Problem 10.1: One-loop structure of QED

Transcription

1 Problem 10.1: One-loo structure of QED In Secton 10.1 we argued form general rncles that the hoton one-ont and three-ont functons vansh, whle the four-ont functon s fnte. (a Verfy drectly that the one-loo dagram contrbutng to the one-ont functon vanshes. There are two Feynman dagrams contrbutons to the three-ont functon at one-loo order. Show that these cancel. Show that the dagrams contrbutng to any n-ont hoton amltude, for n odd, cancel n ars. (b The hoton four-ont amltude s a sum of sx dagrams. Show exlctly that the otental logarthmc dvergences of these dagrams cancel art (a The one-ont functon s gven by the dagram k Π µ 1-t (q2 µ (1 q k Alyng the Feynman rules we have [ Π µ 1-t (q2 Tr d 4 k ( ] k + m e (2π 4 ( eγµ k 2 m 2 0. (2 e + ɛ The trace of m e γ µ and the ntegral d 4 k kν f(k 2 At one-loo, the three-ont functon s vansh gvng the result (2. β β 3-t α k k 3 3 Π αβγ k + + α k k + k 3 (3 γ γ where the momentum flow n the loo s counter-clockwse and the momentum of the external hotons s drected nward towards the loo. The frst grah n equaton (4 contans electrons whle the second grah contans ostrons. In the second grah we use mnus the momentum of the frst grah so that k s the same for both grahs ths s due to the fact that the momentum flow s n the same drecton as n the frst dagrams and that the trace s taken aganst the fermon flow, whch s reversed relatve to the frst dagram. Alyng the QED Feynman rules we fnd Π αβγ 3-t d 4 k ( e 3 3 (2π 4 ((k + 2 m 2 e (k 2 m 2 e ((k 2 m 2 e Tr [ γ α ( k m e γ β ( k + m e γ γ ( k 3 + m e γ α ( k m e γ γ ( k m e γ β ( k + m e ]. (4 1

2 The frst term n the trace of (4 s Tr [ γ α ( k + γ β kγ γ ( k ] + m 2 etr [ γ α ( k + γ β γ γ + γ α γ β kγ γ + γ α γ β γ γ ( k ] (5 whle the second term s Tr [ γ α ( k γ γ kγ β ( k + ] m 2 etr [ γ α ( k γ γ γ β + γ α γ γ kγ β + γ α γ γ γ β ( k + ] Tr [ ( k + γ β kγ γ ( k γ α] m 2 etr [ γ β γ γ ( k γ α + γ β kγ γ γ α + ( k + γ β γ γ γ α] Tr [ γ α ( k + γ β kγ γ ( k ] m 2 etr [ γ α γ β γ γ ( k + γ α γ β kγ γ + γ α ( k + γ β γ γ]. (6 Snce (6 s the exact negatve of (5, the three-ont functon (4 vanshes. Note that n the second lne of (6 we have used the fact that the trace of a roduct of gamma matrces s the same as the trace of the reversed roduct: Recall that the charge conjugaton matrx, C γ 0 γ 2, satsfes Therefore, C 2 1 and Cγ µ C (γ µ T. (7 Tr [γ α1 γ α2... γ αn ] Tr [CCγ α1 CCγ α2 CC... CCγ αn CC] (1 n Tr [ C(γ α1 T (γ α2 T... (γ αn T C ] (1 n Tr [ (γ αn... γ α2 γ α1 T ] (1 n Tr [γ αn... γ α2 γ α1 ] Tr [γ αn... γ α2 γ α1 ] (8 where n the last lne we have used the fact that n must be even to get a non-zero trace. The n-ont functon for n odd s gven by Π α1...αn n-t k 2 k 1 k n + k 2 k 1 k n +... (9... n... n α n α n where k n k and k k + l1 l for 1 n 1. Alyng the Feynman rules we obtan d 4 k ( e n n Π α1...αn n-t (2π 4 (k1 2 m2 e... (kn 2 m 2 e Tr [ ( k n + m e γ αn... ( k 1 + m e γ α1 + γ α1 ( k 1 + m e... γ αn ( k n + m e ] (10 We can show, for odd n, that the trace from the second dagram of (9 s the negatve of the trace from the frst dagram n (9: Tr [ γ α1 ( k 1 + m e... γ αn ( k n + m e ] Tr [ CCγ α1 CC( k 1 + m e CC... CCγ αn CC( k n + m e CC ] Tr [ C( γ α1 T ( k T 1 + m e... ( γ αn T ( k T n + m e C ] [ (( ( 1 n Tr k n + m e γ αn... ( k 1 + m e γ α1 ] T Tr [ ( k n + m e γ αn... ( k 1 + m e γ α1]. (11 2

3 Therefore the frst two terms n the n-ont functon vansh for odd n. There are more dagrams other than those shown exlctly n equaton (9. However, each dagram has a ar where the artcles n the loo are antartcles and vce versa. In exactly the same way that the frst two dagrams of (9 cancel, each other ar of dagrams cancel. Thus, the n-ont functon for odd n vanshes art (b The dagrams for the 4-t functon are gven n Fg. 1. To calculate the dvergent art of each dagram we only need to retan the terms n the numerator wth hghest ower of loo momentum, k. Furthermore, the trace of the dagram wth artcles n the loo s the same as the trace for antartcles n the loo and therefore add. Therefore, the total amltude s M 2(M 1 + M 3 + M 5. (12 α 3 α 3 α 3 k 2 k 1 k 2 k 3 k 1 k 2 k 3 k 1 k k k k α 4 α 4 α 4 α 3 k 1 + α 3 k 1 α 3 k 2 k 1 k + k α 4 k 1 k Fgure 1: Dagrams contrbutng to the 4-t functon. To kee track of the dagrams let us label them by M where 1,..., 6. We begn by evaluatng the dvergent art of M 1 by seres exandng the denomnators and keeng the O(k 4 term [ ] M 1 e 4 d 4 k Tr kγ α4 kγ α3 kγ α2 kγ α1 (2π 4 (k 2 m 2 e 4 + fnte terms. (13 Note that because we know that there wll be no dvergence we can work n 4 dmensons to make k 3 α 4 k 1 k k 3 α 4 3

4 the Drac algebra n the trace smler Tr [ kγ α4 kγ α3 kγ α2 kγ α1] k4 24 (g β 1β 2 g β3β 4 + g β1β 3 g β2β 4 + g β1β 4 g β2β 3 Tr [ γ β4 γ α4 γ β3 γ α3 γ β2 γ α2 γ β1 γ α1] k4 3 (Tr [γα4 γ α3 γ α2 γ α1 ] g α4α2 Tr [γ α3 γ α1 ] 4k4 3 (gα4α3 g α2α1 2g α4α2 g α3α1 + g α4α1 g α3α2. (14 The trace n M 3 s the same as the trace n M 1 wth α 4 α 3 and the trace n M 5 s the same as the trace n M 1 wth α 3. Therefore, M g α4α3 g α2α1 2g α4α2 g α3α1 + g α4α1 g α3α2 + g α4α3 g α2α1 2g α3α2 g α4α1 + g α3α1 g α4α2 + g α4α2 g α3α1 2g α4α3 g α2α1 + g α4α1 g α2α3 0 (15 Problem 10.2: Renormalzaton of the Yukawa Lagrangan Consder the seudo scalar Yukawa Lagrangan, L 1 2 ( µφ m2 φφ 2 + ψ( m ψ ψ g ψγ 5 ψφ (16 where φ s a real scalar feld and ψ s a Drac fermon. Notce that ths Lagrangan s nvarant under the arty transformaton ψ(t, x γ 0 ψ(t, x, φ(t, x φ(t, x, n whch the feld φ carres odd arty. (a Determne the suerfcally dvergence amltudes and work out the Feynman rules for renormalzed erturbaton theory for Lagrangan. Include all necessary counter term vertces. Show that the theory contans a suerfcally dvergent φ 4. Ths means that the theory cannot be renormalzed unless one ncludes a scalar self-nteracton, δl λ 4! φ4, (17 and a counter term of the same form. It s of course ossble to set the renormalzed value of ths coulng to zero, but that s not a natural choce, snce the counter term wll stll be nonzero. Are there any further nteractons requred? (b Comute the dvergent art (the ole as d 4 of each counter term, to the one-loo order of erturbaton theory, mlementng a suffcent set of renormalzaton condton. You need not worry about fnte arts of the counter terms. Snce the dvergent arts must have a fxed deendence on the external momenta, you can smlfy ths calculaton by choosng the momenta n the smlest ossble way. 4

5 Part (a The Feynman rules for the seudo scalar Yukawa theory are lsted n Fgs. 2 and 3. φ φ m 2 φ, ψ ψ m, ψ Fgure 2: Feynman roagators for the Lagrangan (16. gγ 5 Fgure 3: Feynman vertces for the Lagrangan (16. D 2 D 1 D 0 D 0 Fgure 4: All one-loo dagrams along wth ther suerfcal dvergence. Our frst task s to determne all the suerfcally dvergent dagrams at one-loo (Fg. 4. Note that snce the Lagrangan s nvarant under arty, the nteracton cannot change the arty of the ntal state. Ths means that scatterng amltudes wth an odd number of external seudo scalars and no external fermons must vansh. Furthermore, snce there s a dvergent φ 4 dagram n Fg. 4, we must nclude a φ 4 nteracton, δl λ 4! φ4, (18 and a corresondng counter term to renormalze the theory. The new Lagrangan becomes L 1 2 ( µφ m2 φφ 2 + ψ( m ψ ψ g ψγ 5 ψφ λ 4! φ4. (19 5

6 Next, we rescale the felds by the feld strength ψ Z ψ ψ φ Z φ φ (20 to get L 1 2 Z φ( µ φ Z φm 2 φφ 2 λ 4! Z2 φφ 4 + Z ψ ψ( m ψ ψ gz ψ Zφ ψγ 5 ψφ. (21 Defnng Z ψ 1 + δ Zψ Z φ 1 + δ Zφ m 2 ψz ψ m 2 ψ + δ mψ m 2 φz φ m 2 φ + δ mφ Z 2 φλ λ + δλ, g Z φ Z ψ g + δg. (22 where m ψ, m φ, λ and g are now the hyscal masses and coulng constants, the Lagrangan becomes L L 0 + δl, L ( µφ m2 φφ 2 + ψ( m ψ ψ g ψγ 5 ψφ λ 4! φ4, δl 1 2 δ Z φ ( µ φ δ m φ φ 2 1 4! δ λφ 4 + δ Zψ ψ( ψ δ mψ ψψ δg ψγ 5 ψφ. (23 In addton to the Feynman rules of Fgs. 2 and 3, the counter term Lagrangan adds the vertces of Fg. 5. ( δ Zφ δ mφ ( δ Z ψ δ mψ δ g γ 5 δ λ Fgure 5: Counter term vertces. Part (b Now we comute the nfnte art of the counter terms. We wll use an on-shell scheme. We rescale the of the felds by the feld strength n such away that we set the resde of the Fourer transformed 6

7 roagator to 1. Furthermore, to gve meanng to the mass and coulng counter terms we must defne the hyscal mass and hyscal coulng constants. The hyscal mass s defned as the locaton of the ole of the Fourer transformed roagator and the hyscal coulng constants are defned to the magntude of the scatterng amltude at zero momentum m φ + terms regular at m 2 φ, (24 + terms regular at m ψ m ψ, (25 gγ 5, (26 amutated λ at s 4m 2 φ, t u 0. (27 amutated The renormalzaton condtons (24 and (25 can be stated more clearly by defnng the self energes 1PI M 2 (, (28 1PI Σ(. (29 Then the full two-ont functon s gven by the geometrc seres 1PI + 1PI 1PI +... (30 m 2 φ M 2 (, (31 1PI + 1PI 1PI +... (32 m ψ Σ(. (33 7

8 The renormalzaton condtons (24 and (25 can then be stated as Comutaton of δ Zφ and δ mφ At one-loo the scalar self-energy s M 2 ( m 2 0, (34 φ Σ( m ψ 0, (35 dm 2 ( d 0, (36 m 2 φ dσ( d 0. (37 m ψ M 2 ( + +. (38 We evaluate the frst two dagrams above, keeng only the dvergent terms, and then determne the dvergent arts of the counter terms. The frst dagram s (2π d ( λ k 2 m 2 φ λ λ (4π d 2 1 (2π d ( k 2 m 2 φ 1 d 2 Γ(1 d/2 1 m 2 φ λm2 φ 16ɛπ 2. (39 where denotes the equalty of the dvergent terms. The second dagram s [ (2π d Tr (gγ 5 ( k + m ψ k 2 m 2 (gγ 5 ( ] ( k + + m ψ ψ (k + 2 m 2 ψ dg 2 dg 2 (2π d k 2 + k m 2 ψ (k 2 + 2xk + x m 2 ψ 2 d d l (2π d l 2 x(1 x m 2 ψ (l 2 2 (40 8

9 where x(1 x + m 2 ψ. Performng the momentum ntegraton yelds [ dg2 2 + x(1 x + m 2 ψ 4π 2 + ( ( 2 + x(1 x + m 2 4πe γ E ψ log ɛ [ g2 2 + x(1 x + m 2 ψ 16π x(1 x 2m 2 ψ ɛ ] + O(ɛ g2 4π 2 ɛ g2 4π 2 ɛ ( 2 ( 3x(1 x m 2 ψ 2 m2 ψ Therfore, the self energy s + ( ( 2 + x(1 x + m 2 4πe γ E ψ log M 2 ( λm2 φ 16ɛπ 2 g2 4π 2 ɛ ( 2 Next we evaluate the dervatve of the self-energy Then, the renormalzaton condtons mly Comutaton of δ Zψ and δ mψ 2 m2 ψ ] + O(ɛ (41 + ( δ Zφ δ mφ. (42 dm 2 ( d g2 8π 2 ɛ + δ Z φ. (43 δ Zφ g2 8π 2 ɛ, At one-loo the fermon self-energy s δ mφ 1 16ɛπ 2 ( λm 2 φ + 4g 2 ( m 2 φ m 2 ψ. (44 Σ( +. (45 We start by evaluatng the frst dagram gγ 5 ( k + + m ψ gγ 5 (2π d ((k + 2 m 2 ψ (k2 m 2 φ g 2 g 2 k + m ψ (2π d ((k + x 2 + x(1 x xm 2 ψ (1 xm2 φ 2 d d l 1 [(1 x m ψ ] (2π d (l 2 2 (46 9

10 where x(1 x + xm 2 ψ + (1 xm2 φ. Performng the momentum ntegral we obtan g 2 16π 2 g2 16π 2 ( 4π Γ(ɛ [(1 x m ψ ] ( ( ( 4πe γ 2 m ψ ɛ + log E ( 2 m ψ g2 16ɛπ 2 + O(ɛ. (47 Therefore, u to fnte terms, the fermon self energy s ( Σ( g2 16ɛπ 2 2 m ψ + ( δ Z ψ δ mψ. (48 Its dervatve s then dσ( d g 2 32ɛπ 2 + δ Z ψ. (49 The renormalzaton condtons then mly δ Zψ g2 32ɛπ 2, δ mψ g2 m ψ 16ɛπ 2. (50 Comutaton of δ δg Snce the tree level amltude already satsfes the renormalzaton condton, we requre the one-loo contrbuton to the 3-ont ψψφ amltude to vansh + 0. (51 one-loo, amutated We start by evaluatng the followng loo dagram g 3 γ 5 ( k + + m ψ γ 5 ( k + m ψ γ 5 (2π d ((k + 2 m 2 ψ ((k 2 m 2 ψ (k2 m 2 φ g 3 2! dy ( k k + k k + 2m ψ k m ψ ( m 2 ψ (2π d ((k + x y 2 (, ; x, y 3 γ 5 (52 where (, ; x, y 2xy x(1 x y(1 y 2 + (x + ym 2 ψ + (1 x ym2 φ. Shftng the loo momentum to l k + x y we obtan g 3 2! dy l 2 + f(, ; x, y (2π d (l 2 (, ; x, y 3 γ5 (53 10

11 where f(, ; x, y x(1+x +y(1+y 2 +2xy +(1 x y +m ψ (2x 1 m ψ (2y 1 m 2 ψ. Snce the renormalzaton condtons are secfed for on-shell fermons, f s to be sadwtched between the fermon snors, ū( f(, ; x, yu( ū( (( x(4 + x + y 2 2 m 2 ψ + 2(1 x y + xy u(. (54 Furthermore, we are only nterested n the lmt where. Therefore, we take and ū( f(, ; x, yu( ū(f(, ; x, yu( (, ; x, y (, ; x, y ū( (x(2 + x y(2 y m 2 ψu( (55 (x y 2 m 2 ψ + (1 x ym 2 φ. (56 Wth these consderatons we obtan 2g 3 l 2 + f(, ; x, y dy (2π d (l 2 (, ; x, y 3 γ5 [ 2g 3 dy ( 2 d/2 d 1 Γ(2 d/2 (4π d/2 2 (4π d/2 ( 2 d/2 [ 2g 3 Γ(2 d/2 1 d dy (4π d/2 2 f ] (2 d/2 γ 5 [ ( 2g3 2 4πe γ 16π 2 dy ɛ + 2 log E f ] + O(ɛ γ 5 g3 4π 2 ɛ γ5 dy g3 4π 2 ɛ γ5. ( 3 d/2 1 Γ(3 d/2f] γ 5 Thus, the dvergent art of the one-loo contrbutons to the total amltude, evaluated for on-shell fermons at zero momentum, becomes one-loo, amutated The renormalzaton condtons then mly δ g (57 g3 4π 2 ɛ γ5 + δ g γ 5. (58 g3 4π 2 ɛ. (59 Comutaton of δ δλ Snce the tree level amltude already satsfes the renormalzaton condton, we requre the one-loo contrbuton to the 4-ont scalar amltude to vansh 11

12 + + one-loo, amutated 0. (60 We start by evaluatng the followng box dagram 4 g 4 [ Tr γ 5 ( k + m ψ γ 5 ( k m ψ γ 5 ( k m ψ γ 5 ( k m ψ ] (2π d (k 2 m 2 ψ ((k m 2 ψ ((k m 2 ψ ((k + 2 m 2 ψ g 4 [ Tr ( k + m ψ ( k m ψ ( k m ψ ( k m ψ ] (2π d (k 2 m 2 ψ ((k m 2 ψ ((k m 2 ψ ((k + 2 m 2 ψ (61 Followng the advce gven n the queston we secfy to a secfc frame to smlfy the momentum. Let us choose Then ths dagram becomes [ g 4 Tr ( k + m ψ ( k m ψ ( k + m ψ ( k m ψ ] (2π d (k 2 m 2 ψ 2 ((k + 2 m 2 ψ ((k + 2 m 2 ψ g 4 f(k,, 3! (1 x dy (2π d (k 2 + 2(y + (x y k + y 3 + (x y2 1 m2 ψ 4 6g 4 (1 x dy f(k,, (2π d ( 4 (62 (k + y + (x y 2 (, ; x, y where f(k,, 4k 4 + 4k 2 ( k ( + 2m 2 ψ + 8k k 4m 2 ψ (k ( + + m 4 ψ, (63 and (, ; x, y y(1 y 3 (x y (1 (x y 1 + 2y(x y + m 2 ψ. (64 Shftng the loo-momentum to l k + y + (x y, mosng the on-shell condton, and settng as requred of the renormalzaton condton, we obtan where 6g 4 (1 x dy f(l,, ; x, y 4l 4 + m 2 ψl 2 ( 8 16x + 16x 2 d d d l f(l,, ; x, y (2π d (l 2 (, ; x, y 4 ( x + 8x 2 + m 4 ( ψ 5 + 8x 4x 2 8x 3 + 4x 4 (66 12

13 Only the frst term of f contrbutes to the dvergent art of the amltude, 24g 4 24g 4 24g 4 3g4 4π 2 ɛ (1 x (1 x dy (1 x (4π d/2 Next we evaluate the scalar bubble dagram d d l l 4 (2π d (l 2 (, ; x, y 4 d d l l 4 (2π d (l 2 (, ; x, y 4 ( 2 d/2 1 Γ(2 d/2 (67 ( λ( λ (2π d (k 2 m 2 φ ((k m 2 φ λ 2 λ 2 λ 2 λ 2 1 (2π d (k 2 + 2xk ( + + x 1 + x2 2 m2 φ 2 1 (2π d (k 2 + x( x(1 x( x2 m 2 φ 2 d d l 1 (2π d (l 2 + x(1 x( x2 m 2 φ 2 ( 2 d/2 1 (4π d/2 x(1 x( x2 + m 2 Γ(2 d/2 φ λ2 16π 2. (68 Fnally, the dvergent art of the full one-loo contrbuton to the amltude s one-loo, amutated The renormalzaton condtons then mly 3g4 4π 2 ɛ + λ2 16π 2 δ λ. (69 δ λ 3g4 4π 2 ɛ + λ2 16π 2. (70 13