Textbook Problem 4.2: The theory in question has two scalar fields Φ(x) and φ(x) and the Lagrangian. 2 Φ ( µφ) 2 m2

Transcription

1 PHY 396 K. Solutons for problem set #11. Textbook Problem 4.2: The theory n queston has two scalar felds Φx) and φx) and the Lagrangan L 1 2 µφ) 2 M2 2 Φ µφ) 2 m2 2 φ2 µφφ 2, S.1) where the frst 4 terms on the RHS descrbe the free felds whle the ffth term s the nteracton that we treat as a perturbaton. In Feynman rules, the propagators follow from the free part of the Lagrangan, so for the theory at hand there are two dstnct propagators, Φ Φ q 2 m 2 +0 and φ φ q 2 M S.2) Lkewse, there are two knds of external lnes accordng to the speces of the ncomng or outgong partcles for the process n queston. The Feynman vertces follow from the nteracton part of the Lagrangan, whch for the theory at hand s the cubc potental term V 3 µφφ 2. Consequently, all vertces should be connected to three lnes net valence 3), one double lne for the one ˆΦ feld and two sngle lnes for the two ˆφ felds, φ Φ 2µ, S.3) φ where the factor 2 2! comes from the nterchangeablty of two dentcal ˆφ felds n the vertex. 1

2 Now consder the decay process Φ φ + φ. To the lowest order of the perturbaton theory, the decay ampltude follows from a sngle tree dagrams φ 1 Φ S.4) φ 2 Ths dagram has one vertex, one ncomng double lne, two outgong sngle lnes and no nternal lnes of ether knd, hence φ 1 +φ 2 ˆT Φ M 2π) 4 δ 4) p p 1 p 2 ) 2µ 2π)4 δ 4) p p 1 p 2 ), S.5) or n other words MΦ φ 1 +φ 2 ) 2µ. S.6) Ths ampltude s related to the Φ φφ decay rate as Γ M 2 dp S.7) where the phase space factor for 1 partcle 2 partcles decays s dp 1 2E d3 p 1 2π) 3 2E π 2 EE 1 E 2 d3 p 2 2π) 3 2E 2 2π) 4 δ 4) p p 1 p 2 ) d 3 p 1 δe E 1 E 2 ) for p 2 p p 1 and on-shell energes, S.8) cf. 4.5 of the textbook. In the rest frame of the decayng Φ partcle E M, p 2 p 1, 2

3 and E 1 E 2 m 2 +p 2 for equal masses of the two fnal-state partcles), hence dp d 3 p 32π 2 ME 2 δm 2E p )) p 2 dp d 2 Ω 32π 2 ME 2 δm 2E p )). S.9) To remove the remanng δ functon, we ntegrate over the p p : dp δm 2E p )) dp 2dE p) E onshell 2p, S.10) hence dp p E d2 Ω 64π 2 M S.11) where p E 1 m2 E 2 1 4m2 M 2 S.12) snce 2E M by energy conservaton. Altogether, the partal decay rate of a heavy partcle of mass M nto two lghter partcles of equal masses m < 1 2 M s dγ d 2 Ω 1 4m2 M 2 M 2 64π 2 M. S.13) For the problem at hand, M 2µ regardless of drectons of fnal partcles, hence dγ d 2 Ω 1 4m2 M 2 µ2 16π 2 M. S.14) Integratng ths partal decay rate over the drectons of p we must remember that the two fnal partcles are dentcal bosons, so we cannot tell p 1 from p 2 p 1. Consequently, d 2 Ω 4π/2 and therefore Γ 1 4m2 M 2 µ2 8πM. S.15) 3

4 Textbook Problem 4.3a): Smlar to the prevous problem, the Feynman propagators of a theory follow from the free part of ts Lagrangan. Ths tme, we have N scalar felds Φ x) of smlar mass m, hence n momentum space Φ j Φ k δ jk q 2 m S.16) Note the δ jk factor the two felds connected by a propagator must be of the same speces. Graphcally, ths means that both ends of the propagator carry the same speces label j k. Lkewse, the external lnes should also carry a speces label of the ncomng or outgong partcle n queston. For the external lnes, these labels are fxed for a partcular process), whle for the nternal lnes we sum over j 1,2,...,N. The Feynman vertces follow from the nteractons between the felds; for the theory n queston, they come form the quartc potental V 4 λ 4 Φ Φ j Φ j Φ j) 2 j λ ˆΦj ) 4 λ + ˆΦj ) 2ˆΦk ) j<k S.17) Consequently, all vertces have net valence 4, but there are two vertex types wth dfferent ndexologes: 1) a vertex nvolvng 4 lnes of the same feld speces Φ j, wth the vertex factor λ/4) 4! 6λ; and 2) a vertex nvolvng 2 lnes of one feld speces Φ j and 2 lnes of a dfferent speces Φ k, wth the vertex factor λ/2) 2!) 2 2λ. The combnatoral factors arse from the nterchanges of the dentcal felds n the same vertex, thus 4! for the frst vertex type and 2!) 2 for the second type.) Equvalently, we may use a sngle vertex type nvolvng 4 felds of whatever speces, wth the speces-dependent vertex factor Φ j Φ l 2λ δ jk δ lm +δ jl δ km +δ jm δ kl). S.18) Φ k Φ m Now consder the scatterng process Φ j + Φ k Φ l + Φ m. At the lowest order of the perturbaton theory, there s just one Feynman dagram for ths process; t has one vertex, 4

5 4 external legs and no nternal lnes. Consequently, at the lowest order, MΦ j +Φ k Φ l +Φ m ) 2λ δ jk δ lm +δ jl δ km +δ jm δ kl) S.19) ndependent of the partcles momenta. Specfcally, MΦ 1 +Φ 2 Φ 1 +Φ 2 ) 2λ, MΦ 1 +Φ 1 Φ 2 +Φ 2 ) 2λ, S.20) MΦ 1 +Φ 1 Φ 1 +Φ 1 ) 6λ, and consequently usng eq. 4.85) of the textbook) dφ 1 +Φ 2 Φ 1 +Φ 2 ) dω c.m. dφ 1 +Φ 1 Φ 2 +Φ 2 ) dω c.m. dφ 1 +Φ 1 Φ 1 +Φ 1 ) dω c.m. λ 2 16π 2 Ec.m. 2, λ 2 16π 2 Ec.m. 2, 9λ 2 16π 2 Ec.m. 2. S.21) These are partal cross sectons. To calculate the total cross sectons, we ntegrate over dω, whch gves the factor of 4π when the two fnal partcles are of dstnct speces, but for the same speces, we only get 2π because of Bose statstcs. Hence, tot Φ 1 +Φ 2 Φ 1 +Φ 2 ) tot Φ 1 +Φ 1 Φ 2 +Φ 2 ) tot Φ 1 +Φ 1 Φ 1 +Φ 1 ) λ 2 4πEc.m. 2 λ 2 8πEc.m. 2 9λ 2 8πE 2 c.m.,,. S.22) 5

6 Textbook Problem 4.3b): The lnear sgma model was dscussed earler n class. The classcal potental VΦ 2 ) 1 2 µ2 Φ 2 ) λφ2 ) 2 S.23) wth a negatve mass term m 2 µ 2 < 0 has a mnmum or rather a sphercal shell of mnma) for Φ 2 Φ Φ v 2 µ2 λ > 0. S.24) Sem-classcally, we expect a non-zero vacuum expectaton value of the scalar felds, Φ 0 wth Φ 2 v 2, or equvalently, Φ j vδ jn modulo the ON) symmetry of the problem. Shftng the felds accordng to Φ N x) v + x), Φ j x) π j x) j < N), S.25) and re-wrtng the Lagrangan n terms of the shfted felds, we obtan L 2 1 )2 µ π )2 λv 2 +π 2 ) 4 1λ2 +π 2 ) 2 + const S.26) where π stands for the N 1) plet of the π j felds, thus π 2 j πj ) 2. The frst three terms on the RHS of eq. S.26) comprse the free Lagrangan for one massve real scalar feld x) of mass m 2µ and N 1) massless real scalar felds π j x). They are massless because they are Goldstone bosons of the ON) symmetry spontaneously broken down to the ON 1). There are N 1 broken symmetry generators, hence N 1 Goldstone bosons π j x).) Consequently, the Feynman rules have two dfferent propagator types q 2 2µ 2 +0 and π j π k δjk q 2 +0, S.27) and the ππ propagator carres a label j k 1,2...,N 1) specfyng a partcular speces of the pon feld. 6

7 The Feynman vertces follow from the nteracton terms n the LagranganS.26), namely the cubc and quartc potental terms V nt λv ˆ 3 + λv ˆˆπ 2 + λ 4 ˆ4 + λ 4 ˆ2ˆπ 2 + λ 4 ˆπ 2 ) 2. S.28) The fve terms here gve rse to fve types of Feynman vertces, two types of valence 3 and three types of valence 4. The 4-valent types whch follow from the quartc terms n the potental S.28) are just as n part a) of ths problem modulo renamng of the felds, π j π l 2λ δ jk δ lm +δ jl δ km +δ jm δ kl) S.29) π k π m and smlarly π j 2λδ k and 6λ. S.30) π k Lkewse, the 3-valent vertces follow from the cubc terms n the potental S.28); proceedng just as we dd n the prevous problem, we obtan π j 2λvδ jk and 6λv. S.31) π k Ths completes the Feynman rules of the lnear sgma model. 7

8 Textbook Problem 4.3c): In ths part of the problem, we use the Feynman rules we have just derved to calculate the tree-level ππ ππ scatterng ampltudes. As explaned n class, a tree dagram L 0) wth E 4 external legs has ether one valence 4 vertex and no propagators, or else two valence 3 vertces and one propagator. Altogether, there are four such dagrams contrbutng to the tree-level M π j p 1 )+π k p 2 ) π l p 1 )+πm p 2 )) they are shown n the textbook. The dagrams evaluate to: π j p 1 ) π l p 1 ) 2λ δ jk δ lm +δ jl δ km +δ jm δ kl), π k p 2 ) π m p 2 )... π j p 1 ) π l p 1 ) 2λvδ jk ) s 2µ 2 2λvδlm ), π k p 2 ) π m p 2 )... π j p 1 ) π l p 1 ) S.32) 2λvδ jl ) t 2µ 2 2λvδkm ), π k p 2 ) π m p 2 )... π j p 1 ) π l p 1 ) 2λvδ jm ) u 2µ 2 2λvδkl ), π k p 2 ) π m p 2 ) 8

9 where s,t,u are the Mandelstam varables s def p 1 +p 2 ) 2 p 1 +p 2 )2, t def p 1 p 1) 2 p 2 p 2) 2, u def p 1 p 2 ) 2 p 2 p 1) 2. S.33) Each of the three 2-vertex dagrams S.32) comes wth a dfferent combnaton of Kronecker δ s for the pon ndces, j, k, l, whle the 1-vertex dagram comprses all three combnatons. Thus, arrangng the net tree-level scatterng ampltude by the δ s, we obtan M π j p 1 )+π k p 2 ) π l p 1)+π m p 2) ) 2δ jk δ lm λ + 2λ2 v 2 ) s 2µ 2 2δ jl δ km λ + 2λ2 v 2 ) t 2µ 2 2δ jm δ kl λ + 2λ2 v 2 ) u 2µ 2. S.34) Moreover, each of the three terms on the RHS may be smplfes usng a relaton between the cubc and quartc couplngs of the shfted felds and the -partcle s mass 2 2µ 2. Indeed, the quartc couplng s λ and the cubc couplng s λ v for v 2 µ 2 /λ, cf. eq. S.24), hence Thanks to ths relaton, 2µ 2 λ 2λv) 2 S.35) λ + 2λ2 v 2 s 2µ 2 λs 2µ2 λ+2λ 2 v 2 s 2µ 2 λs s 2µ 2 S.36) and lkewse λ + 2λ2 v 2 t 2µ 2 λt t 2µ 2 and λ + 2λ2 v 2 u 2µ 2 λu u 2µ 2. S.37) Consequently, the ampltude S.34) smplfes to M 2λ δ jk δ lm s s 2µ 2 + δjl δ km t t 2µ 2 + δjm δ kl u ) u 2µ 2. S.38) Note that ths ampltude vanshes n the zero-momentum lmt for any one of the four pons, ntal or fnal. Indeed, for the massless pons wth p 1 ) 2 p 2 ) 2 p 1 )2 p 2 )2 0 9

10 we have s def p 1 +p 2 ) 2 p 1 +p 2 )2 +2p 1 p 2 ) +2p 1 p 2 ), t def p 1 p 1) 2 p 2 p 2) 2 2p 1 p 1) 2p 2 p 2), u def p 1 p 2) 2 p 2 p 1 )2 2p 1 p 2 ) 2p 1 p 2), S.39) so whenever any one of the four momenta becomes small, all three numerators n the ampltude S.38) become small M Osmall p). The reason for ths behavor s the Goldstone theorem: Among other thngs, t says that all scatterng ampltudes nvolvng Goldstone partcles such as the pons n ths problem become small as Op π ) when any Goldstone partcle s momentum p π becomes small. A few lnes above we saw how ths works for the tree-level π,π M π,π ampltude S.38); the same behavor perssts at all the hgher orders of the perturbaton theory, but seeng how that works s waaay beyond the scope of ths exercse. To complete ths part of the problem, let us now assume that all four pons momenta are small compared to the -partcle s mass m 2µ. In ths lmt, all three denomnators n eq. S.38) are domnated by the 2µ 2 term, hence M λ µ 2 1 ) p v 2 δ jk δ lm s + δ jl δ km t + δ jm δ kl 4 )) u + O m 2. S.40) For generc speces of the four pons, ths ampltude s of the order Op 2 /v 2 ), but there s a cancellaton when all four pons belong to the same speces whch s unavodable for N 2). Indeed, for j k l m δ jk δ lm s + δ jl δ km t + δ jm δ kl u s + t + u 4m 2 π 0, S.41) hence Mπ j +π j π j +π j ) 1 p 4 )) 0 v 2 + O m 2. S.42) Q.E.D. 10

11 Fnally, let us translate the ampltudes S.40) nto the low-energy scatterng cross sectons: dπ 1 +π 2 π 1 +π 2 ) dω c.m. tot π 1 +π 2 π 1 +π 2 ) E2 c.m. 48πv 4, t 2 64π 2 v 2 s E2 c.m. 64π 2 v 4 sn4 θ c.m. 2, dπ 1 +π 1 π 2 +π 2 ) dω c.m. s 2 64π 2 v 2 s E2 c.m. 64π 2 v 4, S.43) tot π 1 +π 1 π 2 +π 2 ) E2 c.m. 32πv 4, π 1 +π 1 π 1 +π 1 ) Op8 /m 4 ) 64π 2 v 2 s E 6 O c.m. v 4 m 4 ). Textbook Problem 4.3d1): Addng a lnear term V aφ N) to the classcal potental for the N scalar felds explctly breaks the ON) symmetry of the theory. Before we do anythng else, we must fnd how ths term affects the vacuum states of the theory and the masses of the and π felds. Fortunately, we have already done ths calculaton back n homework set 6, problem 1 solutons), so let me smply summarze the results: Instead of a sphercal shell of degenerate mnma, the modfed potental V λ 4 Φ j Φ j) 2 µ 2 2 Φ j Φ j) a Φ N S.44) has a unque mnmum at Φ 0. 0 v for v µ2 λ + a 2µ 2. S.45) Shftng the felds as n eq. S.25) but for the modfed VEV S.45) we arrve at the 11

12 Lagrangan L 1 2 )2 m π )2 m2 π π 2 λv 2 +π 2 ) λ π ) 2 +const S.46) whch looks just lke the old S.26), except for the modfed masses m 2 π λv 2 µ 2 a v a λ µ, S.47) m 2 3λv 2 µ 2 2λv 2 + m 2 π. S.48) In partcular, the pons are no longer massless Goldstone bosons, but they are stll much lghter than the partcle. Textbook Problem 4.3d2): Now consder the Feynman rules of the modfed theory. Snce the nteracton terms n the Lagrangan S.46) are exactly smlar to what we had n eq. S.26) of part b), the Feynman vertces of the modfed sgma model are exactly as n eqs. S.29), S.30) and S.31), wthout any modfcaton except for the slghtly dfferent value of v. On the other hand, the Feynman propagators need adjustment to accommodate the new masses S.47) and S.48), thus π j π k q 2 m 2 +0, δ jk q 2 m 2 π +0. S.49) The tree-level π + π π + π scatterng ampltude s governed by the same four Feynman dagrams as before, thus M π j p 1 )+π k p 2 ) π l p 1 )+πm p 2 )) 2δ jk δ lm λ + 2λ2 v 2 ) s m 2 2δ jl δ km λ + 2λ2 v 2 ) t m 2 2δ jm δ kl λ + 2λ2 v 2 ) u m 2, S.50) exactly as n eq. S.34), except for the new v and new m 2. However, nstead of m2 2λv2 12

13 we now have m 2 2λv2 λv 2 µ 2 m 2 π > 0, S.51) hence λ + 2λ2 v 2 s m 2 λ s m2 +λv2 s m 2 λ s m2 π s m 2 S.52) and lkewse λ + 2λ2 v 2 t m 2 λ t m2 π t m 2, λ + 2λ2 v 2 u m 2 λ u m2 π u m 2. S.53) Therefore, nstead of eq. S.38) we now have M 2λ δ jk δ lm s m2 π s m 2 + δ jl δ km t m2 π t m 2 + δ jm δ kl u m2 ) π u m 2. S.54) In the low energy-momentum lmt p µ m, ths ampltude smplfes to M 2λ m 2 1 v 2 ) δ jk δ lm s m 2 π) + δ jl δ km t m 2 ) π + δ jm δ kl u 2 m 2 ) p 4 ) ) π + O m 2. S.55) In partcular, for the slow pons wth p 0 m whle p m π, we have s E cm 2m π ) 2 4m 2 π whle t,u Op2 ) m 2 π, so the ampltude S.55) becomes M m2 π v 2 3δ jk δ lm δ jl δ km δ jm δ kl). S.56) Ths threshold ampltude does not vansh. Instead, M m2 π v 2 a v 3. S.57) Q.E.D. 13