Textbook Problem 4.2: We begin by developing Feynman rules for the theory at hand. The Hamiltonian clearly decomposes into Ĥ = Ĥ0 + ˆV where

Transcription

1 PHY 396 K. Solutions for problem set #11. Textbook Problem 4.2: We begin by developing Feynman rules for the theory at hand. The Hamiltonian clearly decomposes into Ĥ = Ĥ0 + ˆV where Ĥ 0 = Ĥfree Φ + Ĥfree φ S.1) and ˆV = µ d 3 x ˆΦx)ˆφ 2 x). S.2) In Feynman rules, the propagators follow from the free Hamiltonian S.1): Contracting the creation and annihilation operators comprising the two scalar fields Φx) and φx), we obtain Φx) === Φy) = Φx) Φy) = G F x y) mass=m, φx) φy) = φx) φy) = G F x y) mass=m, S.3) but no mixed φ Φ contractions. In the momentum space, this gives us two distinct propagators, = i q 2 M 2 +i0 and = 1 q 2 m 2 +i0. S.4) The Feynman vertices follow from the perturbation Hamiltonian S.2), which involves one power of the Φ field and two powers of the φ field. Thus, the vertices involve one double line and two single lines net valence = 3), φ Φ = 2iµ S.5) φ where the factor 2 comes from the interchangeability of the two identical ˆφ fields in the vertex. 1

2 Now consider the decay process Φ φ + φ. To the lowest order of the perturbation theory, we have a single diagram φ 1 Φ S.6) φ 2 with one vertex, one incoming double line, two outgoing single lines and no internal lines of either kind. Thus, φ 1 +φ 2 iˆt Φ im 2π) 4 δ 4) p p 1 p 2 ) = 2iµ 2π)4 δ 4) p p 1 p 2 ), S.7) i.e., MΦ φ 1 +φ 2 ) = 2µ. S.8) This amplitude is related to the Φ φφ decay rate as Γ = M 2 dp S.9) where the phase space factor for 1 particle 2 particles decays is dp = 1 2E d3 p 1 2π) 3 2E 1 1 = 32π 2 EE 1 E 2 d3 p 2 2π) 3 2E 2 2π) 4 δ 4) p p 1 p 2) d 3 p 1δE E 1 E 2) for p 2 = p p 1 and on-shell energies, S.10) cf. 4.5 of the textbook. In the rest frame of the decaying Φ particle E = M, p 2 = p 1, 2

3 and E 1 = E 2 = m 2 +p 2 for equal masses of the two final-state particles), 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.11) To remove the remaining δ function, we integrate over the p = p : dp δm 2E p )) = dp 2dE p) = E onshell 2p, S.12) hence dp = p E d2 Ω 64π 2 M S.13) where p E = 1 m2 E 2 = 1 4m2 M 2 S.14) since 2E = M by energy conservation. Altogether, the partial decay rate of a heavy particle of mass M into two lighter particles of equal masses m < 1 2 M is dγ d 2 Ω = 1 4m2 M 2 M 2 64π 2 M. S.15) For the problem at hand, M = 2µ regardless of directions of final particles, hence dγ d 2 Ω = 1 4m2 M 2 µ2 16π 2 M. S.16) Integrating this partial decay rate over the directions of p we must remember that the two final particles are identical 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.17) 3

4 Textbook Problem 4.3a): Similar to the previous problem, the propagators are contractions of the free fields, thus for N distinct fields Φ i of the same mass m we have Φ j x) Φ k y) = Φ j x) Φ k y) = δ jk G F x y) mass=m, S.18) or in momentum space, Φ j Φ k = iδ jk q 2 m 2 +i0. S.19) The vertices follow from the perturbation operator ˆV = d 3 x λ4 Φ Φ) 2 j λ 4 ˆΦ j ) 4 + j<k λ 2 ˆΦ j ) 2 ˆΦ k ) 2), S.20) hence two vertex types: 1) a vertex involving 4 lines of the same field species Φ j, with the vertex factor i λ 4 4! = 6iλ; and 2) a vertex involving 2 lines of one field species Φj and 2 lines of a different species Φ k, with the vertex factor i λ 2 2!)2 = 2iλ. The combinatorial factors arise from the interchanges of the identical fields in the same vertex, thus 4! for the first vertex type and 2!) 2 for the second type.) Equivalently, we may use a single vertex type involving 4 fields of whatever species, with the species-dependent vertex factor Φ j Φ l = 2iλ δ jk δ lm +δ jl δ km +δ jm δ kl). S.21) Φ k Φ m Now consider the scattering process Φ j + Φ k Φ l + Φ m. At the lowest order of the perturbation theory, there is just one Feynman diagram for this process; it has one vertex, 4

5 4 external legs and no internal lines. 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.22) independent of the particles momenta. Specifically, MΦ 1 +Φ 2 Φ 1 +Φ 2 ) = 2λ, MΦ 1 +Φ 1 Φ 2 +Φ 2 ) = 2λ, S.23) MΦ 1 +Φ 1 Φ 1 +Φ 1 ) = 6λ, and consequently using 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.24) These are partial cross sections. To calculate the total cross sections, we integrate over dω, which gives the factor of 4π when the two final particles are of distinct species, but for the same species, we only get 2π because of Bose statistics. 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.25) 5

6 Textbook Problem 4.3b): The linear sigma model was discussed earlier in class. The classical potential VΦ 2 ) = 1 2 µ2 Φ 2 ) λφ2 ) 2 S.26) with a negative mass term m 2 = µ 2 < 0 has a minimum or rather a spherical shell of minima) for Φ 2 Φ Φ = v 2 = µ2 λ > 0. S.27) Semi-classically, we expect a non-zero vacuum expectation value of the scalar fields, Φ 0 with Φ 2 = v 2, or equivalently, Φ j = vδ jn modulo the ON) symmetry of the problem. Shifting the fields according to Φ N x) = v + x), Φ j x) = π j x) j < N), S.28) and re-writing the Lagrangian in terms of the shifted fields, we obtain L = 2 1 )2 µ π )2 λv 2 +π 2 ) 4 1λ2 +π 2 ) 2 + const S.29) where π stands for the N 1) plet of the π j fields, thus π 2 = j πj ) 2. The free part of the Lagrangian S.29) the first 3 terms) describe one massive real scalar field x) of mass m = 2µ and N 1) massless real scalars π j x) which are the Goldstone particles of the ON) symmetry spontaneously broken down to ON 1) thus N 1) broken symmetry generators, forming a vector multiplet of the unbroken ON 1) symmetry). Consequently, the non-zero contractions of the free and π fields are x) y) = G F x y) mass=m, π j x) π k y) = δ jk G F x y) mass=0, S.30) which give us two distinct Feynman propagators in the momentum basis, = π j i q 2 2µ 2 +i0, π k = iδjk q 2 +i0. S.31) The last two terms in the Lagrangian S.29) give rise to the interaction Hamiltonian of 6

7 the linear sigma model, namely ˆV = d 3 x λvˆ 3 + λvˆˆπ 2 + λ 4ˆ4 + λ 2ˆ2ˆπ 2 + λ 4 ˆπ 2 ) 2). S.32) The five terms in this interaction Hamiltonian give rise to five types of Feynman vertices. Proceeding exactly as in part a) of the problem, we obtain π j π l = 2iλ δ jk δ lm +δ jl δ km +δ jm δ kl) S.33) π k π m and similarly π j = 2iλδ ik and = 6iλ. S.34) π k The remaining two vertices have valence = 3 and follow from the cubic terms in the interaction Hamiltonian S.32). The analysis proceeds exactly as in the previous problem and yields π j = 2iλvδ jk and = 6iλv. S.35) π k This completes the Feynman rules of the linear sigma model. 7

8 Textbook Problem 4.3c): In this part of the problem, we use the Feynman rules we have just derived to calculate the tree-level ππ ππ scattering amplitudes. As explained in class, a tree diagram L = 0) with E = 4 external legs has either one valence = 4 vertex and hence no propagators) or two valence = 3 vertices and hence one propagator). Altogether, there are four such diagrams contributing to the tree-level im π j p 1 )+π k p 2 ) π l p 1 )+πm p 2 )) they are shown in the textbook. The diagrams evaluate to: π j p 1 ) π l p 1 ) = 2iλ 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 ) = 2iλvδ jk ) i s 2µ 2 2iλvδlm ), π k p 2 ) π m p 2 )... π j p 1 ) π l p 1 ) S.36) = 2iλvδ jl ) i t 2µ 2 2iλvδkm ), π k p 2 ) π m p 2 )... π j p 1 ) π l p 1 ) = 2iλvδ jm ) i u 2µ 2 2iλvδkl ), π k p 2 ) π m p 2 ) where s,t,u are the Mandelstam variables 8

9 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.37) Each of the three 2-vertex diagrams S.36) comes with a different combination of Kronecker δ s for the pion indices i, j, k, l, while the 1-vertex diagram comprises all three combinations. Thus, arranging the net tree-level scattering amplitude by the δ s, we obtain 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.38) Each of the three terms on the right hand side here may be simplified thanks to eq. S.27) which leads to the relation 2µ 2 λ = 2λv) 2 S.39) betweenthecubicandquarticcouplingsoftheshiftedfieldsandthesigma-particle smass 2 = 2µ 2. Thanks to this relation, λ + 2λ2 v 2 s 2µ 2 = λs 2µ2 λ+2λ 2 v 2 s 2µ 2 = λs s 2µ 2 S.40) and likewise λ + 2λ2 v 2 t 2µ 2 = λt t 2µ 2 and λ + 2λ2 v 2 u 2µ 2 = λu u 2µ 2. S.41) Consequently, the amplitude S.38) simplifies 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.42) Note that this amplitude vanishes in the zero-momentum limit for any one of the four pions, initial or final. Indeed, for the massless pions with 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.43) so whenever any one of the four momenta becomes small, all three numerators in the amplitude S.42) become small = M = Osmall p). The reason for this behavior is the Goldstone theorem: Among other things, it says that all scattering amplitudes involving Goldstone particles such as the pions in this problem become small as Op π ) when any Goldstone particle s momentum p π becomes small. A few lines above we saw how this works for the tree-level π,π M π,π amplitude S.42); the same behavior persists at all the higher orders of the perturbation theory, but seeing how that works is waaay beyond the scope of this exercise. To complete this part of the problem, let us now assume that all four pions momenta are small compared to the -particle s mass m = 2µ. In this limit, all three denominators in eq. S.42) are dominated 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.44) For generic species of the four pions, this amplitude is of the order Op 2 /v 2 ), but there is a cancellation when all four pions belong to the same species this is unavoidable 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.45) hence Mπ j +π j π j +π j ) = 1 p 4 )) 0 v 2 + O m 2. S.46) Q.E.D. 10

11 Finally, let us translate the amplitudes.44) into the low-energy scattering cross sections: dπ 1 +π 2 π 1 +π 2 ) dω c.m. = t 2 64π 2 v 2 s = E2 c.m. 64π 2 v 4 sin4 θ c.m. 2, tot π 1 +π 2 π 1 +π 2 ) = E2 c.m. 48πv 4, 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.47) 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): Adding a linear term V = aφ N) to the classical potential for the N scalar fields explicitly breaks the ON) symmetry of the theory. Before we do anything else, we must find how this term affects the vacuum states of the theory and the masses of the and π fields. Without the linear term, the potential V 0 Φ) = 1 4 λφ2 ) µ2 Φ 2 ) = 1 4 ) 2 Φ 2 µ2 + const λ has a spherical shell of degenerate minima. The linear term V = aφ N) breaks the degeneracy the net potential V = V 0 + V decreases as one goes North along the sphere, and there is a unique global minimum at the North pole Φ = 0,...,0,+v). There is also a small shift of the radial coordinate of the minimum from v 0 = µ 2 /λ to a nearby minimum of the quartic polynomial V Φ = 0,...,0,+v) ) = 1 4 λv4 1 2 λv2 av, S.48) i.e., to the root of the cubic equation dvv) dv = λv 3 µ 2 v a = 0. S.49) For small a, this equation has 3 real roots one near +v 0, one near v 0, and one near zero but we only need the first root. Using the Cardano formula and selecting the right root, 11

12 we obtain v = 4µ 2 3λ cos π arcsin a ) 27λ µ 3 ) µ 2 λ 1 + a λ 2µ 3 3a2 λ 8µ 6 +. S.50) Alternatively, we may let v = v 0 +δv and expand the equation S.49) in power of δv, thus λv 3 0 µ 2 v 0 = 0) + δv 3λv 2 0 µ 2 ) + Oδv 2 ) a = 0 = δv a a2 3λv0 2 = µ2 2µ 2. S.51) Having found the minimum of the potential, let s shift the fields Φ N) x) = v + x), Φ j x) = π j x) for j < N S.52) for the new VEV v and expand the scalar potential in terms of the shifted fields: V = λ v 2 + 2v π 2 ) 2 µ 2 v 2 + 2v π ) 2 av +) = const + λv 3 µ 2 v a = 0) λv 2 µ 2 ) π 2 λv 2 µ 2 ) S.53) + 2 +π 2 ) λv + 2 +π 2 ) 2 λ 4. Of particular importance are the quadratic terms here, which give us the particles masses: M 2 π = λv 2 µ 2 = a v a λ µ S.54) and M 2 = 3λv2 µ 2 = 2µ 2 + 3M 2 π. S.55) Note that the pions are no longer massless but acquire M 2 π > 0. However, for a small linear term with a µ3 λ), the pions a much lighter than the particle. This is a special case of a general rule: When an approximate symmetry is spontaneously broken, we do not get exactly massless Goldstone bosons; instead, we get massive but light particles called pseudo-goldstone bosons whose masses 2 are proportional to the small symmetry-violating parameters such as a. 12

13 In real life, the π mesons are pseudo-goldstone bosons of the approximate SU2) L SU2) R chiral symmetry of QCD. Indeed, the two lightest quark flavors u and d have small but non-zero masses, which explicitly break the SU2) L SU2) R chiral symmetry. Besides this small explicit breaking, the chiral symmetry suffers from a larger spontaneous breakdown to SU2) V, which gives rise to 3 pseudo-goldstone bosons with quantum numbers of the broken symmetry generators isospin = 1 and odd parity and small masses M 2 π = m u +m d ) OSSB mass scale) M 2 othermesons. S.56) Indeed, experimentally M 2 π 0.03M2 ρ. Textbook Problem 4.3d2): The net Lagrangian for the shifted fields x) and π x) is L = 1 2 )2 2 1m π )2 2 1m2 ππ 2 λv 2 +π 2 ) 4 1λ2 +π 2 ) 2 S.57) plus an irrelevant constant). The interaction terms here are exactly similar to those we had in part b) cf. eq. S.29)), hence the Feynman vertices of the modified sigma model are exactly as in eqs. S.33), S.34) and S.35), without any modification except for the slightly different value of v. On the other hand, the Feynman propagators need adjustment to accommodate the new masses S.54) and S.55), thus = π j π k = i q 2 m 2 +i0, iδ jk q 2 m 2 π +i0. S.58) The tree-level π + π π + π scattering amplitude is governed by the same four Feynman diagrams 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.59) exactly as in eq. S.38), except for the new v and new m 2. However, instead of m2 = 2λv2 13

14 we now have m 2 2λv2 = λv 2 µ 2 = m 2 π > 0, S.60) hence λ + 2λ2 v 2 s m 2 = λ s m2 +λv2 s m 2 = λ s m2 π s m 2 S.61) and likewise 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.62) Therefore, instead of eq. S.42) 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.63) In the low energy-momentum limit p µ i m, this amplitude simplifies 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.64) In particular, near the energy threshold s = Ec.m. 2 4m2 π when all the pions are slow in the CM frame), p 0 m π, p i m π, and hence t,u m 2 π, the amplitude S.64) becomes M m2 π v 2 3δ jk δ lm δ jl δ km δ jm δ kl). S.65) This threshold amplitude does not vanish. Instead, M m2 π v 2 = a v 3. S.66) Q.E.D. 14