Applied asymptotic expansions in momenta and masses

258 Pages·2010·1.436 MB·English

by  Vladimir A. Smirnov

1 Introduction
The problem of obtaining asymptotic expansions of Feynman integrals in
various limits of momenta and masses is a typical mathematical problem in
elementary-particlephysics. In this book, it is explained how this problem is
solved.Tocharacterize,briefly,themainstepsofthe solutionletusconsider,
for example, the process of e+e− annihilation, where an incoming electron
and positron produce, according to the quantum-theoretical description, a
virtual photon, e+e− → γ∗, which in turn produces some particles, for ex-
ample a quark–antiquark pair, γ∗ → qq¯, which may then be transformed
into mesons. The process of quark production is described, in perturbative
quantum field theory,by Feynman integrals correspondingto various graphs
generatedbytheFeynmanrules.Oneofthethreeexternallegsofsuchadia-
gramcorrespondstoatriplevertexassociatedwiththequarkvectorcurrent,
and the other two external legs correspond to the external quarks. If we are
interestedinthetotalcross-sectionfortheproductionofthequarkstheprob-
lemreducestothe evaluationofthe imaginarypartofdiagramscontributing
to the vacuum polarizationand containing only two external vertices for the
two vector currents.
Suppose that the Euclideanmomentum squaredofthe intermediate pho-
ton, Q2 = −q2, is much larger than the other parameters involved in the
problem. In operator language, the problem of the large-Q2 behaviour re-
ducestotheanalysisoftheproductofthetwocurrentswhenthemomentum
transfer between the currents is large. The corresponding diagrams are gen-
eratedbythe twoverticesforthe currentsandafamilyofverticesassociated
with interactions via the QCD Lagrangian. At the diagrammatic level, we
arrive at the problem of determining the asymptotic behaviour of such dia-
grams in the limit when Q2 and also the scalar products Q·qi with other
externalmomentaaremuchlargerthantheremainingkinematicalinvariants
and the squares of masses.
Let us take an example of such a diagram expanded in the large-
momentum limit. To simplify the situation let us turn to the scalar massless
φ4 theory and replace the quark currents by two composite operators of the
form φ2. In this case the simplest one-loop diagram with two external ver-
tices for the composite operators and one external vertex generated by the
interaction Lagrangian is that of Fig. 1.1. According to the Feynman rules,
VladimirA.Smirnov:AppliedAsymptoticExpansionsinMomentaandMasses,
STMP177,1–15(2002)
(cid:3)c Springer-VerlagBerlinHeidelberg2002
2 1 Introduction
q
p
Fig. 1.1. One-loop triangle diagram with a large external
momentumq and a small external momentum p
q+p
this diagram is given by the following integral:
(cid:1)
d4k
FΓ(q,p)= k2(q−k)2(p+k)2 . (1.1)
The integration is over four components of the loop momentum k =(k ,k),
0
wherek=(k ,k ,k ).Thesquaresoffour-vectorsareimpliedinthe pseudo-
1 2 3
Euclidean metrics: k2 = k2−k2, etc. This Feynman integral, depending on
0
two four-vectorsp and q, is a scalarquantity,so that it is a function of three
kinematical invariants, q2, p2 and p·q=p q −p·q.
0 0
Let us suppose that the external momentum q is large with respect to
(cid:2)the external momentum p in the Euclidean sense, i.e. |p| (cid:5) |q| where |p| =
p2+p2. Therefore at least one four-component of q is larger than any
0
component of p. We can measure the order of magnitude in powers of the
smallmomentum p so that q2 is large,p·q is smallandof the firstorder,and
p2 is of the second order.
AlthoughFeynmanintegralsaregenerallydivergent,this specific integral
turnsouttobeconvergent,whichcanbeseenbypowercountingatsmalland
large values (in the Euclidean sense) of the four-dimensional momentum k.
Letusnowposetheproblemofexpandingintegral(1.1)inthegivenlimit
withoutcomputingitforgeneralvaluesofthearguments.Moreover,wewould
like to write down a result for the expansion in terms of Feynman integrals
that depend on a lower number of scales. To simplify the situation as much
as possible, let us confine ourselves to the leading asymptotics in this limit.
To further simplify the subsequent discussion let us consider the Euclidean
variant1 of the integral(1.1). The form of the integralstays the same but all
the scalar products are now defined in accordance with Euclidean metrics,
i.e. k2 =k2+k2, k·q =k q +k·q, etc.
0 0 0
Letus observethat ifwe naively takethe limit p→0 in the integrandwe
obtain the integral
(cid:1)
d4k
, (1.2)
(k2)2(q−k)2
1Weshalllaterseewhatwillcauseadditionaldifficultiesinthepseudo-Euclidean
case.
1 Introduction 3
which is divergent at small loop momenta k. This can be seen by introduc-
inggeneralizedsphericalcoordinatesandcountingpowersofthe correspond-
(cid:3)
Λ
ing radial variable, r. The divergence is of the logarithmic type, dr/r.
0
The non-existence of such a naive limit and the character of the divergence
in (only) one of the local coordinates (i.e. r) provide a hint that the true
asymptotic behaviour, when p→0, is of the logarithmic type.
WhenoneissolvingtheproblemoftheasymptoticbehaviourofFeynman
diagrams,the goalcanbe quite different: it canrange fromthe evaluationof
the coefficients of the leading logarithms to the evaluation of the coefficients
of all powers and logarithms. Suppose for a moment that we are interested
in only the leading logarithmic behaviour of our triangle diagram. In this
case the (standard) strategy based on regions can be used: one concentrates
on the contributions of only those regions of the loop momenta which are
responsible for the leading logarithmic behaviour. In our case, the relevant
region is that of small k so that we can restrict ourselves to the integral
(cid:1)
d4k
,
k2(q−k)2(p+k)2
|k|≤Λ
with a cut-off at some non-zerovalue of Λ. Since the logarithmic divergence,
when p → 0, arises only at the point k = 0, can we put k = 0 in the factor
1/(q−k)2, which is independent of p.
Let us choose,for convenience,p=(p ,0), where p =|p|, and introduce
0 0
the spherical coordinates
k =rcosθ,
0
k =rsinθsinψcosφ, k =rsinθsinψsinφ, k =rsinθcosψ,
1 2 3
where d4k = r3dr sin2θdθ sinψdψdφ. Integrating over the first two angles
φ and ψ, we obtain
(cid:1) (cid:1)
π Λ
4π rdr
sin2θdθ . (1.3)
q2 r2+2|p|rcosθ+p2
0 0
The internal integral in r can be evaluated in terms of elementary functions
with a result which has the asymptotic beh(cid:3)aviour −ln|p| for general θ and
Λ. Multiplying this result by the integral πsin2θdθ = π/2, we arrive at
0
the following asymptotic behaviour of the given EuclideanFeynman integral
when p→0:
π2 (cid:4) (cid:5)
FΓ(q,p)∼− ln p2/Λ2 . (1.4)
q2
If we want to obtain not only the leading logarithm but also a constant
term in the leading power, 1/q2, we have to ‘honestly’ take into account
the contribution of the whole integration domain to the loop integral. Let
4 1 Introduction
us choose an intermediate scale Λ, where |p| (cid:5)Λ (cid:5) |q|, and divide2 our
integral into two parts corresponding to integrations over small and large
loop momenta:
FΓ =fsmall+flarge , (1.5)
where
(cid:1)
d4k
f = , (1.6a)
small k2(q−k)2(p+k)2
|k|≤Λ
(cid:1)
d4k
f = . (1.6b)
large k2(q−k)2(p+k)2
|k|≥Λ
Thelimitp→0canbetakensafelyintheintegrandofthesecondintegral.
Onthe otherhand,the terms inthe combination(q−k)2 =q2−2q·k+k2 in
the first integralare of different orders so that is legitimate to take the limit
k →0 there. Thus we have
(cid:1)
1 d4k
f ∼f(0) ≡ , (1.7a)
small small q2 k2(p+k)2
|k|≤Λ
(cid:1)
d4k
f ∼f(0) ≡ , (1.7b)
large large (k2)2(q−k)2
|k|≥Λ
and
FΓ ∼fs(m0)all+fl(a0r)ge (1.8)
in the limit p→0.
Uptonowwehavedealtwithconvergentintegrals.Nowwewanttoarrive
atasimpleresultforthebehaviouroftheinitialintegralinthegivenlimitat
the cost of running into divergences.To handle such divergent quantities we
useastandardprocedurecalled‘regularization’whichmeansthat,insteadof
agivendivergentintegral,weturnourattentiontoaquantitywhichdepends
ona (regularization)parameter,is welldefined insome domainof the values
of this parameter and formally coincides with the initial divergent integral
at some limiting value of the regularization parameter. We shall use dimen-
sionalregularization [28,144],which,formally,reducestoreplacingthespace
dimension 4 by a complex number d ≡ 4−2ε. The integrals regularized in
this manner are denoted by switching from d4 to dd. It will be explained in
thenextchapterhowdimensionalregularizationissystematicallyintroduced
but here we make use of the following practical recipe: evaluate the integral
2IfweweredealingwiththeFeynmanintegralinMinkowskispacethenitwould
be necessary to use the same decomposition with |k| understood to be in the Eu-
clidean metrics. (Still, thiswould look unnaturalbecause thedecomposition is not
Lorentz invariant.) On the other hand it would be necessary to suppose that the
externalmomentumis,forexample,space-like.Thestatementsthatwehavemade
after the decomposition are also true in the Minkowski case but not so obvious as
in theEuclidean case.
1 Introduction 5
in a general integer number of dimensions and then replace the number of
dimensions by the complex number d in the corresponding explicit result.
Let us rewrite the result obtained by extending the integration, in both
f(0) and f(0) , to all values of the loop momentum k and subtracting the
small large
corresponding additional terms:
(cid:1) (cid:1)
1 ddk ddk
FΓ ∼ q2 k2(p+k)2 + (k2)2(q−k)2
(cid:1) (cid:1)
1 ddk ddk
− − . (1.9)
q2 k2(p+k)2 (k2)2(q−k)2
|k|≥Λ |k|≤Λ
Now we may take the limit p→0 in the first integral in the second line and
neglect k with respect to q in the factor 1/(q−k)2 in the second integral.
Thus the sum of the two integrals in the second line reduces, in the given
limit, to the integral
(cid:1)
1 ddk
− ,
q2 (k2)2
which is an integral without scale. Such integrals are generally set to zero
within dimensional regularization in accordance with a well-known conven-
tion.However,asweshallseelater,suchFeynmanintegralsareautomatically
set to zero in the case of limits typical of Euclidean space, without referring
to additional prescriptions. On the other hand, for typical pseudo-Euclidean
limits,thisconventionstilllookslikeanadhocprescription;itwillbeimplied
for any integral without scale. So the second line is zero and we obtain the
following leading-power behaviour:
(cid:1) (cid:1)
1 ddk ddk
FΓ ∼ q2 k2(p+k)2 + (k2)2(q−k)2 . (1.10)
Let us stop for a moment and observe that we have already solved the
problem of the expansion (in the leading order) because the initial quantity
depending onthree kinematicalinvariantscanbe replaced,inthe givenlimit
and order, by two integrals that depend only on one scale each, so that
both terms are homogeneous with respect to the expansion parameter. The
homogeneity degree (e.g. with respect to p) is easily computed by power
counting before integration: it equals −2ε and 0 for the first and second
term, respectively. Moreover, both terms on the right-hand side are nothing
but familiar Feynman integrals, rather than functions from another class,
Feynman integrals with additional integrations, etc. However, we have paid
a (negligible) price to obtain this result by introducing the regularization.
Without it, i.e. for d = 4, both terms would be ill-defined because the first
integral would be divergent at large k and the second integral at small k.
Thetwointegralsin(1.10)areindeedmuchsimplerquantities:inparticu-
lar,theycanbeevaluatedforgeneralvaluesoftheregularizationparameterd.
Following the above recipe, we can compute them at general integer values
6 1 Introduction
ofd.Results fortheseintegrals(whichareofthesametype)canbe obtained
fromthecorrespondingintegralwithgeneralvaluesofpowersofpropagators.
Results of this kind and similar ones are presented in the next chapter.3 In
accordance with the recipe formulated above, when evaluating this general
integral we can proceed in an integer number of dimensions, then arrive at
an explicit result in terms of gamma functions, insert the relevant values of
the powersofpropagatorsand,finally,considerthenumberofdimensions,d,
as a complex number. Thus we obtain
(cid:6) (cid:7)
Γ(1−ε)2Γ(ε) 1 Γ(1−ε)2Γ(ε) 1
FΓ ∼πd/2 Γ(2−2ε) q2(p2)ε − Γ(1−2ε) (q2)1+ε . (1.11)
If we allow d to tend to 4 (i.e. ε → 0), we see that the poles in the two
contributions cancel and we obtain the following leading-power behaviour:
ln(p2/q2)−2
FΓ ∼π2 . (1.12)
q2
Letus now observethat we couldarriveat our result(1.10) and, further,
at (1.11) and (1.12), by use of the following prescriptions:
(i) divide the integration domain into regions of small and large momenta
anduseleadingapproximationsfortheintegrand(i.e.takethezero-order
terms in Taylor series in the small parameters) (see (1.8));
(ii) extendthe integrationtoall loopmomentainthe expandedcontribution
of every region.
In this book, the strategy of expansion by regions will be formulated for
• arbitrary order of expansion,
• a general diagram,
• any limit.
The greatest part of the book is devoted to details and subtleties of the
method, which are explained through numerous characteristic examples.
In our first example, we have justified this strategy. It will be justified
for some other examples also.But mathematical proofs for generaldiagrams
are unknown at present. It turns out, however, that for limits typical of
Euclidean space, expansion by regions leads to the same prescriptions for
expanding Feynman integrals as does another strategy [47, 126, 207, 208],
which we shall call the ‘strategy of expansion by subgraphs’. This second
strategy has been mathematically justified (in this simpler case of limits)
for general diagrams so that we obtain here an indirect confirmation of the
strategy of expansion by regions.
3 Another way to compute this general integral is to observe that it is of con-
volution type, and then apply a well-known theorem which says that the Fourier
transform of a convolution equals theproduct of theFouriertransforms of theini-
tialfactors,andalsoformula(A.51)fortheFouriertransformofthefunction(q2)λ
(taken,for example, from [124]).
1 Introduction 7
Let us illustrate expansion by subgraphs, using the same diagram as in
Fig.1.1.Thekeypointofthisstrategyistoreducetheproblemoftheasymp-
totic expansion to the problem of the R-operation, i.e. diagrammatic renor-
malization.The goalofthe R-operationis to makea Feynmanintegralfinite
by removing its ultraviolet divergences. When a diagram is only overall di-
vergent and does not involve subdivergences, the R-operation is of the form
R=1−M,whereM isasubtractionoperator.Whenthereisalsoadivergent
one-particle-irreduciblesubgraphthenRtakestheformR=(1−M)(1−M ),
1
where M performs subtractions in the subgraph. In a general situation, the
1
R-operationisgivenbytheforestformula[245,247].Ithappensthat,fortyp-
ical Euclideanlimits, the remainder of the asymptotic expansioncanbe also
determinedbythe forestformulawrittenintermsofappropriatesubtraction
operatorswhich,inthiscase,performsubtractionsinacertainfamilyofsub-
graphs associated with the given limit and provide the desired asymptotic
behaviour of the remainder.
In the case of our one-loop triangle diagram, in the leading order of ex-
pansion in the limit of large q, the remainder can be written as
RFΓ =(1−M)(1−M1)FΓ , (1.13)
where the subtraction operators M and M are defined by
1
(cid:1) (cid:1)
1 1 1
MFΓ = ddkk2(q−k)2Tp(p+k)2 ≡ ddk(k2)2(q−k)2 , (1.14a)
(cid:1) (cid:1)
ddk 1 1 ddk
M1FΓ = k2(p+k)2Tk(q−k)2 ≡ q2 k2(p+k)2 , (1.14b)
and T... are operators that perform Taylor expansion in the corresponding
variable.In this example,these are Taylorexpansions of zero order,i.e. they
just set the corresponding variable to zero.
The rule for choosing the operators M is very simple: they are operators
that perform Taylor expansion of the integrand of the Feynman integrals
in parameters that are considered small for a given subgraph. The operator
M performs Taylor expansion with respect to the small parameter of the
wholegraph,i.e.thefour-momentump,andtheoperatorM performsTaylor
1
expansion with respect to the small parameter of the subgraph that consists
oftheline withmomentumq−k.(Bydefinition,alltheloopmomentaofthe
whole graph that are external for a given subgraph (e.g. k in our case) are
considered small.) Note that we have introduced a regularization to make it
possibletoconsiderindividualsubtractionoperatorsandtheirproducts,and
we always choose this regularizationto be dimensional.
The remainder defined in this way can be rewritten as
(cid:1) (cid:6) (cid:7)(cid:6) (cid:7)
1 1 1 1 1
RFΓ = ddkk2 (p+k)2 − k2 (q−k)2 − q2
(cid:1)
1 (p2+2p·k)(k2−2q·k)
≡ ddk . (1.15)
q2 (k2)2(q−k)2(p+k)2
8 1 Introduction
Weobserve,bypowercountingatsmallandlargevaluesofk,thattheremain-
der is both infrared and ultraviolet finite. On the other hand, it manifestly
involves the first power of the small expansion parameter p, so that it be-
haves, when p→0, as |p| modulo logarithms (and, in any case, as |p|1−α for
arbitrarily small α).
Starting from the remainder (1.13) with these properties, we represent
our Feynman integral as
FΓ = [1−(1−M)(1−M1)]FΓ +RFΓ
= MFΓ +M1FΓ −MM1FΓ +RFΓ . (1.16)
The third term,
(cid:1)
1 ddk
−MM1FΓ =−q2 (k2)2 , (1.17)
is a massless vacuum integral, which we set to zero.
Hence the asymptotic expansion is given by the first two terms in the
second line of (1.16), and we recognize our previous result (1.10) obtained
above by expansion by regions. Let us also stress that, this time, all the
arguments used in the derivation of this result are equally applicable to a
Feynman integral in Minkowski space.
In this book, it will be explained how expansion by subgraphs can be
applied to
• arbitrary order of expansion,
• a general diagram,
• any limit typical of Euclidean space.
Itisthisstrategywhichweshallprefertoapplytolimitsofthiskind,withthe
correspondingprescriptionswrittenby meansofa simple formulacontaining
a summation over an appropriate family of subgraphs.
Let us remember that we have arrivedat expansionof the triangle Feyn-
mandiagramfromtheproblemoftheexpansionoftwovectorquarkcurrents.
Let us come back to the operator level. It turns out that the techniques for
expanding individual Feynman diagrams enable us to perform operator ex-
pansions. In this particular case, the prescriptions for expanding Feynman
integralsFΓ(q,p1,...,pn)inthe limitwhereoneofthe externalmomenta,q,
islargeandthe otherexternalmomenta,pi,andallthemassesaresmallcan
be naturally extended to the operator level and reformulated for the expan-
sionofthetime-orderedproductoftwocompositeoperatorsatshortdistances
(large momenta q). For simplicity, let us continue to consider the scalar the-
ory, instead of QCD, and analyse the expansionof the time-orderedproduct
oftwo equalcomposite operatorsJ =(1/2)φ2 in the masslessφ4 modelwith
the interaction LagrangianL =−(g/4!)φ4:
I
TJ(x)J(0)∼C (x)1+C (x)J(0)+... . (1.18)
0 2
1 Introduction 9
The expansion is performed in the limit where the difference between four-
coordinates of the operators tends to zero, so that the terms are ordered
accordingtothestrengthoftheirsingularbehaviourwhenx→0.Ithappens
that in these first terms of the operator product expansion (OPE), only the
unit operator and the operator J itself are present, as the operators with
the lowestpossible canonicaldimensions (zeroand two,respectively,in mass
units).Asiswellknown,theproductsofquantumfieldoperatorsaresingular
atpointswheretheirargumentscoincide,sothatthefirstcoefficientfunctions
(Wilson coefficients) Ci are singular.
The limit x→0 canbe translatedinto momentum space languageas the
limit q →∞ for Fourier transforms. Turning to momentum space, we have
(cid:1)
TJ˜(q)J(0)≡ d4xeiq·xTJ(x)J(0)∼C˜ (q)1+C˜ (x)J(x)+... .
0 2
(1.19)
The Fourier transform TJ˜(q)J(0) is represented in perturbation theory, ac-
cording to the Feynman rules, by diagrams which have two external vertices
for the two operators and a set of φ4 vertices (either external or internal)
generated by the interaction Lagrangian– see Fig. 1.2.
p1
p2 q
(cid:0)q(cid:0)Ppi Fig. 1.2. The set of diagrams contributing to the
pn product of two φ2 operators
It is implied that all the quantities are renormalized. In the diagram-
matic language, this means that we have rules for removing ultraviolet di-
vergencesfromthe diagrams.We shallusually assume a formofdimensional
renormalization [144] which is based on subtracting poles in the parame-
ter ε = (4−d)/2 of dimensional regularization. In addition to the rules for
renormalizingdiagramsconstructedfromtheintegrationLagrangian,wealso
havesimilar prescriptionsfor making finite one-particle-irreduciblediagrams
whichinvolveoneorbothverticescorrespondingtothecompositeoperators.
Attheoperatorlevel,therenormalizationisperformedbyinsertingcountert-
erms into the Lagrangianand redefining bare composite operators and their
products. The renormalization of the Lagrangian (which reduces to a redef-
inition of the coupling constant) is irrelevant in our example up to one-loop
order. The one-loop renormalization of the composite operator J reduces,
in our massless case, to the substitution of the bare composite operator by
a renormalized one: JB → Z−1J, where Z is a renormalization factor. Dia-
10 1 Introduction
p1 p1
p2 p2
Fig. 1.3. Diagrams of zero and first order with two S-matrix external lines con-
tributingto the operator J
grams of zero and first order in g with two S-matrix external lines and one
vertex for the operator J are shown in Fig. 1.3.
Now,inaccordancewiththeFeynmanrules,wetakeintoaccountafactor
i for each propagator,1/(2π)d for the loop integrationand −ig/4! for the S-
matrix vertices. In particular, the value of the first diagram in Fig. 1.3 is
one. The renormalization factor Z is generated, in one loop, by the second
diagram,
(cid:1)
ig ddk
.
2(2π)4 k2(q−k)2
TheMSsubtraction(whereonesubtractsthecombination1/ε+ln(4π)−γ ,
E
rather than the pure pole in ε) gives the value
g 1
Z =1+ . (1.20)
2(16π2)ε
MultiplicationofthesumofthetwodiagramsinFig.1.3bythisZ (equivalent
to the renormalizationof the operator J) then produces a finite result when
ε→0.
The expansion of the product TJ˜(q)J(0) reduces to the expansion of
the corresponding diagrams in the limit where q is much larger than all
the other external momenta,in the Euclideansense. The coefficientfunction
C can be evaluated through diagrams that depend only on the external
0
momentum q and no other external momenta. The problem of the large-
momentum expansion does not arise for C and, since we are dealing with a
0
masslesstheory,thesediagramshaveonlypowerandlogarithmicdependences
on q. In this sense, the coefficient function C is trivial, and we turn now to
0
the Wilson coefficient C .
2
Considerdiagramswithtwo‘S-matrix’externalvertices.Inotherwords,
consider the Green function
(cid:14)TJ(x)J(0)φ(y )φ(y )(cid:15)amp
1 (cid:8) 2 (cid:6) (cid:1) (cid:7) (cid:9)
amp
= TJ(x)J(0)exp i d4xL (x) φ(y )φ(y ) ,
I 1 2
0
where the symbol ‘amp’ denotes the contribution of one-particle-irreducible
diagrams in which the external y lines have been amputated (i.e. the prop-
agators corresponding to these lines are omitted). The vacuum expectation