• No results found

(2)d.) The eld-strength of a Yang-Mills theory transforms homogenously under a loal gauge, F(x)!F 0 (x)=U(x)F(x)U y (x)

N/A
N/A
Protected

Academic year: 2022

Share "(2)d.) The eld-strength of a Yang-Mills theory transforms homogenously under a loal gauge, F(x)!F 0 (x)=U(x)F(x)U y (x)"

Copied!
6
0
0

Laster.... (Se fulltekst nå)

Fulltekst

(1)

NTNU Trondheim, Institutt for fysikk

Examination for FY3464/8914 Quantum Field Theory I

Contat: Mihael Kahelrie,tel. 99890701

Allowed tools: mathematialtables

1. Misellaneous and quiz

a.) Write down A

for (3pts)

A=u(p

2 )

u(p

1 )

b.) Calulate (3pts)

tr[

℄:

.) The ovariant derivative of a Yang-Mills theory transforms under a loal gaugetrans-

formationU(x) as: (2pt)

2 D

!D 0

=D

2 D

!D 0

=U(x)D

2 D

!D 0

=U(x)D

U

y

(x)

2 D

!D 0

=U(x)D

U

y

(x)+ i

g (

U(x))U y

(x)

a.) Starting from

A

=A y

=(u y

(p

2 )

0

u(p

1 ))

y

=u y

(p

1 )

y

0y

u(p

2 );

and using y

= 0

0

and ( 0

) 2

=1,wearrive at

A

=u(p

1 )

u(p

2 ):

b.) Contrating (1) with

gives

2

=2

=8

or

=4. Togetherwithtr(1)=4 we nd

tr[

℄=2

= 244= 32:

.) TheovariantderivativeofaYang-Millstheorytransformshomogenouslyunderaloalgauge,

D!D 0

(x)=U(x)DU y

(x).

(2)

d.) The eld-strength of a Yang-Mills theory transforms homogenously under a loal gauge,

F(x)!F 0

(x)=U(x)F(x)U y

(x).

2. Salar eld.

Consider a real, salar eld with mass m and self-interation g 3

.

a.) WritedowntheLagrangedensityL,explainyourhoieofsignsandpre-fators(when

physially relevant). (6pts)

b.) Write down the generatingfuntional for onneted Greenfuntions. (4pts)

.) Determine the mass dimension in d=4 spae-time dimensions of all quantities in the

Lagrangedensity L. (4pts)

d.) Drawthe divergent one-loop diagramsand determine their superialdegree of diver-

gene D (ind=4spae-time dimensions). (6pts)

e.) Determine the number d of spae-time dimensions for whih the theory is

renormalisable. (4pts)

a.) The freeLagrangian is

L

0

= 1

2

1

2 m

2

2

:

The relative sign is xed by the relativisti energy-momentum relation, the overall sign by the

requirement that the Hamiltonian isbounded from below. The fator 1/2 inthe kineti energy

leads to \anonially normalised" eld, the fator 1/2 for the mass follows then from the rel-

ativisti energy-momentum relation. As the self-interation is odd, adding +

3!

3

or

3!

3

is

equivalent: both hoies will lead to an unstable vauum. In order to reprodue the Feynman

rule,weused asnormalisationthefator 1=3!,

L =L

0 g

3!

3

:

b.) We setm 2

!m 2

i"as dampingtermand add asoureJ oupled linearlyto theeld,

L

e

=L +J: (1)

The generating funtionalZ for disonneted Green funtions is the pathintegralover eldsof

exp(i R

d 4

xL

e g),

Z[J℄= Z

Dexpfi Z

d 4

xL

e g=e

iW[J

; (2)

whileW[J

℄generates onnetedGreen funtions.

.) The ationS = R

d d

xL is(for~=1)dimensionless. Thekinetiterm[() 2

℄=m 4

xesthe

dimensionoftheeldasm 1

,onsistentwiththeinterpretationofm inthemasstermasmass,

[m℄=m 1

. This impliesthattheoupling ghas thedimension[g℄=m 1

.

d.) The primitive divergent diagrams arethedivergent 1-loop diagrams. We an order them by

thenumberE of external(bosoni)legs anddetermine thesuperialdegree ofdivergeneDby

(3)

naive power-ounting, seethe lastpage fortheFeynman diagrams.

E =0and D=4 orresponding a ontributionto theosmologial onstant,

E =1and D=2 orresponding to a tadpolediagram,

E = 2 and D =0 orresponding to the self-energy. Note that the vertex orretion, E =3, is

already nite.

(The vauum graphs (E = 0) are optional { you may prefer to \hide" them by asking for a

properly normalizedgeneratingfuntional.)

e.) We have to ndd suh that [g℄=m 0

: For general d, it is [℄=m (d 2)=2

. Only solutionfor

[ 3

℄=d isthusd=6with[℄=m 2

.

3. Fermion with Yukawa interation.

Consider a Dirafermion with mass m interating with real salar eld with mass M

through a Yukawa interation,

L =

(i= m) ig

5

+ 1

2

1

2 M

2

2

:

a.) Determinethe global(internal) symmetriesof the free,masslessfermioni Lagrangian,

L =

i= , and the resultingNoether urrents. (6pts)

b.) Calulatethe self-energy(p= )atone-loopofafermionwith momentump 2

6=m 2

using

dimensional regularisation. Express (p=)as (12 pts)

(p= )= A

"

+Bln(D=

2

)℄:

.) What is your interpretationof the funtional formof A? (3pts)

d.) What is your interpretation of the dependene of the self-energy onthe parameter ?

[.) and d.): max. 50words explanation.℄ (3pts)

a.) Consider globalphasetransformations: FirstU

V

(1), hange phi to #

(x)! 0

(x)=e i

(x) and

(x)!

0

(x)=e i

(x);

keepthe Lagrangianinvariant,ÆL =0. Noether's theorem (12)leads thenwithÆ =i to

j

= ÆL

Æ(

)

Æ + ÆL

Æ(

) Æ

=

i

i +0: (3)

Thusthe vetor urrent isonserved. Next lookat axialtransformationsU

A (1),

0

(x)!e i

5

(x) and

(x)!

0

(x)=(e i

5

(x)) y

0

=

(x)e i

5

: (4)

The resulting(innitesimal)hange is

L 0

=

0

i= 0

=

(1+i 5

)i=(1+i 5

) =

(1 i)i=(1+i) = (5)

(4)

and thusagainÆL =0(for m=0). With Æ =i 5

to

j

= ÆL

Æ(

)

Æ =

i

i 5

(6)

Thusthe axial-vetor urrentis onserved too (form=0).

b.) Followingthefermionlineand usingtheFeynman rules,we have

i(p=)=( ig) 2

Z

d 4

k

(2) 4

5

i

p

=+k= m

5 i

k 2

M 2

:

We ombinerst thedenominatorsand ompletethenthesquare,

D=[(p+k) 2

m 2

℄z+(k 2

M 2

)(1 z)=k 2

+2pkz+(p 2

m 2

)z M 2

(1 z)= (7)

=(k+zp) 2

+p 2

z(1 z) m 2

z M

2

(1 z)q 2

+a (8)

Next we evaluatethenominatorusingp= 5

=

5

p

= and ( 5

) 2

=1,and substitutethen k!q,

N = 5

(p=+k= m) 5

= (p=+k=+m)= (p= (1 z)+q=+m):

The linear term will vanish after integration and we drop it. Addingthe mass sale 4 n

to g

and using,,,,wend

i(p=)=g 2

( 2

) 4 d

i 1

(4)

!

(2 !)

(2) Z

1

0 dz

(p=(1 z)+m)

a 2 !

From thedimensionlessquantity(a=4 2

)

"

,and expand (")and (a=4 2

)

"

forsmall",

(p=)= g

2

16 2

(p= m=2) 1

"

Z

1

0

dz(p= (1 z)+m)ln(a=(4 2

)

.) The oeÆient of the divergent 1=" termis a polynomialin the externalmomentum. More

preisely,they orrespond to terms

i and m

inthelassialLagrangian, and an thusbe

subtratedbymassand wave-funtion renormalisation.

d.) runningparameters

4. Spin-1 elds.

a.) A massive spin-1 eld A

satises the Proa equation,

(

2

)A

+m

2

A

=0:

UsethetensormethodtodeterminethepropagatorD

(k)ofsuhaeld[don'tareabout

the poles℄. (8pts)

b) Give one argumentwhy this methoddoesnot work setting m=0. (3pts)

(5)

a.) Wewritefristm 2

A

=m 2

A

. ThepropagatorD

foramassivespin-1eldisdetermined

by

(2+m 2

)

D

(x)=Æ

Æ(x): (9)

InsertingtheFourier transformationof thepropagator andthe delta funtiongives

k 2

+m 2

+k

k

D

(k)=Æ

: (10)

We willapplythetensormethodto solvethisequation: Inthisapproah,weuserst alltensors

availableintheproblemto onstrut therequiredtensorof rank2. Inthease athand,wehave

at ourdisposalonlythemomentum k

of thepartile|whihwe an ombine tok

k

|andthe

metritensor

. Thusthetensorstruture ofD

(k)has to be ofthe form

D

(k)=A

+Bk

k

(11)

withtwounknown salarfuntions A(k 2

) and B(k 2

). Insertingthisansatz and multiplyingout,

we obtain

( k 2

+m 2

)

+k

k

[ A

+Bk

k

℄=Æ

;

Ak 2

Æ

+Am

2

Æ

+Ak

k

+Bm

2

k

k

;

A(k 2

m 2

+(A+Bm 2

)k

k

: (12)

Inthelaststep,weregroupedtheLHSintothetwotensorstruturesÆ

andk

k

. Aomparison

of theiroeÆients gives thenA= 1=(k 2

m 2

) and

B = A

m 2

=

1

m 2

(k 2

m 2

) :

Thusthe massive spin-1propagator follows as

D

F (k)=

+k

k

=m 2

k 2

m 2

+i"

: (13)

b.) There'samismathofdegreesoffreedom,3$2,betweenthemassiveandmasslessase/The

longitudinalpart k

k

=m 2

whihblowsup form !0does notontribute to themasslessprop-

agator/Theprojetionoperatorfollowingfrom theMaxwellLagrangian hasan eigenvalue0 and

isthusnot invertible.

Feynman rules and useful formulas

f

;

g=2

: (14)

f

; 5

g=0 and ( 5

) 2

=1: (15)

(6)

= i

2 [

;

℄ (16)

= 0 y

0

(17)

1

ab

= Z

1

0

dz

[az+b(1 z)℄

2

: (18)

Z

d 2!

k

(2) 2!

1

[k 2

m 2

+i"℄

=i ( 1)

(4)

!

( !)

( ) [m

2

i"℄

!

: (19)

f

"=2

=1

"

2

lnf+O("

2

): (20)

(z)= Z

1

0 dte

t

t z 1

(21)

(n+1)=n! (22)

( n+")= ( 1)

n

n!

1

"

+

1

(n+1)+O(")

; (23)

1

(n+1)=1+ 1

2

+:::+ 1

n

; (24)

j

= ÆL

Æ

a Æ

a K

: (25)

f

d h

f e

ig

5

f p

f

i(p=+m)

p 2

m 2

+i"

h k

h h

i

k 2

M 2

+i"

Referanser

RELATERTE DOKUMENTER