Renormalization of the fermion self energy

Transcription

1 Part I Renormalization of the fermion self energy Electron self energy in general gauge

2 The self energy in n = 4 Z i 0 = ( ie 0 ) d n k () n (! dimensions is i k )[g ( a 0 ) k k k ] i /p + /k m 0 use Z i 0 = e d 4 k 0 () 4 k [(p + k) m 0 ] f ( /p + /k + m 0 ) ( a 0 ) k /k( /p + /k + m 0 ) /kg /k( /p + /k + m 0 ) /k = k /p + (pk) /k + k /k + m 0 k = k /p + f[(k + p) m 0 ] k (p m 0 )g /k + k /k + m 0 k The term k /k cancels and the term [(k + p) m 0 ] /k vanishes by symmetrical integration in n dimensions.

3 Then i 0 = e 0 Z d 4 k () 4 k [(p + k) m 0 ] f ( /p + /k + m 0 ) + ( a 0 )( /p m 0 ) + ( a 0 )(p m 0 ) /k k g The general structure of 0 is 0 = m 0 A(p ) + B(p ) /p : with the scalar functions A and B given by A = m 0 4 Sp 0 B = 4p Sp /p 0 (Sp = 4, also in n dimensions).

4 De ne N = ( /p + /k + m 0 ) + ( a 0 )( /p m 0 ) Then + ( a 0 )(p m 0 ) /k k = (n )( /p + /k) + nm 0 + ( a 0 )( /p m 0 ) + ( a 0 )(p m 0 ) /k k : 4 SpN = nm 0 ( a 0 )m 0 4 Sp /pn = (n )[p + pk] + ( a 0 )p + ( a 0 )(p m 0 )pk k Calculation of A(p ): A(p ) = ie 0 Z d n k (n + a 0 ) () n k [(p + k) m 0 ] The integral is evaluated with the help of Feynman s for-

5 mula ab = Z 0 dx [(a b)x + b] with b = k and a = p + pk + k m 0, (a b)x + b = x p + kp m 0 + k = k + p x + pkx m 0 x = (k + px) p x + p x m 0 x! k + p x( x) m 0 x; Then where A(p ) = ie 0 Z 0 dx Z d n k m 0 (n + a 0 ) () n [k R ] R x( x)p + m 0 x We had calculated the momentum integral in the previous lecture,

6 Z d n k I(; 0) = () n [k R ] = i 6 (4)! (R )! (!) Then A(p ) = ie 0 (3! + a i 0) 6 (4)! Z dx h x( x)p + m 0 0 xi! (!) We expand the integrand for small! ( x( x)p + m 0 x)! =! ln x (x ) p + m 0 x, the x integration becomes elemen- As R 0 ln xdx = tary. Result: +! ln x (x ) p + m 0 x +O! 3 A(p ) = 0 4 (3! + a 0) "! + C ( m 0 p ) ln m 0 p m + 0 #

7 with 0 (e 0! ) 4 C + ln 4 ln m 0 and we put ( )! =! ln +:::. Note how a scale enters in this way into dim. reg. We expand also the rest for small!: A(p ) = " 0 4 (3 + a 0)! + C ( m 0 p ) ln m 0 p m ( ) + # Calculation of B(p ): p B(p ) = ie 0 Z d n k () n k [(p + k) m 0 ] () [ (n )(p + pk) + ( a 0 )p + ( a 0 )(p m 0 )pk k ]

8 Use pk = (k + p) k p Then we get B(p ) = ie 0 Z d n k () n k [(p + k) m 0 ] or f (n ) [(k + p) m 0 + m 0 + p k ] + ( a 0 )p + ( a 0 )(p m 0 )(k + p) m 0 + m 0 k p k ] B(p ) = ie 0 + Z d n ( k n () n k + ( a 0 ) (p m 0 ) k 4 k [(p + k) m 0 ](n 3 + a 0)(p + m 0 ) [(p + k) m ) 0 ](n ) k 4 [(p + k) m 0 ]( a 0)(p m 0 ) The rst two terms are tadpole integrals which vanish.

9 We need the following integrals Z d n k () n = i 6! k [(p + k) m 0 ] () "! + C ( m 0 p ) ln m 0 p # m + 0 Z d n k () n [(p + k) m 0 ] = Z d n k () n = i 6! m 0 [k m 0 ] (3)! + C + Z d n k () n k 4 [(p + k) m 0 ] (4) " = i 6! p m C ( + m 0 0! IR p ) ln m 0 p m 0 The last integral is UV convergent, but IR divergent. It exists for n = 4! IR > 4, while the UV divergent integrals exist for n = 4! UV < 4. It makes sense #

10 to take n complex, then one can set! IR =! UV!. This IR divergence is not real (s. Eq.()), it was created by introducing the tadpole integrals which a cancellation between IR and UV singularities. With these integrals, we obtain where p B(p ) = 0 4 f (! + a 0)(p + m 0 ) [! + C + ( m 0 p )L] + (!)m 0 [! + C + ] [ ( a 0 )(p m 0 )! L ln m 0 p m 0 C ( + m 0 p )L]g ; C = ln 4 ln m 0

11 Expanding some more for small!, we get p B(p ) = 0 4 f ( + a 0)(p + m 0 ) [! + C + ( m 0 p )L] + (p + m 0 ) + m 0 [! + C + ] m 0 Collecting terms, ( a 0 )(p m 0 )[! C ( + m 0 p )L]g p B(p ) = 4 f a 0p 0 (! + C) + a 0 (p + m 0 )( m 0 p )a 0L a 0 (p + m 0 )g

12 Mass renormalization in O(): To rst order, we can put Then 0 ( /p; m 0 ; e 0 ; a 0 ;!) 0 ( /p; m; e; a;!) m = 0 ( /p = m; m; e; a;!) = ma(m ) + mb(m ) Substituting for A(m ) and B(m ) gives m = m 4 3! + C Note: m is gauge invariant and IR nite. This is true in all orders of perturbation theory. From the de nition m m m 0 = m( Z m ) we get Z m = m m, i. e. Z m = 4 3! + C :

13 Details: (3 + a)(! + C + ) A(m ) = 4 B(m ) = 4 f a( + C) 4ag! = 4 m (! + C)( a) a 3 Wave function renormalization to O(): For the renormalization of the wave function we need the derivative of ( /p).. :

14 We work in the Feynman gauge a = for simplicity: To lowest order, we can put e 0 = e and m 0 = m. We had with A(p ) = ie Z 0 dx Z d n k () n [k R ] n n = 4! ; R x( x)p + m. After the integration over k we get Then A(p ) = (e! ) n(4)! (!) 6 Z " x dx p (p m = (e! ) 6 (4)! n (!)(!) Z 0 If we expand dx x( x) #! : " x p (p m )x! (!) = ( +!) = (! + :::) ; # (+!)

15 = p =m 4 Z m n(!) m!! 4 dx( x)x (+!) : 0 We use now the formula Z for = 0 dxx ( x) = B(; ) Z! and = to obtain 0 dx( x)x (+!) = = = (!) () () () ( + ) ; (!) (!) () (!) (!) (!) () (!)( +!) The integration over the Feynman parameter has introduced a new singularity (!) =! + :::.This is an IR singularity which is regularized by staying away from n = 4 keeping! < 0 (n = 4! IR > 4).

16 Better: view dimensions in the complex n- plane. We = m p =m 4 m(4!)(!)! ln 4 ( +!)(!)! = " 4 m 4 +! IR = " 4 m C 3 #! IR : ln m 4 We write! IR to indicate the origin of the divergence,! # n = 4! IR or UV The infrared divergence arises because of the vanishing of the photon mass. One can regularize this divergence

17 by giving a mass to the photon. In the Feynman gauge i 0 = e 0 Z d 4 k () 4 h k i [(p + k) m 0 ] f ( /p + /k + m 0 ) For! 0 we obtain as above A(p ) = Z! 0 dx ln x p (p m + )x + 4 The derivative is = p =m Z 0 dx x( x) x m x + = m ln m + konst: comparison with dim. reg. gives! IR! ln m + konst: ) :

18 : Again we limit ourselves to O() and to the Feynman gauge (e 0! e ; m 0! m ). We had, B(p ) = ie 0 Z d n k () n k [(p + k) m ] [ (n )(p + pk)] and introduce a trick by writing p + pk = [(p + k) m + p + m k ]: The term (p + k) m results in a tadpole integral which vanishes. Then we get B(p ) = ie n 0 If we compare with A(p ) = ie 0 Z Z d n k p + m k () n k [(p + k) m ] d n k () n (n + a 0 ) k [(p + k) m ]

19 we see that we have related B(p ) to A(p ), B(p ) = n n (p + m )A(p ) Z + ie n d n k 0 () n (p + k) m In the second integral, we can let (p + k)! k. Then it becomes independent of = n n " (p + + A(p ) Therefore we obtain for the derivative at p = m = n p =m n 4 C 3! IR 4(! + C + ) #! + 4 = n n 4 " # + C + 3! IR

20 Now we can write Z to O(), Z = = " " = " where we /p F (p ) = /p /p=m Result: with /p 0( /p; m 0 ; e /p 0( /p; m; e) # # /p=m + B 4 + O( F (p ) = m /p=m p =m " + # + 3C + 4! F (p ) C ln 4 ln m 0 ; n = 4! IR;UV : If one calculates in an arbitrary gauge, one nds Z = 4 " a! UV + (3 a)! IR + 3C + 4 # (5) : p =m

21 (Exercise) Remarks:. 9 a gauge where Z is UV- nite, a = 0, the Landau gauge... 9 a gauge where Z is IR- nite, a = 3, the Yennie gauge (only to O()). 3. Z is gauge invariant for! IR =! UV =!.

22 4 The renormalized propagator: The renormalized propagator is given by or S R ( /p) = Z S 0( /p) ; /p m R ( /p) = Z [ /p m 0 0 ( /p)] (6) where = ma(p ) + B(p ) /p : Comparing the coe cients of m and /p in Eq.(6) we get: We write + A R = Z m 0 m ( + A 0) = Z Z m ( + A 0 ) B R = Z ( B 0 ) Z = + z :

23 and obtain to O() + A R = ( + z + z m )( + A 0 ) = + z + z m + A 0 or A R = z + z m + A 0 and B R = ( + z )( B 0 ) = + z B 0 or B R = B 0 z ; We had to O() A(p ) = (3 + a) " 4! + C ( m p )L 3 + a + #

24 B(p ) = 4 where f a(! + C) + a( + m p )( m p )al m a( + p )g L ln m p m ; C = ln 4 ln m 0 With these we nally obtain the die renormalized self energy ( R = m (a + 3) 4 a + 3 ( ) + C) (3 a) 8! RI + /p 4 ( +La a m p + m 4! p 4 + 4! ) L m! p + + (! RI + C) (3 a)!

25 Details: (without the factor 4 m ; z = z ; x = z m ) A = (3 + a)(! + C ( m p )L 3+a + ) B = ( a(! + C) + a( + m p )( m p )al a( + m p )) z = ( a! + (3 a)! IR + 3C + 4) x = 3(! + C ) B z = a m p + + (! RI + C) (3 a) +La m p + m p + 4 z + x + A = (a + 3) a+3 L m p + (! RI + C) (3 a) 8

26 5 Mass renormalization to O( ) We change the notation slightly and write for the unrenormalized self energy of the electrons 0 (p) = m 0 A( m 0 p ; e 0) + ( /p m 0 )B( m 0 p ; e 0) (7) The A and B are now dimensionless. If we work to second order then A and B are of the form A = A + A ; B = B + B where A = O(), A = O( ) etc. The mass renormalization is calculated from Z m = m 0 m = m 0( /p; m 0 ; e 0 ) We get to O( ) /p=m 0 (m) = m 0 [A ( m 0 m )+A ( m 0 m )]+(m m 0)B ( m 0 p );

27 where we have noted that (m m 0 ) is of O() and have suppressed the dependence on e 0. We expand in powers of m 0 =m # 0 (p)j /p'm'm0 = m 0 "A () + ( m 0 m )A 0 () + A () where A 0 = We then obtain for Z m : Z m = + ( /p m 0 )B () + O[( m 0 m ) ] + O( 3 ); ( with 0 =m ) A ( m 0 m ; e 0) : m 0 " A + ( m 0 m m )A 0 + A m 0 m )B ; # A A (); etc.

28 With Then m0 m m 0 m = A + O( ) = A + O( ) Note Z m = A + A + A A 0 A A B Z m = + 0 Z () m where Z m () and Z m () m). + 0 Z() m + ::: are pure numbers (independent of The mass itself changes correspondingly m m m 0 = m( Z m ) : We have de ned the renormalization constants as functions of 0 (e 0! ) 4 0 (e 0! ) 4. Often one carries out the charge renormalization e 0 = Z e e to obtain the Z as function of m and e, e.g. Z m = Z m (e 0 (m; e))

29 We prefer to substitute for 0 (; m) at the end (S-matrix element, Green s function) 6 Renormalization of the wave function to O( ) The renormalization of the wave function is obtained from We use with Z /p /p /p /p 0( /p; m 0 ; e # /p=m 0 (p) = m 0 A( m 0 p ; e 0) + ( /p m 0 )B( m 0 p ; e 0)

30 and get Z = ( m A ( /p m 0 B B ) /p=m = A (m B Calculational A 0 @(m 0 =p ) A(m 0 =p ) B = m 0 p 4 A0 (m 0 =p ) and similarly for B(m 0 =p ). Then we get

31 = m 0 m A ( m m B Z B = + m3 0 m 3A0 (m 0 =m ) m + ( 0 m )m 0 m B0 (m 0 =m ) Now use B A 0 (m 0 =m ) = A 0 () + m 0 m! A 00 () + A 0 () + O( ) = A 0 () + ( A ) A 00 () + A0 () + O( ) with B 0 (m 0 =m ) = B 0 () + O() we get B( m 0 m ) = B() + (m 0 m )B 0 () + O( 3 ) = B () + B () + ( A )[B 0 () + B0 ()] + O( 3 )

32 and m0 m 0 m = A + O( ) = A + O( ) Then we get Z m m0 m 3 = 3A + O( ) = + ( 3A )(A 0 A A 00 + A0 ) + A B 0 + A ( A )B + B + ( A )(B 0 + B0 ) = + B + A 0 + A0 4A 00 A + A B 6A A 0 A B 0 + O(3 ) where A A () etc. With the series expansion + x = x x + ::: we obtain the nal result for Z Z = B A 0 A 0 + 4A00 A A B + 6A A 0 + A B 0 4A 0 + O(3 )

33 Again we have to substitute at the end e 0 = e 0 (e; m), with e 0 = Z e e, see below. Z = Z (e 0 (e; m))