2 Lagrangian and Green functions in d dimensions

Transcript

1 Renormalzaton of φ scalar feld theory December 6 Pdf fle generated on February 7, 8. TODO Examne ε n the two-pont functon cf Sterman. Lagrangan and Green functons n d dmensons In these notes, we ll use η =.. Lagrangan, counterterms, and Feynman rules Consder the scalar lagrangan wth cubc nteracton L = φ m φ 6 d gµ 6 φ. where φ s the renormalzed fnte feld, and m and g renormalzed fnte parameters, but not drectly dentfable as physcal parameters. It also depends on d to regularze the loop ampltudes, and a scale µ to make the couplngs dmensonless n d dmensons. The counterterm lagrangan L c.t. = δz φ δm φ 6 d δgµ 6 φ δτφ. contans poles n ɛ and also depends on arbtrarly chosen fnte constants c. Addng the counterterm Lagrangan to the orgnal lagrangan, we obtan the full Lagrangan L = L + L c.t. = + δz φ m + δm φ 6 d g + δgµ 6 φ δτφ. Renormalzed ampltudes are fnte as ɛ, but depend on m and g, as well as µ and c. The choce of c s determned by the renormalzaton scheme. If we choose an on-shell scheme, m and g wll be drectly related to physcal propertes. The c wll nvolve log µ, and the ampltudes wll be ndependent of µ. Alternatvely, n an MS or MS scheme, the c only cancel poles and do not depend on µ. Physcal couplngs are functons of m, g, and µ. Ignorng the tadpole counterterm for now, we can defne φ = + δz / φ, m = + δz m + δm, g = µ 6 d + δz / g + δg.4 and then rewrte L = φ m φ 6 g φ.5 showng that physcal quanttes depend only on two parameters m and g. Ths then can be used to derve renormalzaton group equatons.

2 The propagator and three-pont vertex are gven by p m + ε, wth counterterms δm + δz p, 6 d gµ.6 δg µ 6 d.7. Two-pont functon We now construct the two-pont Green functon T φφ = p m + Σ δm + δz p p m p m + = + Σ + δm δz p + p m p m = Σ + δm δz p p m p m = p m Σ δm + δz p = Γ p where Σ s the self energy dagram, and Γ p the PI two-pont functon. Expandng.8 Σp = Σm p + Σ m pp m p + Op m p.9 about the physcal mass m p, we fnd that the denomnator of the two-pont Green functon s m p m Σm p δm + δz m p + Σ m p + δzp m p + Op m p. The physcal mass s determned by the solutons of and the two-pont functon at the pole s gven by m p m Σm p δm + δz m p =. T φφ = R p m p where the fnte feld renormalzaton s gven by R = Σ m p + δz.. Now make the one-loop approxmaton, settng m p = m n the terms that are already frst order, to fnd m p = m + Σm + δm δz m.4 R = + Σ m δz.5

3 . Self energy The one-loop self energy s Σp, m = gµ 6 d d d l π d l m + ε l + p m + ε.6 evdently UV dvergent for d 4 and IR dvergent for m = for d. Σp, m = g µ 6 d I = g µ 6 d = g µ 6 d m d 4 d 4 Γ 4 d 4π d/ Γ 6 d 4π d/ = g µ 6 d p d 4 Γ 6 d d 44π d/ dx m x xp ε d 4 dx x x p m dx x x m d 4 p d 4.7 wth the expected UV poles n d = 4, 6,. In the massless lmt we have d 4 dxx x = Γ d /Γd so Σp, = g µ 6 d p d 4 d 4 Γ 6 d d Γ 4π d/ Γd.8 wth the IR pole at d =. We also obtan p p, m = g µ 6 d m d 6 Γ 6 d 4π d/ = g µ 6 d p d 6 Γ 6 d 4π d/ dx x x x x p m d 6 dx x x x x m p d 6.9 In the massless lmt we have d 6 p p, = g µ 6 d p Γ 6 d 4π d/ Γd Γ d.

4 .4 Three-pont functon The three-pont functon s Γ p, p = g + δgµ 6 d + gµ 6 d = g + δgµ 6 d + d d l π d l + p m + ε l + p + p m + ε l m + ε gµ 6 d I p, p, p p. Ths has a UV dvergence for d 6. From the loop ntegral notes, we have so I p, p, p = Γ p, p = µ 6 d 6 d Γ 4π d/ g + δg + g µ 6 d Γ 6 d 4π d/ wth the expected UV poles at d = 6, 8,. dβ dβ dβ δ β m β β p β β p + p β β p d/. dβ dβ dβ δ β m β β p β β p + p β β p In the massless m = lmt, and wth two legs onshell p = p =, we have Γ p, p = µ 6 d = µ 6 d = µ 6 d g + δg + g µ 6 d d 6 p + p 4π d/ Γ 6 d dβ β d 6 g + δg + g µ 6 d d 6 p + p 4π d/ Γ 6 d d 4 g + δg + g µ 6 d Γ 6 d d 4 d Γ Γ 4π d/ p + p d 4Γd β d/. dβ β d 6 dβ β d 6 β d 4.4 For l = l + p, the denomnator goes as l l p l p, gvng an IR dvergence n d = 4. In fact, looks lke a double pole, suggestng also a collnear dvergence?.5 Four-pont functon The PI four-pont functon has contrbutons from three topologcally-dstnct box functons Γ 4 = Bs, t + Bt, u + Bu, s.5 4

5 where where Bs, t = = gµ 6 d gµ 6 d I 4 s, t = 4 d d l π d l m + ε l + p m + ε l + p + p m + ε l + p + p + p m + ε 4 I4 s, t.6 8 d Γ 4π d/ 4 j= δ 4 d j c jβ j j j β dβ j m j β j <j β β j P j 4 d/.7 We see the expected UV dvergence for d 8. Choosng c =, and lettng external legs be on-shell P = P = P 4 = P 4 = m I 4 s, t = 8 d Γ 4π d/ 4 j= dβ j δ j β j m β β β β β β 4 β 4 β β β t β β 4 s 4 d/.8 In the massless lmt I 4 s, t = 8 d Γ 4π d/ 4 j= dβ j δ j β j β β t β β 4 s 4 d/.9 If the external legs are on-shell, then n the l lmt, the denomnator goes to l l p l p 4 and thus has an IR dvergence n d = 4..6 Four-pont scatterng ampltude Frst we compute the sum of all one-loop truncated dagrams M = Γ 4 + Γ s Γ s Γ s + Γ t Γ t Γ t + Γ u Γ u Γ u = Γ 4 Γ s Γ s Γ t Γ t Γ u Γ u The one-loop external leg correctons yeld 4 Rm p Γ 4 Γ s m Γ s Γ t Γ t Γ u Γ u =.. 5

6 LSZ tells us to multply ths by 4 = p m Rm. to obtan the scatterng ampltude M = Rm Γ 4 Γ s Γ s Γ t Γ t Γ u Γ u. 6

7 Renormalzaton n sx dmensons For convenence, defne F x x p m.. Self energy n sx dmensons Evaluate eq..7 near sx dmensons d = 6 ɛ, expandng n ɛ to wrte Thus usng we obtan Σp, m = Σp, m = In the massless lmt m so g Γ + ɛ 4π ɛ ɛ = g 4π 4πµ ɛ dx m F ɛ m ɛ γ + + log dx m F = dx 4πµ m dx m F ɛ log F + Oɛ. m x xp = m p g m 4π p 4πµ ɛ γ + + log m m dx F log F p Σp, = dx x x g 4π Also evaluatng eq..9 near d = 6 ɛ p p, m = g Γ + ɛ 4π ɛ log p m p ɛ γ log 4πµ ɛ dx x xf ɛ m + logx x = p log 4πµ p dx F log F + Oɛ = g 4πµ 4π ɛ γ + log dx x x ɛ log F + Oɛ m = g 4πµ 4π ɛ γ + log + m whch we could also have obtaned by takng the dervatve of eq..4 above...4 p 5 6 p m.5.6 dx x x log F + Oɛ.7 7

8 The massless lmt can alternatvely be obtaned by evaluatng eqs..8 and. near sx dmensons Σp, = g µ ɛ p ɛ ΓɛΓ ɛ ɛ 4π ɛ Γ4 ɛ = g p 4π ɛ γ + 8 p log 4πµ p p, = g µ ɛ p ɛ ΓɛΓ ɛ 4π ɛ Γ4 ɛ = g 4π ɛ γ + 5 p log 4πµ.8.9. Sx dmensonal wavefuncton and mass counterterms To render the feld renormalzaton R = + Σ m δz fnte, we choose the counterterm δz = g 4π ɛ + c φ. n order to absorb the UV pole n Σ m cf. eq..7. We evaluate where to obtan and thus dx x x log F = p m, m = g dx x x log x + x = π 7 8 g 4π R = + 4π F F p =m = x + x. ɛ γ + log c φ + γ log 4πµ m 4πµ m + π π note that π 6.. R s now UV fnte, but stll has an IR dvergence when m. Ths means we wll not be able to do on-shell renormalzaton for the massless theory. Equaton.4 mples the mass counterterm must satsfy δm /ɛ = Σm /ɛ + m δz /ɛ so δm = g m 4π ɛ + c m.5 Then usng eq.. and = dx log F = dx F log F = log x + x dx = π x + x log x + x dx = π

9 we evaluate Σm, m = m g 4π 5 ɛ γ + log 4πµ m + π From ths we can evaluate the physcal mass m p = m + Σm + δm δz m namely m p = m { + g 4π c φ c m + 5 γ + log 4πµ m } + π Renormalzed two-pont functon n sx dmensons Fnally, we evaluate the renormalzed two-pont functon n sx dmensons T φφ = where Γ p = p m + g m 4π p δz = Usng eq.. and m g 4π p m Σ δm + δz p = dx x x log F + Γ p 4πµ m dx F log F + c m + γ log m 4πµ p c φ + γ log, m 4πµ c φ + γ log dx F log F = m π p π m.9.. we can also wrte these expressons as Γ p = p m + g m F dx F log + c 4π m + γ + π 4πµ m F log m c φ + γ + π 4πµ p log, m 4πµ p δz = g 4π F dx x x log + c φ + γ + π 7 F log m. 9

10 In the massless m lmt m F dx F log so that F p = p log dx x x log p m p m + 4 π Γ p p g log p + c 4π 4πµ φ + γ 8 Wth on shell renormalzaton.8 ths agrees wth Srednck eq... x x + log x + x..4 Alternatvely, usng eqs..8 and.9 n the massless lmt, wth δz gven by eq.. and δm absent Γ p = p g p log + c 4π 4πµ φ + γ 8 p δz = g p log + c 4π 4πµ φ + γ 5.5 In the MS scheme, we have c m = γ + log 4π and c φ = γ + log 4π so cf Srednck eq. 7.4 Γ p = p m + g m µ m dx F log F + log 4π m p p δz = g dx x x log F µ 4π log.6 m For the massless theory, MS gves Γ p = p g p δz = g 4π p log 8 4π µ p log 5 µ.7 On-shell renormalzaton, only avalable for m, mples R = 4πµ c φ + γ log = 7 π m Thus p δz = g 4π = g 4π dx x x log F + 7 π 6 F dx x x log F.8.9

11 whch vanshes at p = m. On-shell renormalzaton also mples m p = m so c φ + γ log 4πµ m c m + γ log 4πµ m 4πµ c m + γ log so that the nverse propagator wth on-shell renormalzaton s Γ p = p m + g m π dx F log F + 4π = p m + g m 4π F dx F log F m = π 4 6 = 5 6π 8 m + + p m agreeng wth Srednck eq. 4.4 whch manfestly vanshes at p = m. Accordng to Srednck, p. 4, dx F log F = m p m S tanh whch I verfed numercally. Mathematca gves dx F log F = p 5y 4 8 whch agrees numercally wth Srednck. Hence Γ p = p m + g π m + 4π + 4 y / tan y 4 y y.4 Three-pont functon n sx dmensons. π 4 p 6., S = 4m. S p π 9 p p 6 6 S tanh y = p m..4 S Evaluatng the three-pont functon n d = 6 ɛ, we get Γ p, p = µ ɛ g + δg + g µ ɛ Γɛ dβ dβ dβ δ β 4π ɛ ɛ m β β p β β p + p β β p = µ ɛ g + δg + g 4π ɛ + Oɛ.5 Choose the counterterm g δg = 4π ɛ + c g.6

12 to cancel the UV dvergence Γ p, p = µ ɛ g + g cg + Oɛ 4π.7 Instead evaluate the massless three-pont functon.4 wth two on-shell legs n sx dmensons Γ p, p = µ ɛ g + δg + g 4π ɛ log p + p γ + 4πµ = µ ɛ g + g c 4π g log p + p γ +.8 4πµ MS renormalzaton wth c g = γ + log 4π gves Γ p, p = µ ɛ g + g log p + p 4π µ Four-pont functon n sx dmensons Recall that Γ 4 = Bs, t + Bt, u + Bu, s.4 where Bs, t = g 4 I 4 s, t.4 and I 4 s, t = 4π 4 j= dβ j δ j β j m β β β β β β 4 β 4 β β β t β β 4 s.4 In the massless lmt, ths s accordng to Srednck.7 gven by Bs, t g4 4π 4 j= dβ j δ j β j β β t β β 4 s = g4 4π s π + log s + t t.4 I checked ths result numercally..6 Four-pont massless scatterng ampltude n sx dmensons The scatterng ampltude s M = Rm Γ 4 Γ s Γ s Γ t Γ t Γ u Γ u.44

13 where Rm = + g 4πµ c 4π φ + γ log + π 7 = + g m m 6 4π log + const µ.45 Rm has an IR dvergence as m but the expresson n brackets n eq..44 does not. Thus we can evaluate t n the massless lmt n MS renormalzaton Γ s log g s Γ s g 4π µ = s log g g g s 8 s 4π log sµ π µ and Bs, t Thus the scatterng ampltude s { M = Rm g g s 4π g t g u g4 π 4π u + s log.47 t log sµ + t log + π u 5 9 g 4π log t + u µ log + π s 5 9 g 4π log uµ + s log + π t 5 } 9.48 Ths s consstent wth Srednck.9. Fnally, wth s t u, and neglectng constants, ths together wth eq..45 gves M = g s g t g + g u 4π log sµ + 6 m log.49 µ.7 Infrared dvergences from collnear partcles Srednck very clearly explans how the emsson of collnear scalar partcles gves rse to IR dvergences such that eq. 6.4 M obs = M + g δ 4π log s.5 m Combnng ths wth the result above M obs = g s g t g + g u 4π 6 log sµ + m log µ + δ log s m.5

14 we see that the log m dependences cancels gvng M obs = g s g t g + g u 4π log sµ + log δ.5 whch s fnte n the m lmt, but depends on the resoluton of the detector. Ths equaton can be used to derve the one-loop beta functon. 4

15 4 Renormalzaton n four dmensons 4. Four dmensonal self energy As before, defne F x x p 4. m Evaluate eq..7 near four dmensons d = 4 ɛ, expandng n ɛ to wrte Thus Σp, m = g µ Γ + ɛ 4π ɛ If p <, we can recast = g µ 4π Σp, m = g µ 4π 4πµ ɛ dx F ɛ m ɛ γ + log ɛ γ + log 4πµ dx ɛ log F + Oɛ 4. m 4πµ m dx log F + Oɛ 4. p F = x x m p = m p m 4 S y, y = x, S = 4m > 4.4 p then p dx log F = log + Thus for p <, dx log F = m / / dy log S + y + log S y p = log + y=/ S + y log S + y S y log S y y m y= / p S S + y S y=/ = log + m log + y log S y 4 y y y= / p S + S = log + S log + log m S 4 S + = S log 4.5 S 4m p 4m + p log 4m p 5 { log p /m, p p < p /6m, p 4.6

16 If < p < 4m, we can analytcally contnue S = p 4m p = eα T, α = ± π, T = mplyng that ndependently of α + S dx log F = S log = e α + e α T log T = e S α T T p 4m p, < T < 4.7 arctan T 4.8 Thus for < p < 4m 4m dx log F = p p arcsn p 4m { p /6m, p + + π m 4m p, p 4m 4.9 In the massless m lmt so dx log F dx log Σp, = g µ 4π p m + logx x = log ɛ γ + + log Equvalently we could evaluate eq..8 at d = 4 ɛ p 4. m 4πµ + Oɛ 4. p Σp, = g µ 4πµ ɛ Γ + ɛγ ɛ 4π p ɛγ ɛ 4. and expand to obtan the same result. Also evaluatng eq..9 near d = 4 ɛ p p, m = g µ 4π m = g 4π µ m 4πµ ɛ Γ + ɛ dx x xf ɛ dx m x x F 4. whch s the dervatve of the expresson above. Takng the massless lmt we get p p, = g µ 4.4 4π p 6

17 4. Four dmensonal wavefuncton and mass counterterms Snce Σ p s UV fnte, so s the counterterm δz = g µ c φ 4π m 4.5 For m =, we must choose c φ =. Evaluatng x x x x dx = dx F x + x = π we obtan p m, m = g µ 4π m π so the feld strength renormalzaton R = + Σ m δz s R = + g µ 4π m π c φ Equaton.4 mples the mass counterterm must satsfy δm /ɛ = Σm /ɛ + m δz /ɛ so Next we evaluate dx log F = δm = g µ 4π ɛ + c m 4.9 dx log x + x = π to obtan Σm, m = g µ 4π ɛ γ + log 4πµ + π + Oɛ 4. m Now we can evaluate the physcal mass m p = m + Σm + δm δz m to be m p = m + g µ c 4π φ + c m + γ log 4πµ m + π 4. 7

18 4. Renormalzed two-pont functon n four dmensons Fnally, we evaluate the renormalzed two-pont functon T φφ = p m Σ δm + δz p = Γ p 4. where Also Γ p = p m + g µ 4πµ p c 4π m γ + log + c m φ m p δz = g µ c 4π m φ dx x x F dx log F + Oɛ In the massless lmt, wth c φ =, we have usng logx x dx = Γ p = p + g µ 4πµ c 4π m γ + + log + Oɛ 4.6 p On-shell renormalzaton mples R = so Thus p δz = whch vanshes at p = m. c φ = π g µ + π 4π m On-shell renormalzaton also mples m p = m so 4πµ c m + γ log m dx x x F = c φ π + = 5π so that Γ p = p m + g µ 4π whch vanshes at p = m. 5π + π p m dx log F + Oɛ 4. 8

19 4.4 Three-pont functon n four dmensons The three-pont functon n d = 4 ɛ s Γ p, p = µ 6 d g + δg + g µ 6 d Γ 6 d 4π d/ = µ +ɛ g + δg + g µ +ɛ 4π Γ + ɛ ɛ dβ dβ dβ δ β m β β p β β p + p β β p dβ dβ dβ δ β m β β p β β p + p β β p d/ +ɛ 4. Ths s fnte for m so we can set ɛ = : Γ p, p = µ g + δg + g µ dβ dβ dβ δ β 4. 4π m β β p β β p + p β β p In the massless lmt, and wth p = p =, we have Γ p, p = µ +ɛ g + δg + g µ +ɛ Γ + ɛ 4π ɛ p + p = µ +ɛ g + δg + g µ 4π ɛ p + p Ths s also the result obtaned from eq Four-pont functon n four dmensons +ɛ ɛ γ γ ɛ + π β dβ β ɛ dβ β ɛ 4. The PI four-pont functon has contrbutons from three topologcally-dstnct box functons where Γ 4 = Bs, t + Bt, u + Bu, s 4.4 Lettng external legs be on-shell P = P = P 4 = P 4 = m I 4 s, t = Γ 4π 4 j= Bs, t = gµ +ɛ 4 I4 s, t 4.5 dβ j δ j β j m β β β β β β 4 β 4 β β β t β β 4 s 4.6 9

20 In the massless lmt I 4 s, t = 8 d Γ 4π d/ 4 j= dβ j δ j β j β β t β β 4 s 4.7 If the external legs are on-shell, then n the l lmt, the denomnator goes to l l p l p 4 and thus has an IR dvergence n d = 4. Need to study ths!