a nonperturbative method for the scalar field theory

Introduction Currentlyverymuchisknownabouttheperturbative behaviorofmanytheorieswithorwithoutgaugefields.Betafunctionsforthe Φ4theoryandQEDare knownuptothefifthorderwhereasforQCD, uptothef

Contents lists available atScienceDirect

Physics Letters B

www.elsevier.com/locate/physletb

A nonperturbative method for the scalar field theory

Renata Jora

National Institute of Physics and Nuclear Engineering, PO Box MG-6, Bucharest-Magurele, Romania

a r t i c l e i n f o a b s t r a c t

Article history:

Received 20 August 2014

Received in revised form 28 November 2014

Accepted 23 December 2014

Available online 30 December 2014

Editor: A Ringwald

WecomputeanallordercorrectiontothescalarmassintheΦ4theoryusinganewmethodoffunctional integrationadjustedalsotothelargecouplingsregime

©2014TheAuthor.PublishedbyElsevierB.V.ThisisanopenaccessarticleundertheCCBYlicense

(http://creativecommons.org/licenses/by/4.0/).FundedbySCOAP3

1 Introduction

Currentlyverymuchisknownabouttheperturbative behaviorofmanytheorieswithorwithoutgaugefields.Betafunctionsforthe

Φ4theoryandQEDare knownuptothefifthorderwhereasforQCD, uptothefourthorder[1–7] However, thereisalimited knowledge regardingthenonperturbative behaviorofthesametheories.Recentlyattempts[8]havebeenmadefordeterminingtheexistenceinsome renormalizationschemeofallorderbetafunctionsforgaugetheorieswithvariousrepresentationsoffermions.Itisratherusefultosearch foralternativemethods, whichmayrevealeitherthehigherordersofperturbationtheoriesoreventhenonperturbative regime

HereweshallconsiderthemassiveΦ4 theoryasalaboratoryforimplementingamethodthatcanbefurtherappliedtomore compre-hensivemodels.Thereisan ongoingdebatewithregardtothebehavioroftherenormalizedcouplingλ atsmallmomentareferredtoas

“thetrivialityproblem”[9–11] Withthehopethatourapproachmightshedlightevenonthisproblem, weintroduceanewvariablein thepathintegralformalismwhichallowsforamoretractablefunctionalintegrationandseriesexpansion.Thenwecomputeinthisnew methodthecorrectionsto themassofthe scalarinall orderofperturbation theory.Thisapproach shouldbe regardedasan alternate renormalizationprocedure.Sincethe corresponding massanomalous dimension γ (m2) =d ln m2

d μ2 hasthefirst order(oneloop)universal coefficient, weverifythatthefirstordercorrectioniscorrect.However, weexpectthatthenextordersaredifferent

2 The set-up

Weshallillustrateourapproachforasimplescalartheory,givenbytheLagrangian:

L = L0+ L1, L0=1

2(∂μΦ)



∂μΦ 

−1

2m

2Φ2, L1= − λ

Forconvenience,wewillworkbothintheMinkowskiandEuclidianspace

ThegeneratingfunctionalintheEuclidian spaces hastheexpression:

W[J] =



dΦexp



−



d4x



1 2



∂Φ

∂ τ

2 +1

2(Φ)

2+1

2m

2

0Φ2+ λ

4! Φ

4+ JΦ



(2)

andcanbewrittenas

W[J] =exp

 

d4xL1



δ

δJ



where

E-mail address:rjora@theory.nipne.ro

http://dx.doi.org/10.1016/j.physletb.2014.12.050

0370-2693/©2014 The Author Published by Elsevier B.V This is an open access article under the CC BY license ( http://creativecommons.org/licenses/by/4.0/ ) Funded by SCOAP 3

W0[J] =



dΦexp

 

d4x( L0+ JΦ)



FromEq.(3)isclearhowtheperturbativeapproachcanwork.Ifλisasmallparameter, onecanexpandtheexponentialintermsofλ

andsolvesuccessive contributionsaccordingly.However, weareinterestedintheregime whereλislargeandonecannotusetheabove expansion

Wewillillustrateourapproachsimplyonasimplefunction.Assumethat wehavethefollowingone-dimensionalintegralwhichcannot

besolvedanalytically:

I=



dx exp

where f ispolynomialofx.Fora small, theexpansionina makessense.Fora→ ∞, theTaylorexpansionuses:

lim

a→∞

d nexp[−af(x) ]

whichdoesnotleadtoacorrectanswer

Weshall use,however, asimple trick.We replaceinthepolynomial f someofthe variables x witha newvariable y (forexample

x4→x2y2).Thenwewrite:

I=



dxdyδ(x−y)exp

−af(x,y) =



dxdydz exp

−i(x−y)z exp

−af(x,y) =



dxdydz exp

−i(x−y)z−af(x,y) (7)

Thisdoesnothelptoomuchinthepresentform However, if f(x,y) =x2y2 oranyother functionthatcontains x2, wecanformthe perfectsquare:

−ixz−ax2y2= − √

axy+ iz

2√

a y

2

− z2

IntroducedinEq.(7)thisleads:

I=const



d√1

a y dz exp



− z2

4ay2



Thenexpansionin 1a makessenseandonecanwrite:

I=const



dxdz√1

a y



1− z2

4ay2+



Thisexpansionmayseemilldefinedandhighlydivergent.Forexample, ifoneintegratesoverz,thenone alreadyencountersinfinities However, inthefunctionalmethodoneisdealingwithfunctionsinsteadofsimplevariablesandone encountersdivergencesalsointhe usualexpansioninsmallparameters.We willconsidertheaboveapproachasourstartingpointandsolvetheproblemofdivergencesas theyappear

WewillstartwiththesimplepartitionfunctionforaΦ4 theorywithoutasource:

W[0] =



dΦexp



i



d4x[ L0+ L1]



(11)

WeconsidertheextendedfunctionalδdefinedintheMinkowskispaceas(seeAppendices Aand B):

δ(Φ) =const



dK exp



i



d4x M KΦ



(12)

whichintheEuclidian spacebecomes:

δ(Φ) =const



dK exp



−



d4xKΦ



(13)

WethenrewriteEq.(11)in Minkowskispaceas

W[0] =



dΦdΨ δ(Φ − Ψ )exp



i



d4x



L0− λ

8Φ

2Ψ2



=const



dΦdΨdK exp



i



d4xK(Φ − Ψ )



exp



i



d4x



L0− λ

8Φ

2Ψ2



=const



1

√

λdΦdK exp



i



d4x2

λK

2



exp



i



d4xKΦ2



exp



i



d4xL0



(14)

Inordertoobtainthisresult, wemadethefollowingchangeofvariableinthesecondlineofEq.(14):K→KΦ,Ψ → Ψ

Φ√

λ.Notethat theλtermgetsrescaledby3 suchthattotakeintoaccountthevariouscontributionoftheFouriermodes

WewillestimatethefirstorderoftheintegralinEq.(14)givenby:

 1

√

λdΦdK

1 exp



i



d4xKΦ2



exp



i



d4xL0



(15)

Inordertosolvetheintegral, wewrite:



d4x

KΦ2+ L0 =



d4x



1

V2

k n+k m+k p=0

(Re K m+i Im K m)(ReΦn+i ImΦn)(ReΦp+i ImΦp)

− 1

2V

k n



m20−k2n

WedenotethebilinearformintheexponentialinEq.(16)by:

Φ



K

V2− 1

2V



2K0

V − m20−p n2

[δ2n+1, 2n+1+ δ2n+2, 2n+2]



wherethecountingstartsfromn=0 andwe arrangedforexampletheReΦn andImΦn componentsinthe2n+1,respectively, 2n+2 columnsofaninfinitelydimensionalvector

ThentheintegralinEq.(15)canbesolvedeasilyasaGaussian integral:

const



1

√

λdΦdK exp



i



d4xKΦ2



exp



i



d4xL0



=



det[K

V2+ 1

2V[2K0

V − (m20−p2) [δ2n+1, 2n+1+ δ2n+2, 2n+2]]]1/2.

(18)

NotethatonecanwritealsoaresultforthefullpartitionfunctioninEq.(14):

const

 1

√

λdΦdK exp



i



d4x2

λK

2



exp



i



d4xKΦ2



exp



i



d4xL0



=



dK exp



i



d4x2

λK

2



1 det[K

V2+ 1

2V[2K0

V − (m20−p2) [δ2n+1, 2n+1+ δ2n+2, 2n+2]]]1/2. (19)

Thenextstepistodeterminethepropagatorthroughthisprocedure

3 The propagator

Thepropagatorisgivenby:

Ω|TΦ(x1)Φ(x2) |Ω = lim

T→∞

dΦΦ(x1)Φ(x2)exp[iT

−T d4xL ]

dΦexp[iT

ForourpartitionfunctionEq.(20)isrewrittenas

Ω|TΦ(x1)Φ(x2) |Ω =

 1

√

λ dΦdKΦ(x1)Φ(x2)exp[i

d4x2λ K2]exp[i

d4xKΦ2]exp[i

d4xL0] 1

√

λ dΦdK exp[i

d4x2λ K2]exp[i

d4xKΦ2]exp[i

d4xL0]

= 1

V2

m

exp

−ip m(x1−x2) i V

δ δ( m2−p2m )

dΦdK exp[i

d4x2λ K2]exp[i

d4xKΦ2]exp[id4xL0]

dΦdK exp[i

d4x2λ K2]exp[i

d4xKΦ2]exp[id4xL0] (21) Note that the first line in Eq (21) is the standard definition of the two-point function The second line in Eq (21) needs some clarification.From thefirstlineintheequation itcanbe seenthat thescalartwo-point function mayreceive contributionseitherfrom thekinetictermorfromthetermsthatcontain K Weneedtoshowthatalsothesecondlineisjustified.Inordertoseethat, oneshould considerthesimplefunctionalintegralinEq.(21)andtreatitindependentlywithout anyreferencetotheFeynmandiagrams Thenthe firstlineofEq.(21)leadsalsoto:

Ω|TΦ(x1)Φ(x2) |Ω

=

 1

√

λ dΦdKΦ(x1)Φ(x2)exp[i

d4x2λ K2]exp[i

d4xKΦ2]exp[i

d4xL0] 1

√

λ dΦdK exp[i

d4x2λ K2]exp[i

d4xKΦ2]exp[i

d4xL0]

= 1

V2

m

n

exp[−ip m x1]exp[−ip n x2]i V2

dΦdK exp[i

d4x2λ K2] δ

δ K (−p m−p n )exp[i

d4xKΦ2]exp[id4xL0]

dΦdK exp[i

d4x2λ K2]exp[i

d4xKΦ2]exp[id4xL0]

= 1

V2

m

n

exp[−ip m x1]exp[−ip n x2](−i)V2

dΦdK δ K (−p δ m−p n )exp[i

d4x2λ K2]exp[i

d4xKΦ2]exp[id4xL0]

dΦdK exp[i

d4x2λ K2]exp[i

d4xKΦ2]exp[id4xL0]

= 1

V2

m

n

exp[−ip m x1]exp[−ip n x2]



−2

λ



V

dΦdK K(p n+p m)exp[i

d4x2λ K2]exp[i

d4xKΦ2]exp[id4xL0]

dΦdK exp[i

d4x2K2]exp[i

d4xKΦ2]exp[id4xL0] (22)

Weshallattempttoestimatetheintegraloverthemodes K(p)inEq.(22).Forthatweexpand theexponentialofthetrilinearterm In firstorder, wegettheterm

which evidently brings contributiononly forq= −r, p m= −p n sothe only Kmode that contributes isthe zero mode.In third order (secondorderiszero), weobtaintermsofthetype:

K(p m+p n)K( −q1−r1)K( −q2−r2)K( −q3−r3)Φ(q1)Φ(r1)Φ(q2)Φ(r2)Φ(q3)Φ(r3) (24)

Ifanyoftheq i= −r i, wearebacktothepreviouscasewhereonlyK0modecontributes Assumewithoutlossofgeneralitythatq1= −q2,

r1= −r3,q3= −r2.Thissettles theintegraloverΦ whereasfor K weobtain:

There are three possibilities for this integral: (1) p m+p n=q1+r1, q1−r2= −r1−r2, (2) p m+p n= −q1+r2, q1+r1=r1+r2,

(3) p m+p n= −r1−r2,q1+r1=q1−r2.All ofthesepossibilities leadto p n= −p m.Thisargument continuesforhigherorders inthe expansionsuchthatquitejustifiedwecanexpressthepropagatorfromthebeginningasthederivativewithrespecttom2−p2

m Sincethequantitym2−k m2 appearsonlyinthedeterminantinEq.(19), wecancompute:

δ

δ(m20−p m2)



det



K

V2+ 1

2V



2K0

V − m2−p2

δ2n+1, 2n+1+ δ2n+2, 2n+2

−1/2

= −1

2



det



K

V2− 1

2V



2K0

V − m2−p2

δ2n+1, 2n+1+ δ2n+2, 2n+2

−1/2

×Tr



1

K

V2 + 1

2V[2K0

V − (m20−p2)(δ2n+1, 2n+1+ δ2n+2, 2n+2) ] ( −1)



1

2V(δ2m+1, 2m+1+ δ2m+2, 2m+2)



=const1

2



det



K

V2− 1

2V



2K0

V − m20−p2

δ2n+1, 2n+1+ δ2n+2, 2n+2

−1/2 2

2

V K0− (m20−p m2)

=const



det



K

V2− 1

2V

2K0− m2−p2

δ2n+1, 2n+1+ δ2n+2, 2n+2

−1/2

1

2K0

V − (m2−p2

In Eq.(26) the first three lines are the simple resultof differentiating a determinant The first factor in the third line of Eq.(26)

contains theFourier modesofthe field K with momenta differentthan zero(pμ=0) denoted simplyby K and those withmomenta

pμ=0 denotedby K0.However, themodeswith pμ=0 areirrelevantforthereasonweshalloutlinebelow.First, letusconsider K(x)

asasquareintegrablefunctionintheHilbertspacewhichsatisfies:



d4xK2(x) = 1

V

p

whereMisaquantitylargebutfinite.Thismeansthat K ( V p )<

√

M

√

V.Incontrast, K0

V isfiniteasisgivenby:

K0



Wecould havedropped fromthebeginningthe factor V1 fromits expression butit helpswithdimensional analysis.Thus althoughwe shallkeep K(p=0) intheexpression atsome pointthe limit V → ∞ willbe takensuch thatall thesetermsin thedeterminantwill cancelandtheintegraloftheexponentialoftheK(p n)termsinthenumeratorwillgetcanceledbythatinthedenominator.Inconclusion thezeromodeisusedasasubstituteforallthe othermodesandsumsupalltheircontribution

Themode K0 actslikeanadditionalcontributiontothescalarmassandneedstobemaintainedandintegratedover.Inconsequence

inallcalculationsthatfollow, oneshouldconsideronlythemodesK0 whichsimplifiesthecalculationsconsiderably

ThenEq.(21)becomes:

Ω|TΦ(x1)Φ(x2) |Ω = 1

V2

m

exp

−ip m(x1−x2) i V

×

V K0−(m2−p2

m )exp[i

det[K

V 2+1

2V[2K0 V −( m2−p2)(δ 2n+ 1, 2n+ 1+δ2n + 2, 2n+ 2)]]1 

dK exp[i

det[K

V 2+ 1

2V[2K0

V −( m2−p2)(δ 2n+ 1, 2n+ 1+δ2n + 2, 2n+ 2)]]1

(29)

Wedenote:

1

λ

2

V =ba0, m20−p2m=c2, 2

V =a0,

det



K

V2+ 1

2V



2K0

V − m20−p2

(δ2n+1, 2n+1+ δ2n+2, 2n+2)



dK dK0exp[2iba0K20+ d4x2ibK2] 1

( a0K0−c2)[det[a0K0+B]]1 

dK dK0exp[ibK2+ d4xibK2] 1

[det[a0K0+B]]1

= −1

c2

dK dK0exp[2iba0K02+ d4x2ibK2][1+a0K0

c2 +a2c K42+ ] 1

[det[a0K0+B]]1 

dK dK0exp[ibK02+ d4xibK2] 1

[det[a0K0+B]]1

(31)

Weextractedafactorof V1 fromthedeterminantanddropped thecorresponding constantfactoreverywhere.Inorderto determine theratioinEq.(31), weevaluateeachtermintheexpansioninthedenominator:

I n=



dK0dK(a0K0)

n

c 2n exp

2iba0K02 exp

 

d4x2ibK2

det[a0K0+B] −1/2

=



dK0dK 1

a0

d[( a0K0) n+ 1

c 2n ( n+1)]

2iba0K20 exp

 

d4x2ibK2

det[aK0+B] −1/2

= −



dK0dK 1

a0

(a0K0)n+1

c 2n(n+1) (4iba0K0)exp

2iba0K02 exp

 

d4x2ibK2

det[aK0+B] −1/2

−



dK0dK (a0K0)

n+1

a0c 2n(n+1)exp

2iba0K02

k

a0 a0K0−c k2



exp

 

d4x2ibK2

det[a0 K0+B] −1/2

= − 4ibc4

a0(n+1)I n+2−



dK dK0

(a0K0)n+1

c 2n(n+1)

k

1

c k2



1+a0K0

c k2 + (a0K0)2

c k4 +



×exp

2iba0K02 exp

 

d4x2ibK2

det[a0K0+B] −1/2. (32)

Hereweusedtheformulaofdifferentiationofadeterminant

FromEqs.(32)and(39), weobtainthefollowingrecurrenceformula:

(n+1)I n+I n+2c4



4ib

a0 + 

k

1

c4

k

 +I n+1c2

k

1

c2

k

+ +I n+r c 2r

k

1

c 2r k

First, wemultiplythewholeEq.(33)by V1 andthenintroduce I n c 2n= J n togetthenewrecurrenceformula:

1

V(n+1)J n+ J n+1

1

V

k

1

c2k + J n+2



2ib+ 1

V

k

1

c4k



Finally, sincewedenotedthepartitionfunctionbyI0 fromEqs.(29)and(34), onecanderive:

Propagator= −1

c2

n

I n/I0= −1

c2

n

1

where J0=I0isthefullpartitionfunction

Beforegoingfurther, weneedtodeterminethecoefficientsinEq.(34).Forthatwefirststate

1

V

k

1

c k 2r = 1

V

k

1

(m20−p2k)r= (−1)r



Notethat onlytheintegrals withk=1,2 are divergentwhereastheotheronesarefinite.Weshalluseasimplecut-offto regularize themuponthecase.Thenweget:

q1=i 1

16π2



Λ2−m2ln



Λ2

m20



16π2



−1+ln



Λ2

m20



, q n , n >2=i 1

16π2

(m2)2−n

4 Discussion and conclusions

Theterms J n inthetwo-point functioninEq.(35)correspondtovariousloopcorrectionsandonecancuttheseriestoobtainresults

invariousorders ofperturbationtheory.However, weshallnot attempttodothishere Wewillratheraimto obtainifpossiblean all orderresultforthecorrectiontothemassofthescalar.Wedothiswiththehopethattheapproachinitiated herecanbeextendedeasily

totheorieswithspontaneous symmetrybreaking andevento thestandard model.Itis clearthat anapproach that couldestimate the correctiontotheHiggsbosonmasscould beofgreat interest Onecanwritequitegenerallyan exactexpressionforthepropagatorofa scalar:

wherem isthephysicalmassandM2(p2)istheone particleirreducibleself-energy (forsimplicity, we rename p2m=p2 fortherestof thepaper).Inourapproachthepropagatorisgivenby:

i

p2−m20

n

( −1)n J n

I0

1

NowifweidentifyEq.(38)withEq.(39)andexpandthefirstequationinseriesin 1

( p2−m2) n, weobtain:

i

p2−m2

n

( −1)n J n

I0

1

p2−m2−M2(p2) ( −1)n J n

I0 = m2−m20−M2

p2 n+Y n

p2

J n

I0 = m20−m2−M2

p2 n+ (−1)n Y n



p2

whereY n(p2)isanarbitraryserieswiththeproperty:

n

Y n



Nowweshallconsiderthefollowingrenormalizationconditionswhichstate:

M2

p2

p2=m2=0, dM

2(p2)

dp2

WeapplythefirstconditiontoEq.(40)todeterminethat:

( −1)n J n

I0 p2=m2= m2−m20n

+Y n

m2

= m2−m20n

1+ Y n (m2−m2)n



(43)

Wewillshow thattheterm Y n(m2)inEq.(43)should besetto zero.Forthat wefirstnote thatfromEq.(41)onecan deducethat thereisatleastone n forwhich Y n ( m2)

( m2−m2) n<0.Thenthereisasolutionm forwhich(m2−m2)n= − 1

Y n ( m2)= α (n)n with α (n)real.This solutionisazeroofthecorresponding J n.But J n(m2)hastheexpression:



dK0K0n 1

sohasapoleata0K0= α (n)insteadofazero.Weobtainacontradictionwhichmeansthatthereisnon suchthat Y n ( m2)

( m2−m2) n <0 sothe seriesinEq.(41)hasallthetermsY n(m2) =0

Thenwesimplytake:

J n

Wedenote

andsumintherecurrenceformulainEq.(34)alltermswiththeindicesn+k, k≥3 for p2=m2:

k≥3

J n+k

I0 q k=X n

k

i

16π2m4



X

m20

n

1

(n−1)(n−2) = i

16π2X n+1



X+ m2−X

ln

m2−X

m20



(47)

Thentherecurrenceformulabecomes:

(n+1)a0X n+q1X n+1+



2i

λ +q2



X n+2+ i

16π2X n+1



X+ m20−X

ln

m2

0−X

m20



=0

(n+1)a01

X +q1+



2i

λ +q2



16π2



X+ m20−X

ln



m20−X

m20



=0

q1+



2i

λ +q2



16π2



X+ m20−X

ln



m20−X

m2 0



Hereinthelastlinewetookthelimita0= 1 →0

Notethat although we used the conditions in Eq.(42), we should not consider our approach equivalent withany ofthe standard renormalizationprocedures

ThenEq.(48)willbecome:

q1+ m20−m22i

λ +q2



16π2



m20−m2

+m2ln



m2

m20



Theequation above determinesthe physicalmassin termsofthe baremassandof thecut-offscale Insteadwe observethat fora largecut-offscaleonecandivideEq.(49)byq1 andretainthefirstandsecondterms Then,

m2≈m20+ 2i q1

λ +q2≈m20+

Λ2−m20ln[Λ2

m2]

1+ λ

32π2[−1+ln[Λ2

m2]]

λ

Notethat thisresult leads tothe same firstorder coefficient ofthe massanomalous dimension asin the standard renormalization procedures

Acknowledgements

The work of R.J was supported by a grant of the Ministry of National Education, CNCS-UEFISCDI, project number PN-II-ID-PCE-2012-4-0078

Appendix A

Inthefollowing, wewillshowthattherelationinEq.(13)makesenseperfectsenseinthefunctionalapproach.Westartwith:



dK exp



−



d4xKΦ



k0>0



d Re K n d Im K nexp



−1

V(ReΦn+i ImΦn)(Re K n+i Im K n)



k0>0



d Re K n d Im K nexp



−1

V ReΦn Re K n− i

V ImΦn Re K n



exp



1

V ImΦn Im K n− i

V ReΦn Im K n



Next, letusconsideraregularintegralofthetype:



dx exp[−ipx−ap]dx=



dx



1− (ap) +1

2(ap)

2+



exp[−ipx]

=



dx



1− (−i)aδ

δ +a21

2( −i)2 δ

2

δ 2+



exp[−ipx]

=



1− (−i)aδ

δ +a21

2( −i)2δ

2

δ 2+



LetusapplythisresulttoEq.(A.1)withthevariable a replaceddependingonthecasebyReΦn orImΦn:



dK dΦf(Φ)exp



−



d4xKΦ



=const 

k0>0

d ReΦn d ImΦn f(ReΦk,ImΦk) ×



1−ReΦn( −i) δ

δIm+



δ(ImΦn)

×



1−ImΦn( −i) δ

δReΦn+



δ(ReΦn)

Wewillprovethatbyconsideringafewtermsintheaboveexpansion.Thezerothordertermcontainstwodeltafunctionsandclearly leadsto f(0,0).Anotherpossibletermis:

 

k0>0

d ReΦn d ImΦn f(ReΦk,ImΦk)ReΦn

δ

δImΦn δ(ImΦn)δ(ReΦn) =0 (A.4)

byvirtueoftheδ(ReΦ )function.Anotherpossibletermis:

 

k0>0

d ReΦn d ImΦn f(ReΦk,ImΦk)ReΦn δ

δImΦnδ(ImΦn)ImΦn

δ

δReΦnδ(ReΦn)

= − 

k0>0

ReΦnImΦn

δ

δReΦn



f(ReΦk,ImΦk)ReΦn

δ

δImΦn δ(ImΦn)ImΦn

δ

δReΦn



δ(ReΦn)

= − 

k0>0

ReΦnImΦnImΦn

δ

δImΦn δ(ImΦn)f(0,ImΦn) =f(0,0) (A.5)

Itcanbeshownthatallothertermsareeitherzeroorproportionalto f(0,0)whichconcludesourproofthattheintegralinEq.(A.3)

givesawelldefineddeltafunction

Appendix B

Weshallpresenthereanapproximateestimateofthepropagatorforanarbitraryregularizationschemeforthelimitoflargecouplingλ

WestartfromtherecurrencerelationinEq.(34)whichwerewritehereforcompleteness:

1

V(n+1)J n+J n+1

1

V

k

1

c2

k

+ J n+2



2ib+ 1

V

k

1

c4

k



Wedenote:

1

V

k

1

wheres n maybeconsideredinanyregularizationscheme.First, wewillmakeachangeofvariableK0=√K0

λandrewrite J n= S n

λ n/2 where

S n represents the samequantity as J n, this timein thevariable K0.Notethat withthischangeof notationfor λ verylarge thefactor exp[2iba0

K0

λ ]inEq.(32)becomesnegligible

TherecurrenceformulabecomesintermsofS n:

1

V(n+1)S n/λn /2+S n+1/λ( n+1)/2s1+S n+2/λ( n+2)/2[2ib+s2] +S n+3/λ( n+3)/2s3+ =0 (B.3)

ForlargeV andλ, oneobtainsinfirstorder:

S n+2= −S n+1

√

λs1

todetermine S n= (−1)n−1S1[ √λ s1

2ib+s2]n−1= (−1)n−1x n−1λ( n−1)/2S1

Inordertodetermine S1, weconsiderthezerothorderrecurrencerelation:

1

V S0+ 

n=3

S1( −1)n−1x n−1λ( n−1)/2/λn /2s n=0 (B.5)

Thisyields:

S1= −1

V

√

n=3

( −1)n−1x n−1s n



(B.6)

AccordingtoEq.(35), thepropagatorisgivenby:

Propagator= −1

c2

n

1

c 2n



S nλ−n /2

/S0= −1

c2



S0+ 

n=1

1

c 2n( −1)n−1x n−1λ( n−1)/2λ−n /2S1

S0

= −1

c2



S0+ √ 1

λ(c2+x)S1



S0= −1

c2



1− 1

c2+x

1

V[ n=3( −1)n−1x n−1s n]



wherex= s1

2ib+s2 (see thenotation inEq.(30)).A straightforwardcomputationforthequantitiesinEq.(B.7)inthelimitoflargeλleads to:

Propagator≈ −1

c2



1− Λ2 1

1

c2+ Λ2



whereΛ2=function of(Λ2) < Λ2.Notethatinthenotationinthepaper, c2=m2−p2

Thisisaparticularcaseof“triviality” knowntobe afeatureofthe Φ4 theoriesinthelimit whereλ → ∞.Toshow this, werewrite

Eq.(B.8)as

Propagator= 1

p2−m20



1− Λ2 1

1

m20+ Λ2−p2



= Λ2− Λ21

Λ2

1

p2−m20+ Λ21

Λ2

1

p2−m20− Λ2 (B.9)

Accordingto[12]atheoryistrivialinthestrongcouplingregimeifthepropagatorcanbewrittenas

Propagator= 

n

Z n

where Z n (allofthemcanbezeroexceptone)aretheweightsandm n arethespectra inthelargecouplinglimit.Asitcanbeobserved easilyourresultinEq.(B.9)isaparticularcaseoftrivialitywiththemassesm21=m20andm22=m20+ Λ2

References

[1] A.A Valdimirov, D.I Kazakov, O.V Tarasov, Sov Phys JETP 50 (3) (1979) 521.

[2] S.G Gorishny, A.L Kataev, S.A Larin, L.R Surguladze, Phys Lett B 256 (1991) 81.

[3] T van Ritbergen, J.A.M Vermaseren, S.A Larin, Phys Lett B 400 (1997) 379, arXiv:hep-ph/9701390.

[4] J.A Vermaseren, S.A Larin, T van Ritbergen, Phys Lett B 405 (1997) 327–333, arXiv:hep-ph/9703284.

[5] H Kleinert, J Neu, V Schulte-Frolinde, K.G Chetyrkin, S.A Larin, Phys Lett B 272 (1991) 39, arXiv:hep-th/9503230.

[6] A.L Kataev, S.A Larin, JETP Lett 96 (2012) 61, arXiv:1205.2810 [hep-ph].

[7] P.A Baikov, K.G Chetyrkin, J.H Kuhn, J Rittinger, J High Energy Phys 1207 (2012) 017, arXiv:1206.1284 [hep-ph].

[8] C Pica, F Sannino, Phys Rev D 83 (2011) 11601, arXiv:1011.3832.

[9] K.G Wilson, J.B Kogut, Phys Rep 12 (1974) 75.

[10] I.M Suslov, arXiv:0804.0368, 2008.

[11] D.I Podolsky, arXiv:1003.3670, 2010.

[12] M Frasca, Int J Mod Phys A 22 (2007) 2433–2439, arXiv:hep-th/0611276.

inallcalculationsthatfollow, oneshouldconsideronlythemodesK0... 1

( a< /small>0K0−c2)[det[a< /small>0K0+B]]1... n)termsinthenumeratorwillgetcanceledbythatinthedenominator.Inconclusion thezeromodeisusedasasubstituteforallthe othermodesandsumsupalltheircontribution

Themode