Quantum Field Theory Homework 5

Transcription

1 Quantum Feld Theory Homework 5 Erc Cotner February 19, 15 1) Renormalzaton n φ 4 Theory We take the φ 4 theory n D = 4 spacetme: L = 1 µφ µ φ 1 m φ λ 4! φ4 We wsh to fnd all the dvergent (connected, 1PI (1-partcle rreducble)) 1-loop dagrams of the theory. Then, usng renormalzaton, evaluate the dvergent parts of these dagrams, usng ths to fx the counterterms to cancel the dvergences at ths order. We can see that loops begn appearng n the expanson of the two- and four-pont correlaton functons: = + Fgure 1: -pont correlator to 1 loop = Fgure : 4-pont correlator to 1 loop The -pont loop dagram s k -dvergent, whle the 4-pont loop dagrams are log-dvergent. In order to cancel the dvergences, we ntroduce counter-terms n the Lagrangan: L = 1 µφ µ φ 1 m φ λ 4! φ4 = Z φ 1 µφ µ φ Z m 1 m φ Z λ λ 4! φ4 The quanttes subscrpted wth a are the un-renormalzed or bare quanttes, whereas the unsubscrpted ones are renormalzed. All the dvergences have been absorbed nto the correcton factors (the Z s), whch have the followng relatons to the bare quanttes: φ = Z φ φ, m Z φ = Z m m, λ Z φ = λz λ We can then expand the correcton factors as Z X = 1 + δx, whch gves us the renormalzed Lagrangan: L = 1 µφ µ φ 1 m φ λ 4! φ4 + δφ 1 µφ µ φ δm 1 m φ δλ λ4! ] φ4 1

2 The terms n the brackets are counterterms, and gve rse to the followng Feynman rules: = (p δφ m δm) = λ δλ Fgure 3: Counterterm Feynman rules We can then see that the frst counterterm s gong to be needed to cancel the dvergence from the loop correcton to the propagator, and the 4-pont counterterm s needed to cancel the dvergences of the three loop dagrams n the 4-pont correlator. Startng wth the 4-pont dagrams we can see that the three 1-loop dagrams are just the same dagram but wth the Mandelstam varables s, t and u nterchanged. Calculatng the s-channel dagram: V (s) = ( λ) (π) d k m (k p) m where p = p 1 + p. We can then use Feynman parameters to consoldate the denomnator: = λ = λ (π) d 1 x(k m ) + (1 x)((k p) m )] (π) d 1 (k + (x 1)p) (x(x 1)p + m )] we shft the ntegraton varable k by completng the square: l = k + (x 1)p. = λ d d l (π) d 1 l ] ; = x(x 1)p + m We can then use a wck rotaton to rotate the ntegral nto Eucldean space then evaluate t usng sphercal symmetry. Or, we could use one of the Mnkowsk-space ntegral tables n the back of Peskn to save some tme: = λ Γ( d/) (4π) d/ ( ) 1 d/ Then lettng d = 4 ɛ and λ λµ ɛ/ (so the couplng remans dmensonless) and takng the lmt ɛ : = λ µ ɛ Γ(ɛ/) (4π) ɛ/ λ 16π 1 ɛ λ 3π ( 1 ln ) ɛ/ ( e γ 4πµ ) + O(ɛ)

3 We have successfully solated the dvergent part of ths dagram: λ/8π ɛ. Now we know that the counterterm (Fg. 3) δλ must be chosen so as to cancel ths dvergence, so we can mpose the renormalzaton condton that the 4-pont correlator must be exactly λ at zero momentum (s = 4m, t = u = ), whch s fnte. After addng n the counterterm, ths tells us φ(p 1 )φ(p )φ(p 3 )φ(p 4 ) = λ + ( V (4m ) + V () + V () ) λ δλ + O(λ 3 ) = = δλ = 1 ( V (4m ) + V () ) = 3λ 1 λ 16π ɛ λ ( e γ (x(x 1)s + m ) ( ) e γ m )] 3π ln 4πµ + ln 4πµ Now movng on to the correctons to the propagator, we can see that there s only one dagram wth a loop n Fg. 1: M = ( λ) (π) d k m = = λµɛ Γ(ɛ/ 1)m ɛ (4π) ɛ/ = λm 16π 1 ɛ λm 3π ln ( 4πe γ 1 m µ λ ( ) 1 1 d/ Γ(1 d/) (4π) d/ m ) + O(ɛ) Addng the -pont counterterm to ths and demandng t be fnte and zero when p = m, we have M + (p δφ m δm) p =m = The smplest way to satsfy ths condton s f δφ =, δm = M m = λ 1 16π ɛ λ ( 4πe γ 1 16π ln m ) µ The theory s now completely fnte at 1-loop, λ order. Usng the nformaton from the counterterms, we can now compute the beta functon. The bare couplng has been determned to behave lke λ = µ ɛ λz λ Z φ Actng on ether sde wth µ d dµ : ɛ λ(1 + δλ) = µ (1 + δφ) = µɛ (λ + λδλ) + O(λ 3 ) ( = ɛµ ɛ (λ + λδλ) + µ ɛ 1 + δλ + λ dδλ ) µ dλ dλ dµ µ dλ dµ ɛ(λ + λδλ) = 1 + δλ + λ dδλ = ɛ dλ (λ + 3λ 16π ɛ 3λ 8π ɛ = 3λ 16π + O(λ3, ɛ) = ɛ λ + 3λ 16π ɛ 1 + 3λ Therefore, the beta functon for φ 4 theory at 1-loop level s β(λ) = 3λ 16π 3 8π ɛ ) + O(λ 3 ) = ɛ

4 Ths beta functon has some nterestng behavor; we can explctly solve ths to get λ(µ) = λ 1 3λ 16π ln(µ/µ ) Where λ = λ(µ ), the couplng at some convenent energy scale. As µ µ exp(16π /λ ), the couplng dverges. We can assume the theory breaks down at ths scale (whch s extremely hgh f λ 1). In the opposte lmt, the couplng rapdly approaches zero as µ, and the theory s practcally free. Otherwse, λ s remarkably constant, varyng only by 1% over hundreds of orders of magntude of µ. ) Renormalzaton n Yukawa Theory We take the Yukawa theory n D = 4 spacetme: L = 1 µφ µ φ 1 m φ φ + ψ(/ m ψ )ψ + gφ ψψ Extractng the dvergent parameters from the bare Lagrangan gves us the renormalzed Lagrangan wth counterterms: L = 1 µφ µ φ 1 m φ φ + ψ(/ m ψ )ψ + gφ ψψ Ths leads to the followng Feynman rules: + δφ 1 µφ µ φ δm φ 1 m φ φ + ψ(δψ / m ψ δm ψ )ψ + gδgφ ψψ = p m φ = (p δφ m φδm φ ) = g = ( /p+m ψ ) p m ψ = (/pδψ m ψ δm ψ ) = g δg Fgure 4: Yukawa theory Feynman rules From these rules, we can construct all the 1PI 1-loop dagrams of the theory: = + + Fgure 5: -pont fermon correlator = + + Fgure 6: -pont scalar correlator 4

5 = + + Fgure 7: 3-pont correlator = Fgure 8: 4-pont fermon correlator = Fgure 9: 4-pont ψψ φφ correlator = Fgure 1: 4-pont scalar correlator Not all of the above loop dagrams are dvergent; for example dagrams 6, 7 and 8 n Fg. 8 are UV fnte. We have also only lsted up to 4-pont correlators, but there are more dvergent dagrams at hgher orders that we don t really care about f we re workng to lowest order n g at 1-loop. Actually, all of the 4-pont loop dagrams are rrelevant f we only want to work at O(g 3 ); 5

6 I ve just ncluded them for completeness. In ths case, we only need to consder the correctons to the propagators and vertex factors from Fgs. 5, 6 and 7. We can see that f the dvergences n these dagrams are to be canceled that we must have δψ, δm ψ, δφ, δm φ, δg g to lowest order. We can now go about computng the dvergent part of the three dagrams mentoned. For the correcton to the fermon propagator: M ψ (p) = (g) ( 1)! (π) d ūs (p) (/k + m ψ) k m u s (p) ψ (p k) m φ Consoldatng the denomnator wth Feynman parameters: = g ū s (p)(/k + m ψ )u s (p) (π) d ( )] x(k m ψ ) + (1 x) (p k) m φ Expandng the denomnator, then shftng the ntegraton varable k l + (1 x)p: = g d d l (π) d ū s ( ) (p) / l + (1 x)/p + m ψ u s (p) l x ( )] (x 1)p + m ψ m φ We can see the term wth the /l wll vansh due to the ntegrand beng odd, so all that remans (where (x, p ) s the strng of (l-ndependent) constants n the denomnator and A(x, p) s what remans n the numerator) s = g d d l (π) d A l ] = g (4π) d/ Γ( d/) A d/ Now lettng d = 4 ɛ and g gµ ɛ/ to retan dmensonalty, then takng the lmt ɛ, we are left wth ( = g 8π A(x, p) + g e γ ) ɛ 16π ln 4πµ A(x, p) + O(ɛ) We can see that the dvergent part of ths dagram s M ψ = g 8π ɛ ū s (p)((1 x)/p + m ψ )u s (p) + fnte Ths ntegral looks lke t would be qute complcated to evaluate n general, so I ll just leave t as s. We can then go on and compute the correcton to the scalar propagator (dagram n Fg. 6): M φ (p) = (g) (/k + m ψ ) (/k + /p + m ψ ) ( 1)Tr! (π) d k m ψ (k + p) m ψ ] = g = g (π) d (π) d Tr k + /k/p + m ψ /k + m ψ/p + m ψ ] x(k m ψ ) + (1 x)((k + p) m ψ ) 4k + 4k p + 4m ψ k + (1 x)p k (m ψ + (x 1)p ) ] 6

7 Now we shft the ntegraton varable to k l + (x 1)p: = 4g d d l l + (x 1)l p + (x x + 1)m ψ (π) d ] l (x x + 1)m ψ The l p term vanshes due to symmetry, and so our ntegral becomes = 4g d d l l + (π) d l ] = 4g d (4π) d/ Γ(1 d/) 1 1 d/ + Now we let d = 4 ɛ, g µ ɛ/ and take the lmt ɛ : M φ (p) = 3g π ɛ (x) + g 4π ln ] (4π) d/ Γ( d/) 1 1 d/ ( 4πµ e γ Evaluatng the ntegral n the dvergent part s easy ths tme, and we get )] + O(ɛ) M φ = 15g m φ 1π + fnte ɛ Fnally, movng on to the trangle dagram (dagram, Fg. 7), we have M = (g)3 (3 ) 3! (π) d vs (p ) (/k + m ψ) (/p 3 /k + m ψ ) k m ψ (p 3 k) m u s (p 1 ) ψ (p k) m φ = g 3 v s (p )(/k + m ψ )(/p dy 3 /k + m ψ )u s (p 1 ) (π) d ] 3 x(k m ψ ) + y((p 3 k) m ψ ) + (1 x y)((p k) m φ ) = g 3 dy v s (p )(/k + m ψ )(/p 3 /k + m ψ )u s (p 1 ) (π) d ] 3 k + ((x + y 1)p yp 3 ) k + ((1 x y)(p m φ ) + yp 3 (x + y)m ψ ) We then make the change of coordnates l = k + ((x + y 1)p yp 3 ). All odd powers of l n the numerator drop out, and we have: ) )] = g 3 d d l vs (p ) l ((1 x y)/p + y/p 3 + m ψ ((1 y)/p 3 + (x + y 1)/p u s (p 1 ) dy )] (π) d l (((x + y 1)p yp 3 ) + (x + y)m ψ ym φ + (x + y 1)(m ψ m φ ) I m gong to smplfy these expressons by renamng the l-ndependent quanttes n the ntegrand ) ) A = v s (p )u s (p 1 ), B = v s (p ) ((1 x y)/p + y/p 3 + m ψ ((1 y)/p 3 + (x + y 1)/p u s (p 1 ) = ( ((x + y 1)p yp 3 ) + (x + y)m ψ ym φ + (x + y 1)(m ψ m φ )) B and do depend on x and y however, so we can t pull them out of the x, y ntegral. Wth ths smplfcaton, the ntegral becomes M = g d d l Al B dy (π) d l ] 3 = g 3 d Γ( d/) A dy (4π) d/ Γ(3) d/ ] Γ(3 d/) B (4π) d/ Γ(3) 3 d/ 7

8 Lettng g gµ ɛ/ and d = 4 ɛ: = g3 µ 3ɛ/ (4π) ɛ/ dy ( ɛ/)γ(ɛ/) A B + Γ(1 + ɛ/) ɛ/ 1+ɛ/ ] Now takng the lmt ɛ : = g3 4π ɛ A + g3 16π dy B(x, y) ( e γ )] A(x, y) ln 4πµ 3 + O(ɛ) There s some ssue wth the dmensonalty of the argument of the logarthm (/µ 3 ] = 1), but just lookng at the dvergent part, we can see that t s M = g3 4π ɛ vs (p )u s (p 1 ) + fnte That completes the analyss of all the O(g 3 ), 1-loop dagrams. We can go a lttle further and compute the β functon for ths theory as well. Lookng back at our renormalzaton scheme, we see that our bare couplng constant s related to the renormalzed one by g = Z ggµ ɛ/ Z ψ Zφ = gµɛ/ (1 + δg) (1 + δψ) 1 + δφ gµɛ/ (1 + δg)(1 δψ)(1 1 δφ) In order to cancel the dvergences n the three dagrams we computed, we need to constran the counterterms. To cancel the dvergence of the trangle dagram, we see that we need δg = M g = Ag 4π ɛ For the two correctons to the propagators, f we demand that the two dagrams are (p m ) 1 at p = m, we have the freedom to absorb the dvergences nto the δm counterterms (lke n problem 1), settng δψ = δφ =. Ths vastly smplfes the problem and we are left wth Actng on ether sde wth µ d dµ only terms up to O(g 3 ): ( ) g = gµ ɛ/ 1 Ag 4π ɛ and takng the lmt ɛ, we can solve for the β functon, keepng µ dg = β(g) = Ag3 dµ π + O(g4 ) 8