduction
0
/
h The occurrence of divergencies in perturbative quantum field theory has made
p
it clear that although measurable quantities should be finite, this need not
-
p be the case for the parameters of the theory. To handle this, divergencies are
e regularized. Toobtainanorder-by-orderfiniteperturbationexpansion,itshould
h
be specified what to do with divergencies as they occur at each loop level.
:
v This introduces an arbitrary constant for each loop level. Although the full
i perturbation series should not depend on these constants, the truncated one
X
does. Thisisaproblemifoneistryingtoapproximateaphysicalquantitythat,
r
a by the definitionof “physicalquantity”,shouldnot depend onthese unphysical
parameters.
It has been pointed out in [10] that scheme invariant quantities can be con-
structed out of the scheme-dependent ones. Therefore, the natural thing to do
would be to try to rewrite the perturbation series into a series expansion with
respectto the quantityX , the invariantthatcanbe calculatedatthe one loop
1
level. Actually, that would be a series expansion in 1/X , since 1/X is the
1 1
small quantity if we are in the perturbative regime.
An expansion in 1/X will, however, contain logarithms in the expansion
1
coefficients. At high energies, these logarithms will dominate all other con-
tributions to the expansion coefficients. We will resum these logarithms and
investigate whether this gives more reliable results.
∗[email protected]
†[email protected]
1
Several attempts to remedy this problem of renormalization scheme depen-
dence have been proposed. They all suffer from a great deal of arbitrariness
and/or the mathematics involved is more complicated than that of simply ma-
nipulating power series, as we will be doing.
To obtain our results, we used the simple case of a one-parameter theory.
We did not yet consider more complicated cases, but will do this later. When
doingexplicitcalculations,weuseQCDwithfivemasslessquarks,asaconcrete
example of a one-parameter theory.
2 Renormalization Scheme Invariants
In this section we outline how renormalization scheme-independent quantities
can be combined into scheme invariant ones. This is also explained in [10] and
[6]. For self-containedness of this paper and also to make clear what our con-
ventions are, we will repeat this in this section. The coupling constant will
be denoted by a. The key idea is that the consistency of perturbation theory
requires that a result up to terms of order an should only differ from the exact
answeruptotermsoforderan+1. Ifweconsiderthecasen=1,weseethatthe
running of the coupling makes the use of a as an expansion parameter scheme
dependent. Thisindeedisaneffectthatstartsatordera2. I.e.,thelowestorder
term in the beta function is of order a2.
The renormalization scheme can be specified by giving the renormalization
scale s and the scheme-dependent beta function coefficients β ,β ,β ,... . The
2 3 4
coupling constant depends on the scale via
∂a
=β(a)=β a2+β a3+β a4+ , (1)
0 1 2
∂s ···
whereds=dµ/µwithµthemassscaleusedindimensionalregularization. This
differential equation can be integrated to give
1 β β +β a a 1 1
1 0 1 ′
s= + log + da . (2)
−β a β2 β a β(a′) − β (a′)2+β (a′)3
0 0 1 Z0 (cid:18) 0 1 (cid:19)
This solution implies the choice of a boundary value. Our choice is similar to
the one in [10]. Taking the set of variables that consists of s and the scheme-
dependent beta function coefficients as independent, we can derive that
∂a a (a′)i+2
′
=β(a) da . (3)
∂β β(a′)2
i Z0
As an example, we imagine that we have calculated a physical quantity R
uptofourthorder. Fromthisexample,wehope,itwillbeclearhowthis canbe
generalized to arbitrary order. Having calculated R up to fourth order means
that we have
R r a+r a2+r a3+r a4. (4)
0 1 2 3
∼
Consistency of perturbation theory requires that
dR
a5. (5)
dscheme ∼
2
To be more concrete, independence of s implies that
∂a ∂r ∂r ∂r
r +2r a+3r a2+4r a3 + 1a2+ 2a3+ 3a4 a5. (6)
0 1 2 3
∂s ∂s ∂s ∂s ∼
(cid:0) (cid:1)
From this, equations for second, third and fourth order can be extracted. Sub-
stituting the beta function for ∂a/∂s, we find
∂r
1
r β + =0;
0 0
∂s
∂r
2β r +r β + 2 =0; (7)
0 1 0 1
∂s
∂r
3
3β r +2β r +r β + =0.
0 2 1 1 0 2
∂s
Toobtaintheequationsthatfollowfromindependenceofβ ,weexpand∂a/∂β
2 2
up to fourth order. We have
∂a 1
= a3+O(a5). (8)
∂β β
2 0
Using this and then demanding that the third and fourth order of ∂R/∂β are
2
zero, we get the equations
r ∂r
0 2
+ =0;
β ∂β
0 2 (9)
r ∂r
1 3
2 + =0.
β ∂β
0 2
Finally, independence of β gives the equation
3
r ∂r
0 3
+ =0. (10)
2β ∂β
0 3
Integrating the equations for r , r , and r , we find
1 2 3
r =X β r s;
1 1 0 0
−
r2 r β β r
r =X + 1 0 2 + 1 1;
2 2 r − β β (11)
0 0 0
3r X 5β r2 r β 2r β r3
r =X + 1 2 + 1 1 0 3 1 2 + 1,
3 3 r 2β r − 2β − β r2
0 0 0 0 0 0
where the quantities X are the renormalization scheme invariants. They arise
i
because the values of the r are obtained from differential equations and hence
i
need constants of integration. The values of the X can be calculated from the
i
values of r and β once these have been calculated. It should be made sure
i i
that the r and β have been calculated in the same renormalizationscheme, of
i i
course, and then it will turn out that the values of X no longer depend on the
i
scheme that was used to obtain r and β .
i i
The acute reader will have noticed that we actually expressed r and r
2 3
into r instead of in s. Hence, we actually label our renormalizationscheme by
1
r ,β ,β ,... rather than by the set of parameters that we mentioned earlier.
1 2 3
The reason that we do this is that if we are going to re-expand into 1/X we
1
would rather have r than s in the expansion coefficients because in a suitable
1
3
renormalization scheme r will be a number of order unity while s will be of
1
order 1/a, which is large. In the end it will hopefully turn out that the result
will not depend on what kind of “suitable renormalization scheme” we used,
however, we are not yet at this point so we should still make sure that we do
not have large expansion coefficients.
3 Expansion in 1/X
1
For anexpansionto work,we needto know thatwe areexpanding with respect
to a small parameter. The idea of perturbation theory is that the coupling
constant, a, is a small quantity. Equation (2) can be expanded in powers of a.
We have
1 β β β2a β a β3a2 β a2 β β a2
s + 1 log 0 + 1 2 1 3 + 1 2
∼−β a β2 β a β3 − β2 − 2β4 − 2β2 β3
0 0 (cid:18) 1 (cid:19) 0 0 0 0 0 (12)
β a3 β2a3 β β2a3 2β β a3 β4a3
4 + 2 2 1 + 3 1 + 1 .
− 3β2 3β3 − β4 3β3 3β5
0 0 0 0 0
From this series we see that if a is small, s must be large. Indeed, in the case
of QCD, s is given by
µ2
s=log , (13)
Λ2
QCD!
where µ is the renormalization scale as it occurs in dimensional regularization
and Λ is of the order of the energy range where the coupling is large and
QCD
perturbation theory is not accurate. Therefore, the perturbation series can
also be written as an expansion in 1/s. For this we use the inverse of the
expansion 12. We have
1 β β2 β β2 β2
a + 1 L + 1 2 1 L 1 L2
∼−β s β3s2 s β5s3 − β4s3 − β5s3 s− β5s3 s
0 0 0 0 0 0 (14)
β3 β 2β3 3β β β3 5β3
1 + 3 1 L + 2 1L + 1 L3+ 1 L2,
− 2β7s4 2β5s4 − β7s4 s β6s4 s β7s4 s 2β7s4 s
0 0 0 0 0 0
where L =log( β /(β2s)).
s − 1 0
Nowweassumethatasuitablerenormalizationschemehasbeenchosen,and
hope to obtain results that turn out not to depend on our “suitable renormal-
ization scheme” and hence might also have been calculated if we had started
out with an unsuitable renormalization scheme. In a suitable renormalization
scheme, we expect that expansion coefficients are not large. In particular r is
1
expectedtobeoforderunity. Fromequation(11)itfollowsthatX =r +β r s.
1 1 0 0
We conclude that X must be a large quantity, because s is. Hence, the expan-
1
sion
1 β r β r β r r β r r2 β r r3
= 0 0 0 0 + 0 0 1 + 0 0 1 + 0 0 1 (15)
s X r ∼ X X2 X3 X4
1− 1 1 1 1 1
isagoodexpansion. Wesubstitutethisequationintoequation(14)obtainingan
expansionofain1/X . Thisexpansionissubstitutedintoequation(4). Wethen
1
obtainanexpansionofthe physicalquantityR into 1/X . Furthermore,inthis
1
expansion, we substitute for r ,r ,r ,... the values as given by equation (11).
2 3 4
4
It should be noted that although X does not depend on the renormalization
1
scheme, it is dependent upon the physical quantity under consideration. There
is,however,nothingwrongwithusingadifferentexpansionparameterforevery
different physical quantity. The expansionthat is obtained by making all these
substitutions is
r2 r3β r3X r4β2
R 0 + 0 1 L 0 2 + 0 1 (1 L L2 )
∼−X β X2 X − X3 β2X3 − X − X
1 0 1 1 0 1 (16)
r4X 3r4β X r5β3
+ 0 3 + 0 1 2L + 0 1 1 2L + 5L2 +L3 ,
X4 β X4 X β3X4 −2 − X 2 X X
1 0 1 0 1
(cid:0) (cid:1)
where L is given by L =log( r β /(β X )). We see that in this expansion
X X 0 1 0 1
−
all scheme-dependent terms have canceled. These scheme-dependent terms are
the ones involving r and β with i 2. Because of the cancellation of these
1 i
≥
terms, we have obtained a scheme-independent perturbation series.
For this method to work for any order in perturbation theory, we must
prove that the cancellation of scheme-dependent terms happens at every order
and not just up to fourth order. For this purpose, note that our expansion is
an expansion with respect to the variables 1/X ,1/β ,β ,β ,β ,...,r ,r ,X ,
1 0 1 2 3 0 1 2
X ,X ,...,L . If we refer to the order of a term in the series, we mean the
3 4 X
orderin 1/X . We shouldprovethat actually the variablesr ,β ,β ,...do not
1 1 2 3
occur in this seriesexpansion. Letus assume that sucha variable actually does
occur at some order n in the expansion.
We introduce v as an alias for one of the offending variables (there might
be several offending variables) that occurs at order n. This means that ∂R/∂v
is of order n. We consider what happens if we re-expand the expression for
R in a again, and then differentiate with respect to v. To do this we first
have to expand 1/X = 1/(β r s+r ) in 1/s and then use equation (12) to
1 0 0 1
expand this in a again. X was defined in such a way that it actually does not
1
depend on scheme-dependent quantities such as v. Therefore, we know that
the entire series of 1/X in a does not depend on v. Furthermore, we note
1
that during re-expanding and differentiating the order of a term in 1/X , 1/s
1
or a, whichever applies, never decreases. Therefore, all the invariant terms up
to ordern after differentiationcauseterms that areat leastofordern+1. The
only terms that can give a contribution of order n are the scheme-dependent
terms. However, the terms obtained by differentiating a, using the chain rule,
are of higher order, so this does not contribute. The conclusion is that the
derivative with respect to v up to order n is the same before re-expanding as
afterre-expanding,exceptthat1/X istobereplacedby a/r andL istobe
1 0 X
−
replacedby log(β a/β ). Hence,the physicalquantityR upto ordern depends
1 0
on the scheme at order n. This is a contradiction with the starting point of
section 2. Therefore the expansion in 1/X must have renormalization scheme
1
independent coefficients.
4 Resumming the Logarithms
Inthis section, wewillresumalllogarithmsL thatoccur inequation(14). We
s
start out by observations that have been made from computer algebra experi-
mentation, but in the end we will prove our results to be correct to all orders.
5
By looking at the expansion (14), we observe that the leading logarithms, i.e.,
terms of the order Ln−1/sn can be summed into the quantity 1/s′ defined by
s
1 1 1
= . (17)
s′ s1+β L/(β2s)
1 0
After this resummation we have the expansion
a(s′,L) 1 β2 + β12 (1 L)+ β13 1 +L L2
∼−β s′ − β4(s′)3 β5(s′)3 − β7(s′)4 −2 − 2
0 0 0 0
(cid:16) (cid:17)
β β4
+ 3 + 1 7 +2L 1L2 1L3 (18)
2β5(s′)4 β9(s′)5 −6 − 2 − 3
0 0
3β2β (cid:0)5β2 β β (cid:1)β
+ 1 2 (1 L) 2 + 1 3 4
β8(s′)5 − − 3β7(s′)5 6β7(s′)5 − 3β6(s′)5
0 0 0 0
Looking at the highest order logarithms in this expansion, we recognize the
expansion of β /(β3s2)log(1 β L/(β2s′)). We therefore define a quantity L′
1 0 − 1 0
by
β
′ 1
L =log 1 L . (19)
− β2s′
(cid:18) 0 (cid:19)
′
Itnowturnsoutthatafterrewritingtheexpansionforawithrespectto1/s and
′ ′
L, we recover the original expansion where 1/s has been replaced by 1/s and
′
L by L. Becauseofthis we caniterate this procedureandobtaina sequenceof
values 1/s and L from the iteration
n n
1 1 1
= ;
sn+1 sn1+ ββ02s1nLn (20)
1
L =log ,
n+1 1+ β1 L
β02sn n
where we have rewritten L in terms of s and L instead of in terms of
n+1 n n
s and L . We note that this iteration increases the order of L with respect
n+1 n
to 1/s. Therefore, in the perturbative regime, the iteration should make L
converge to zero. This resums all logarithms into a new value for 1/s.
Acurvecanbedrawnthroughthesequenceofpoints(1/s ,L ). Thiscurve
n n
is given by
1 1 1
= ;
s(x) s∞1+ β1 x
β02s∞ (21)
1
L(x)=log x,
1+ β1 x −
β02s∞
wherexparameterizesthe curveanddifferentcurves(fordifferentinitialvalues
of 1/s and Ls) are labeled by 1/s∞. That this is correct, can be seen by
checkingthattheiterationforthepair(1/s,L)isrecoveredifxisiteratedusing
the prescription
1
x =log . (22)
n+1 1+ β1 x
β02s∞ n
We seethatalsoforthe x the propertyholdsthatx is ofhigherorderthan
n n+1
x , hence in the perturbative regime it should converge to zero. In this case
n
6
we note that 1/s(x) 1/s∞, which explains our notation “1/s∞” to label the
→
different curves.
We must still prove our assertionthat the iteration preserves the expansion
to all ordersin 1/s. It is sufficient to show that the value of a is constant along
the curves introduced above. The infinitesimal form of the curves is
β
1
δs= δx;
β2
0 (23)
β
1
δL = 1 δx
s − − β2s
(cid:18) 0 (cid:19)
Proving that this is a symmetry of the expansion (14) is also sufficient to see
that the iteration works. This is what we will do in the next section.
5 Proof of the Iteration
Here wewillprovethat the transformation(23) isa symmetryofthe expansion
equation(14). This,atthe sametime,showsthatthe iterationofequation(20)
works to all orders and that the logarithms in the expansioncan be made zero.
We introduce a quantity a¯ that is a power series in 1/s and L . This quantity
s
has the definition
∂a¯ β s
= s2+ 1 β(a¯);
∂(1/s) − β2
(cid:18) 0 (cid:19)
∂a¯ β1 (24)
= β(a¯);
∂L β2
s 0
[sa¯(1/s,L =0)] = 1/β .
s s→∞ − 0
Wewillshowthatthisquantitya¯isactuallythesameasa. Firstnotethatthese
differential equations are consistent because ∂/(∂(1/s)) and ∂/∂L commute.
s
Secondly, if we confine ourselves to the surface L = log( β /(β2s)) we have,
s − 1 0
as is verifiable by using the chainrule for differentiation, da¯/ds=β(a¯), so,also
using the boundarycondition,we seethatonthis surfacea anda¯ arethe same.
Wenowshowthatforeveryorderin1/s,thereisafinitenumberofterms. This
ensures that also away from the surface L = log( β /(β2s)) these functions
s − 1 0
are the same, because it is impossible to express s and L into each others in a
s
finite number ofterms. The pre-factorsof the expansionin 1/sandL of a¯ can
s
be obtained by setting a¯s = 1/β in the derivatives ∂n+ma/(∂Lm∂(1/s)n).
− 0 s
At first sight, it may seem that problems could be caused by terms of the form
a¯msn where n > m. However, the fact that the differential equations have a
solution that is an expansion in Lm/sn with n > 0 and m 0, ensures that
s ≥
these problematic contributions will cancel if we substitute a¯s 1/β . Still,
0
→−
beforethissubstitutionismade,therewillbetermsoftheforma¯p(1/s)p+q,with
p>0 and q 0. If we consider the quantity dna¯/d(1/s)n, the maximum value
≥
ofq forwhichthistypeofmonomialwilloccurisequalton 1. Differentiations
−
with respect to L increase the order in a, hence, in ∂n+ma/(∂Lm∂(1/s)n), the
s s
maximum value for q will be n 1 m. Hence, if m n we will only have
− − ≥
terms containing a¯p(1/s)q with p > q. If we substitute a¯s = 1/β , taking
0
−
s , these terms will become zero andthus do not contribute. Fromthis we
→∞
see that the maximum order in L that occurs in the coefficient of 1/sp in the
s
7
expansion in s is p. Hence, this coefficient of 1/sp contains a finite number of
terms, as we set out to show. We conclude that the quantity a¯ indeed has the
same expansion with respect to 1/s and L as a.
s
Furthermore,fromequations(23)and(24),itfollowsthatda/dx=0. There-
forethesymmetry(23)isindeedasymmetryoftheexpansionofain1/sandL ,
s
and the iteration in equation (20) keeps the value of a constant to all orders
in 1/s.
6 Resumming L
X
The symmetry of equations (23) can be turned into a symmetry of the expan-
sion of a physical quantity in 1/X and L . This will enable us to perform
1 X
resummation of logarithms in such an expansion. From X = r β s+r , it
1 0 0 1
follows that
r β
0 1
δX =r β (δs)= δx. (25)
1 0 0
β
0
Turning L into L is done via
s X
X
1
L =L +log . (26)
s X
X r
1 1
−
Applying δ on both sides gives
β r
1 1
1 δx=δL δX . (27)
X 1
− − β s − X
(cid:18) 0 (cid:19) 1
Equation 25 then gives
r β
0 1
δL = 1+ δx. (28)
X
− β X
(cid:18) 0 1(cid:19)
UsingthissymmetryitispossibletoturnL intozero,therebyriddingourselves
X
of logarithms. The value of X that is obtained while turning L to zero will
1 X
be called X˜ . Integrating equations (25) and (28), we obtain the equation
1
r β r β
X =X˜ 0 1 log− 0 1. (29)
1 1− β0 β0X˜1
ThiscanbeexpressedintheLambertW-function. Thisfunctionisbydefinition
the solution to W(z)eW(z) =z. We have
X˜1 = r0β1W eβr00Xβ11 . (30)
β −
0
(cid:16) (cid:17)
Hence, the conclusion is that we turned the standard perturbation theory into
an expansion in the quantity 1/X˜ . The expansion looks as displayed in equa-
1
tion(16)withX replacedbyX˜ andalltermsthathaveanL removed. Since
1 1 X
this reduces the number of terms considerably, let us display a few more. We
have
r2 r4β2 r3X r5β3 r4X 7r6β4 3r5β2X r5X
R 0 + 0 1 0 2 0 1 + 0 3 0 1 + 0 1 2 0 4
∼−X˜ β2X˜3 − X˜3 − 2β3X˜4 X˜4 − 6β4X˜5 β2X˜5 − X˜5
1 0 1 1 0 1 1 0 1 0 1 1
17r7β5 3r6β3X 4r6β2X r6X
+ 0 1 0 1 2 0 1 3 + 0 5.
12β5X˜6 − 2β3X˜6 − β2X˜6 X˜6
0 1 0 1 0 1 1
(31)
8
highest order not using symmetry using symmetry
1/s 91.5 91.5
1/s2 282.2 249.0
1/s3 247.5 249.3
1/s4 248.6 249.0
Table 1: Values for Λ in MeV obtained by solving equation (12) to various
QCD
ordersnumerically,while usingornotusingthesymmetryinequation(23). We
used a number of quarks equal to five to obtain this result.
7 Determining Λ
QCD
In their Review of Particle Physics [8] the Particle Data Group suggests using
equation (14) to define Λ . To be fully accurate, their definition is not
QCD
completely the same. Instead of our L = log( β /(β2s)) the PDG uses L =
s − 1 0 s
logs. This amounts to a shift in the parameter s. A way to see this is from
−
equation (2). Adding a factor β /β2 inside the logarithm in this equation,
− 1 0
turns our expansion into the one of the PDG. This is equivalent to adding a
constant to the right-hand side of this equation. This constant can then be
movedtotheleft-handside,soweseethatsisindeedshifted. Theconsequence
is that the Λ that we use differs by a multiplicative constant from the one
QCD
of the PDG. We have
β2 β1/(2β02)
ΛPDG =ΛOURS 0 , (32)
QCD QCD −β
(cid:18) 1(cid:19)
where the beta function coefficients are given in our conventions. In the rest of
the paper we use our conventions, hence we will be writing Λ for ΛOURS
QCD QCD
andnevermentionΛPDG again. Note however,thatthe expansionwith respect
QCD
to 1/X and L becomes different if we choose a different prefactor inside the
1 X
logarithm. Wewillseeinthenextsectionthatusingthesymmetryofequations
(25) and (28) resolves this ambiguity.
The suggestionof the PDG to use equation (14) to define Λ is not very
QCD
practical. In the first place it would seem to be easier to use equation (2).
Secondly, ifone is goingto use equation(14), the symmetry fromequation(23)
is useful to obtain a series that converges faster by transforming L to zero.
s
ThePDGgivesα (M ) 0.1187. Intable1wecomparethevalueobtainedfor
s Z
≈
Λ using the symmetryandnotusing it. The usedbeta function coefficients
QCD
werecalculatedin[9]andrecentlyconfirmedin[2]. Thenumberofflavourswas
set to five. We indeed see faster convergence.
Below we will be using Λ . The value that we use comes from equa-
QCD
tion (14) to fourth order where we use our symmetry to get rid of the L ’s.
s
If one uses standard perturbation theory, one has to pick a suitable value for
µ and specify the renormalization scheme. In that case, the consistent way to
proceed is to only use the beta function up to the loop level to which the rest
of the calculation is done. The difference with our case is that while in the
standardapproachα (M ) 0.1187appearsasafundamentalconstant,inour
s Z
≈
approach Λ would be the fundamental quantity. Determining the value of
QCD
a fundamentalconstantfromanotherone is better done with asmuchaccuracy
as possible. This is the reason that we use the beta function up to four loop
level.
9
8 How Invariant is Invariant, Really?
Our method attempts to remedy the arbitrarinessin choosing the renormaliza-
tionprescription,soweshouldnowaskthequestiontowhatextendthismethod
itselfisarbitrary. Afirstpossiblesourceofarbitrarinessisthechoiceofvariables
to parameterize the renormalization prescription. We choose the set of param-
eters r ,β ,β ,β ,... to parameterize the renormalization prescription. If one
1 2 3 4
chooses a different set of variables, r¯ = r +∆ ,β¯ = β +∆ ,β¯ = β +∆ ,
1 1 1 2 2 2 3 3 3
wherethe∆’sareconstants,theintegrationconstantsX fromsection2alsobe-
i
comedifferent. However,ifwehave∆ =0,itturnsoutthatintheendthefinal
1
expansion coefficients still have the same value, so this is not an arbitrariness
of our method.
This shows that the small invariant quantity that we decide to use for ex-
pansion is more important than the precise definition of the other invariants.
The question that might arise is what would happen if we would expand with
respect to some arbitrary function of X instead of with respect to X . We
1 1
could, for instance, expand with respect to the sine of X . The possibility of
1
expressingthecouplingconstantinoneschemeasapowerseriesinthecoupling
constant in another scheme is a possibility that has been mentioned in liter-
ature, for instance in [10]. Our point of view is that the possibility to use an
arbitrarypowerseriesisnotafundamentalarbitrarinessofperturbationtheory.
Note that if it were, it would apply to any perturbative method in any branch
ofphysics. Smallquantitiesthatareusedforexpansionshouldbe theonesthat
come naturally with the problem under consideration, not the ones that can
be used to show that any approximation method can be made to give wrong
answers. In field theory, the situation is that no renormalization scheme is a
prioribetter thanany other, andthis ambiguity canbe parameterizedby using
an arbitrary power series in the coupling constant. This is the reasonarbitrary
power series of the coupling constant can be useful to consider. Considering
the fact that the first scheme invariant arises naturally from the demand that
a physical quantity should not depend on the scale, it does not make sense to
consider arbitraryfunctions ofthis, perhaps with the exceptionofa translation
in de definition of X . I.e., a non-zero value of ∆ .
1 1
If ∆ = 0, we obtain a different invariant variable on the one loop level,
1
6′ ′
namely X =X +∆ . If we were to expand in 1/X , we would indeed obtain
1 1 1 1
a different expansion. Here, it appears, we have finally found arbitrariness.
However,weactuallyalreadymentionedthis. Anexampleofthisarbitrarinessis
givenintheparagraphthatcontainsequation(32). Theprefactorthatwechoose
inside the logarithm L is equivalent to choosing a value for ∆ . Resumming
s 1
the logarithms yields the same result for X˜ and, hence, also the same result
1
for the physical quantity R.
9 Hadronic-R
In this section we consider massless QCD-corrections to hadronic R . This
had
quantity is by definition given by
σ(e++e− hadrons)
R = → , (33)
had σ(e++e− µ++µ−)
→
10