problem set solution 7

8/13/2019 Problem Set Solution 7

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18.324 QuantumFieldTheoryII

ProblemSet7Solutions

1. (a) Changingthevariable =()wecanfind usingthechainruleofdifferentiation. Wefind,()=d

d =dd

dd =()

dd .

Thus transformslikeacontravariantvector. Forexample,recallhow inGRV transformslikeV(x)=V(x)x

x .

(1)

(2)(b) Using(1),togetherwith

=+a22 +a33 +... (3)and

=b22 +b33 +b44 +..., (4)wefind

=(b22 +b33 +b44..)(1+2a2+3a32 +...), (5)whichtofourthordergives

=b22 +(b3 +2a2b2)3 +(b4 +2a2b3 +3a3b2)4 +O(5).Wemustexpress intermsof. Insteadoffindingintermsofwecalculate

2 =2 +2a23 +(2a3 +a22)4 +...3 =3 +3a24 +..4

=4.andrecallingthat=b22 +b33 +b44 +...wefindthat

(6)

(7)

b2 =b2, b3 =b3, and b4 =b4a2b3 +a3b2a22b2.sothatwecanmake b4 anythingwewantbytweakingtheas.The fixedpoint of a renormalizationflow is defined by (f)=0 which implies (f)=0 (ifexist,whichweassume). Finally

()= dd

()d

d

= dd

()d

dd

d =()+()dd

d2d2 .

dd

(8)

and dd

(9)so

that,

at

a

fixed

point

(f)=(f). (10)

(c) Wehavedg

d =bg2cg3dg4 +... . (11)Forconvenience,redefiningthecouplingandthescale

bg, and tlog(/1), (12)


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+

2werewritethedifferentialequationas

d c ddt =2b23b34. . .. (13)

Wethenrearrangeasd 1 d c d

dt= (14)+. . .=d2 2 b2

2+

. . .

c 1 +

+

2 3b b

Thisequationcanbeintegratedtoyield1 c

t=constant+ + ln+O() +.., (15) b2

wheretheconstant is independentoft(i.e. ). Thismeansthattheconstant isRG invariant. Rewritingtheresult intheoriginalvariables,andabsorbing ln1 intothedefinitionoftheconstant,wehave

1 cln=constant+ + log[bg()]+O(g()). (16)

bg() b2SettingtheconstantequaltotheRG invariantscale,weget

1 cln = + log[bg()]+O(g()). (17) bg() b2

= g(),m1(d) In a generic QFT one would have where m1 is a parameter of the theory with, . . .[m1]=1. Contrarytothemostgeneralform,inatheorywithoutmassparameters,wehave

dg(g) = (18)

dwhich implies

dg d= (19)

(g) integratingbothsidesgivesomefunctionofgontheleft(whichcontainsnodimensionfulparameters),anda logwithauniversalconstantscaleof integrationontherightgiving

F(g)=log (20)

Invertingthisrelationgivesthatg issomefunctionofthe log, i.e.

g=f log . (21)

2. TheLagrangian is

L=21 (0)2

21m0220 +g6006

1 1 A B g2C2g()2 m22 + 6 ()2 22 6 (22)+= m

2 2 6 2 2 6with

2(1+A)2(1+C)Thustocalculatethebetafunction,weneedonlycalculateAandC. This involvescalculatingthemomentumdependent part of the two point function, and the three point function. Both of these relevant computations

3g0 =g (23)


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3have been done in class and/or in previous problem sets. In class, we found that the two point function was(withD=m2 +x(1x)p2)

g2(2d) 12

d1

Ap2m2Bdx= dD22 22(4) 0 1

DdxAp2Bm2=finite 0

=finiteorp independentp2Ap2 (24)6

thus A= ina minimalsubtraction scheme. Similarly, inthe previous homework we foundthat thethree6

pointfunctionwas2

=g+gC+finite+g dxdydz(x+y+z1)

=finite+g C+ (25)thus inMS,C=

. To lowestnontrivialorder3 2

1 122

g0 =g=g

63

14 (26)

Usingthescale independenceofbarequantities,we logdifferentiateeachside,whichgives9 g 3

20 =g 1 14 (27)+4 2

Multiplyingby(1+94)andtakingthe0limitgives

g =34g (28)

Usingthechainruleandthedefinitionofgives =3

22 +O(3) (29)whichimpliesthatthetheoryisasymptoticallyfree.

3. (a) Wehave

L= 12

(1)2 +(2)2

4!(41 +42)24!(2122) (30)

Wecanusemostoftheresultswehavederivedpreviously inthecoursewithregardto4 theory,theonlysubtlety here is with symmetry factors. Once again each vertex comes with a factor ofi. However,becausetherearefewercombinatorialsymmetries,eachvertexcomeswithafactorofi42 =i4! 3.Let us renormalize the vertex. For concreteness, consider 1 external lines (it does not matter which).Wewillhaves,t,uchannelcontributionswithbothtypesofparticlesrunningintheloop. Theresultwillgivesomething likePS(10.21)

2=ii(2 + )(V(t) +V(s) +V(u))i (31)

9AllVshavethesameinfinitepartindimensionalregularization(seeegPS(10.23))

1V

162+finite (32)


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4which implies,intheMSscheme

2 3 = 2 + (33)

9 162We thus have (there is no wavefunction renormalization coming from the two point function at one-looporder)

0 =(+)2 3= + 2 + (34)9 162

which implies2 3 2 3

0 =(1+2) + + 2 + + (35)9 162 9 162

Rearranging,andusingthefactthatthebetafunctionsareO(,),to lowestorderthisimplies2 3 2 3

+ + (36) = 29 162 9 162

We now calculate the renormalization of . Two of the loop diagrams are O() and have one type ofparticlerunningintheloop. TheothertwoareO(2)andhavetwotypesofparticlesintheloop. Wemustmultiplytheselasttwodiagramsbyafactoroftwotoaccountforthechangeinsymmetryfactor. Intotalweget

2 =finitei 2Vi 4Vi

3 9 3=finite

3

2i

16

12 i2

4

9 1612 i

3 (37)which implies,intheMSscheme

2 = +

82 1220 = 1 + +2 (38)

82 3Once

again,

differentiating

and

manipulating

to

lowest

order,

we

get

1 2 1

= 1 +82 3 82 (39)

Substituting(39)into(36)to lowestorderandtakingthe 0limitgives3 2

= 2+ (40)162 9

substitutingintheoppositewaygives = 1 +22 (41)

82 3(b) Usingthechainrule,weget

1 =2()

=3 1 (42)482

whichhaszerosat/={0,1,3}. Wewrotetheresult inthis form,becausestability impliesthat >0,while it is harder to say anything definite about thesign of . This beta function ispositive for /


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5(c) Wehavealreadyestablishedtheexistenceofthesefixedpoints; onlytheasymptoticallysymmetricone is

IR stable. In 4 dimensions, we can read off the beta functions from above (before taking the 0limit)

2 3 =+ 2 + (43)

9 162 =+ 1 +22 (44)

82 3Yoursketchshould involvea simultaneousplotof and , ora line of / withsomearrows to indicatetheflowtotheIR.Atfinitebutsmall,youshouldfindthatthereisastableIRfixedpointatfinite,.Inmyplotthefixedpointsare(obtainedfromthezerosofthe functions):

A = 0 A = 0 AisIRunstable (45)162

B = B = 0 B hasoneunstabledirection (46)3

82C = C = 82 C hasoneunstabledirection (47)

3242 242

D = D = D isIRstable. (48)5 5

We conclude that we get an emergent O(2) symmetry asymptotically in the IR, if we start from theappropriateparameterrange. Thebasinof attraction is calleduniversalityclass inthe statisticalphysicsliterature. Bytuningwecangetalargerangeofenergy,wherethefixpointsBorCdeterminethebehaviorofthe theory. Thereare other universality classes in this spaceofcouplings thatwe cannot explore withtheexpansiontechniques,astheylieinthestrongcouplingregimeofthetheory.

0 20 40 60 80 100

0

20

40

60

80

100

A B

C

D

FIG.1. TheRGflowofthetheory.


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8.324 Relativistic Quantum Field Theory IIFall2010

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problem set solution 7

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