Problem 1(a): The scalar potential part of the linear sigma model s Lagrangian (1) is. 8 i φ2 i f 2) 2 βλf 2 φ N+1,

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1 PHY 396 K. Solutions for problem set #10. Problem 1a): The scalar potential part of the linear sigma model s Lagrangian 1) is Vφ) = λ 8 i φ i f ) βλf φ N+1, S.1) where the last term explicitly breaks the ON +1) symmetry of the first term down to ON). To find the minimum of this potential, let s first find the stationary points where all the first derivatives V/ φ i are zero: V for i = 1,...,N, = λ φ i j φ j f) φ i = 0, V and = λ φ N+1 j φ j f) φ N+1 βλf = 0. S.) S.3) Fromeq.S.3)weimmediatelyseethatatanystationarypoint φ f ) 0, henceeqs. S.) tell us that φ 1 = = φ N = 0. In other words, all the stationary points lie on the φ N+1 axis in the N+1) dimensional space of the scalar field values. And in this space, eq. S.3) becomes a simple cubic equation φ 3 N+1 f Φ N+1 βf = 0. S.4) For small β f, this cubic equation has 3 real solutions, approximately φ N+1 1 β, φ N+1 f + β, φ N+1 3 +f + β. S.5) Now let s find out which of the three stationary points is a minimum or at least a local minimum) by looking at the second derivatives of the potential S.1). Along the φ N+1 axis in the field space, the second derivatives amount to 3φ N+1 f ) for i = j = N +1, V φ i φ j = λ 0 for i N, j = N +1 or j N, i = N +1, φ N+1 f ) δ ij for i,j N. S.6) Evaluating these derivatives for the 3 stationary points S.5) while assuming small β > 0 1

2 gives φ N+1 1 φ N+1 φ N+1 3 : V < 0 while other φ N+1 ) V > 0 while other φ N+1 ) V > 0 while other φ N+1 ) V < 0 = maximum, φ i ) V < 0 = saddle point, φ i ) V > 0 = minimum. φ i ) S.7) Thus, the potential S.1) has a unique minimum at φ 1 = = φ N = 0, φ N+1 = +f + β + Oβ /f). ) Quod erat demonstrandum. Problem 1b): Let s shift the fields as in eq. 3). In terms of the shifted fields, T def = i φ i f = π + σ+ φ N+1 ) f = π + σ + φ N+1 σ + φ N+1 f ), S.8) whereπ isashort-handforn-vectorπ 1,...,π N )ofthepionfields, thusπ = π 1 ) + +π N ). Therefore, expanding the scalar potential S.1) into powers of the shifted fields, we obtain V = λ 8 T βλf σ + φ N+1 ) = λ 8 π + σ ) λ φ N+1 + σ π + σ 3) + λ φ N+1 σ + λ φ N+1 f ) 4 π +σ ) λ φn+1 + φ N+1 f ) βλf ) σ + const. S.9) On the last line here, the coefficient of σ vanishes thanks to φ N+1 obeying the cubic equation S.4). For the same reason, the coefficient of π +σ ) on the line before the last may be

3 simplified as λ φ N+1 f ) 4 = βλf φ N+1. S.10) Altogether, we have Vσ,π ) = λ 8 π + σ ) κ + σ 3 + σπ ) + M σ σ + M π π + const, S.11) where quartic coupling λ = λ, cubic coupling κ = λ φ N+1 λf +β), pion mass M π = βλf φ N+1 βλf, sigma mass M σ = M π + λ φ N+1 λff +3β). S.1) Let s take a closer look at the pion s mass, Mπ β λf. In the β = 0 limit, the pions are massless in accordance with the Goldstone theorem. Indeed, for β = 0 the sigma model s Lagrangian has an exact SON + 1) symmetry which is spontaneously broken down to an SON) subgroup; there are N spontaneously broken generators, so there should be N massless Goldstone bosons. But for β 0, the SON +1) symmetry of the Lagrangian is only approximate, and its explicit breaking by the βλf φ N+1 term spoils the Goldstone theorem. Thus, instead of exactly massless Goldstone bosons we should get light but not quite massless pseudo-goldstone bosons; to the first order in β, their mass should be proportional to β. And indeed, in the linear sigma model Mπ β λf. Still, for β f, the pions should be much lighter than the sigma particle. And indeed, according to eqs. S.1), M π M σ βλf λf = β f 1. S.13) 3

4 Problem 1c): Back in homework#8 problem ), we had a very similar setup to the shifted fields of the linear sigma models: N +1 scalar fields σx) and π i x), with the Lagrangian L = 1 µσ) + 1 µπ ) V, Vσ,π ) = λ ) π + σ + f σ 8 9.) = λf σ + λf σ 3 + σπ ) + λ 8 σ + π ). In particular, there is a mass term for the σ field but not for the pions, which are exactly massless exactly as in the present sigma model with β = 0. The cubic and quartic terms in the potential 9.3) also have exactly the same form as in eq. S.11), and the quatic coupling λ and the cubic coupling κ = λf are related to the σ field s mass as κ = λ M σ. S.14) For the present sigma model, we have exactly similar relation for β = 0. Indeed, according to eq. S.1), for β = 0, κ = λf, M σ = λf = κ = λ M σ. S.15) Therefore, the ππ ππ scattering amplitudes in the linear sigma model for β = 0 come out to be exactly as homework#8: At the tree level, Mπ j +π k π l +π m ) = λ + which in light of the relation S.14) becomes κ ) s Mσ δ jk δ lm λ + λ + κ ) t Mσ δ jl δ km κ ) u Mσ δ jm δ kl, S.16) Mπ j +π k π l +π m ) = λs s M σ δ jk δ lm λt t Mσ δ jl δ km λu u Mσ δ jm δ kl. S.17) When any of the 4 pions energy becomes small, we get s,t,u Mσ, and the scattering 4

5 amplitude becomes small as M λ M σ s δ jk δ lm + t δ jl δ km + u δ jm δ kl) = O λe cm M σ ). S.18) Problem 1d): For β 0, the quartic and the cubic couplings of the σ and π i to each other has similar overall form to what we had back in homework#8, but the overall coefficients λ and κ of those couplings are no longer related to the σ particle s mass by eq. S.14). Instead, eq. S.1) gives us κ = λ φ N+!, M σ = M π + λ φ N+! = κ = λ M σ M π). S.19) Now consider the ππ ππ scattering. At the tree level, we have exactly the same 4 diagrams for such scattering as in the homework#8, namely π j p 1 ) π l p 1 ) π j p 1 ) π l p 1 ) π k p ) π m p ) π k p ) π m p ) S.0) π j p 1 ) π l p 1 ) π j p 1 ) π l p 1 ) π k p ) π m p ) π k p ) π m p ) Altogether, these diagrams yield the scattering amplitude exactly as in eq. S.16), but for the 5

6 β-modified couplings and masses. Consequently, in light of eq. S.19) instead of S.14), we have λ + κ s M σ = λs λm σ +κ s M σ = λs λm π s M σ S.1) and likewise λ + κ t M σ = λt M π) t M σ and λ + κ u M σ = λu M π) u M σ, S.) so the amplitude S.16) becomes Mπ j +π k π l +π m ) = λs M π ) s M σ δ jk δ lm λt M π ) t Mσ λu M π ) u Mσ δ jl δ km δ jm δ kl. S.3) When the pions energies become low compared to M σ or in Lorentz-invariant terms, when s,t,u M σ we may simplify this amplitude by approximating all the denominators as M σ, thus Mπ j +π k π l +π m ) λ M σ 1 ) s M π ) δ jk δ lm + t Mπ f ) δjl δ km ). 4) + u Mπ ) δjm δ kl What happens to this amplitude when one of the pions momentum becomes very small? Alas, for β 0 the pions are massive, so we cannot take all 4 components of a pion s p µ to zero. The best we can do is to take p 0 while p 0 m, which is the non-relativistic limit. However, if only one pion is non-relativistic while the other 3 pions have E M π but E M σ ), we generally have s,t,u = OE M π ) Mπ albeit s,t,u M σ ), and the scattering amplitude becomes ) E Mπ M = O f 0. S.4) The strongest low-energy limit we an take for massive pions is to make all four pions nonrelativistic. In this limit, s = Ecm 4M π while u,t = Op ) Mπ, so the scattering 6

7 amplitude 5) becomes Mπ j +π k π l +π m ) λm π M σ βλ ) 3δ jk δ lm δ jl δ km δ jm δ kl). 5) f This amplitude is suppressed by the factor β/f, but it does not vanish! And even if all 4 pions belong to the same species, the scattering amplitude does not vanish in the non-relativistic limit, Mπ 1 +π 1 π 1 +π 1 ) λβ f 0, S.5) unlike what we had back in homework#8. Problem a): given Φ e +iθ U L ΦU R, 8) likewise we have Φ e iθ U R Φ U L, hence Φ Φ U R Φ Φ)U R, Φ Φ ) UR Φ ΦU R U RΦ ΦU R = U R Φ Φ ) U Φ Φ ) n UR Φ Φ ) n U R n = 1,,3,..., S.9) R, S.6) S.7) S.8) and therefore all traces tr Φ Φ ) n ) are invariant under symmetries 8), S.30) thanks to the cyclic invariance rule for traces, tr U R XU ) R = tr XU R U ) R) = tr X for any X = Φ Φ ) n. Consequently, the scalar potential 7) is invariant under symmetries 8). For the global symmetries where e iθ, U L, and U R do not depend on x, the kinetic term 7

8 in 6) is also invariant. Indeed, for constant e iθ, U L, U R, µ Φ e +iθ U L µ Φ)U R, µ Φ e iθ U R µ Φ )U L, S.31) µ Φ µ Φ U R µ Φ µ Φ)U R, and tr µ Φ µ Φ ) is invariant. Altogether, the whole Lagrangian 6) is invariant, Q.E.D. Problem ): The kinetic termin 6) andthe last two terms inthe potential 7)have a much bigger symmetry than G = SUN) SUN) U1), namely the SON ) which does not care for the matrix structure of the Φx) and treats it as N real component fields. Indeed, tr Φ Φ ) = i,j Φ j i = i,j ) ReΦ j i + ImΦ j i ) ) S.3) is invariant under all SON ) rotations of the components, and so is the kinetic term. On the other hand, the tr Φ ΦΦ Φ ) in the potential does depend on the packing of N real components into a complex N N matrix, and it is this term which reduces the internal symmetry group of the theory to G = SUN) SUN) U1). Proving that all the SON )/G symmetries are broken by the quartic trace term is a non-trivial exercise in group theory rather than field theory. You do not have to do it as part of this homework set, and I am not writing down the proof here. Problem b): Given the eigenvalues κ 1,...,κ N ) of the Φ Φ matrix, the invariant traces S.30) obtain as tr Φ Φ ) n ) = N κ n i. i=1 S.33) 8

9 Consequently, the scalar potential is ) V = α κ i + β κ i + m i i i κ i. S.34) Now let s minimize this potential. Since the matrix Φ Φ cannot have any negative eigenvalues, we are looking for a minimum of Vκ 1,...,κ N ) under constraints κ i 0. This requires i = 1,...,N, either κ i 0 and V κ i = 0, or else κ i = 0 and V κ i > 0, S.35) where V κ i = ακ i + m + β j κ j. S.36) These derivatives are linear functions of the eigenvalues κ i, so all the non-zero eigenvalues must obey the same linear equation α κ i = m β j κ j, same for all κ i 0, whichmeansthatallnon-zeroκ i have thesamevalue. Thus, uptoapermutationofeigenvalues, κ 1 = = κ k = C, κ k+1 = = κ N = 0, S.37) for some k = 0,1,,...,N, and C obtains from α C + m + β kc = 0 C = m α+kβ. S.38) To make sure that the solution S.37) is a minimum rather that a maximum or a saddle point, we need C = m α+kβ m + βkc = αm α+kβ > 0 unless k = 0, > 0 unless k = N. S.39) Depending on the signs of α, β and m parameters, this limits the solutions to the following: 9

10 For α > 0, β > 0, and m > 0, the only solution is k = 0, which means κ 1 = κ N = 0 and hence Φ = 0. For α > 0, β > 0, and m < 0, the only solutions is k = N, which means κ 1 = = κ N = C = m α+nβ > 0, 10) and hence Φ = C a unitary matrix. We shall focus on this regime through the rest of this problem. For α < 0 or β < 0, the situation is more complicated: When α +β < 0 or α +Nβ < 0, the scalar potential 7) is unbounded from below and the theory is sick. When α > 0 and β < 0 but α+nβ > 0, the solutions are similar to the β > 0 case: For m > 0 all κ i = 0, while for m < 0 the κ i are as in eq. 9). When β > 0 and α < 0 but α+β > 0: for m > 0 the only solution is k = 0, meaning Φ = 0, but for m < 0 all the solutions S.37) with k = 1,,...,N are good local minima. To find the global minimum, we compare the potentials at the local minima, Vminimum#k) = α kc4 + β kc ) + m kc = kα+k β = m4 k kβ +α. m 4 α+kβ) + km m α+kβ) S.40) Since α < 0 but α+β > 0, the deepest minimum obtains for k = 1, thus κ 1 = m α+β, κ = = κ N = 0. S.41) 10

11 Problem c): Let s act with some SUN) L SUN) R U1) symmetry 8) on the vacuum expectation values 10): Φ = C 1 N N e iθ U L Φ U R = C eiθ U L U R. S.4) Clearly, to keep the VEVs Φ invariant, we need e iθ U L U R = 1 N N S.43) and hence U R = e iθ U L. S.44) Moreover, since the U L and U R matrices have unit determinants, this requires ) det e iθ 1 N N = 1 = N θ = 0 mod π). S.45) Such a phase can be absorbed into the U L SUN), so without loss of generality we need e iθ = 1 and U L = U R SUN). S.46) In other words, the unbroken symmetry group is SUN) which acts on the scalar fields as Φx) UΦx)U, U SUN). S.47) 11

12 Problem d): In terms of the shifted fields 11), µ Φ = 1 µ ϕ 1 + i µ ϕ ), µ Φ = 1 µ ϕ 1 i µ ϕ ), S.48) hence the kinetic term in the Lagrangian becomes tr µ Φ µ Φ ) = 1 tr µϕ 1 µ ϕ 1 ) + 1 tr µϕ µ ϕ ). S.49) As to the potential terms, we have Φ Φ = C 1 N N + C δφ +δφ) + δφ δφ = C 1 N N + C ϕ ϕ ϕ + i [ϕ 1,ϕ ] S.50) and consequently tr Φ Φ ) = NC + Ctrϕ 1 ) + 1 trϕ 1 ) + 1 trϕ ), S.51) tr Φ Φ ) = N C 4 + NC 3 trϕ 1 ) + C tr ϕ 1 ) + NC trϕ 1 ) + trϕ )) + Ctrϕ 1 ) trϕ 1) + trϕ ) ) trϕ 1 ) + trϕ ) ) S.5) Φ tr Φ ) ) = NC 4 + C 3 trϕ 1 ) + 3C trϕ 1) + C trϕ ) + Ctrϕ 3 1) + Ctrϕ 1 ϕ ) trϕ4 1) trϕ4 ) + 3 trϕ 1ϕ ) 1 trϕ 1ϕ ϕ 1 ϕ ). S.53) Plugging all these formulae into the potential 8) and expanding in powers of ϕ 1 and ϕ, we obtain V = const + V 1 + V + V 3 + V 4 S.54) where V 1 = C m + βnc + αc ) trϕ 1 ) = 0 S.55) because m + α+nβ)c = 0 1

13 V = βc tr ϕ 1 ) + 1 m + βnc + 3αC ) trϕ 1) + 1 m + βnc + αc ) trϕ ) = βc tr ϕ 1 ) + αc trϕ 1 ) + 0, S.56) V 3 = βc ) trϕ 1 ) trϕ 1) + trϕ ) + αc ) trϕ 3 1) + trϕ 1 ϕ ), S.57) ) V 4 = β trϕ 8 1) + trϕ ) + α ) trϕ 4 8 1) + trϕ 4 ) + 6trϕ 1ϕ ) trϕ 1 ϕ ϕ 1 ϕ ). S.58) Combining the quadratic part S.56) of this potential with the kinetic terms S.49), we arrive at L = 1 tr µϕ 1 µ ϕ 1 ) βc tr ϕ 1 ) αc trϕ 1 )+ 1 tr µϕ µ ϕ ) S.59) which gives us the mass spectrum of the theory: the ϕ 1 x) matrix of fields is massive and the ϕ x) matrix is massless. Each matrix is N N and hermitian, so it contains N independent real scalar fields, which give rise to N particles. Altogether, the spectrum comprises: N massless particles from the ϕ x) matrix. N 1 massive particles with M = αc from the traceless part of the ϕ 1 x) matrix. One more massive particle with M = α+nβ)c = m from the trace of ϕ 1 x). To see where the values of the masses come from, let s decompose the ϕ 1 matrix into the pure trace plus the traceless part, ξx) def = trϕ 1x)) N and ϕ 1 x) def = ϕ 1 x) ξx) N 1 N N = tr ϕ 1 ) 0. S.60) Consequently, tr ϕ 1 ) = N ξ, trϕ 1 ) = ξ + tr ϕ 1 ), S.61) 13

14 likewise tr µ ϕ 1 µ ϕ 1 ) = µ ξ µ ξ + tr µ ϕ 1 µ ϕ 1 ), so the free Lagrangian S.59) becomes L = 1 µξ) βc N + αc ) 1 ξ + 1 tr µ ϕ 1 ) ) αc 1 tr ϕ 1 ) S.6) + 1 tr µ ϕ ) ) 0 1 trϕ ) where all the masses are manifest. Problem e): The unbroken SUN) symmetry acts on the scalar fields according to Φx) U Φx) U. S.47) andsincethevev10)isinvariant, theshiftedfieldsδφx) = Φx) Φ alsotransformaccording to δφx) U δφx) U. S.63) Moreover, unitary transforms like these preserve hermiticity, so when we decompose δφx) into a hermitian matrix ϕ 1 x) and an antihermitian matrix iϕ x), the transforms S.63) do not mix the ϕ 1 and ϕ with each other. Instead, they transform like ϕ 1 x) Uϕ 1 x)u, ϕ x) Uϕ x)u, S.64) which means that ϕ 1 and ϕ comprise separate SUN) multiplets. Furthermore, the transforms S.64) preserve trances trϕ 1 ) and trϕ ), so to make the SUN) multiplet structure manifest, let s decompose both ϕ 1 and ϕ into their traceless parts and pure traces along the 14

15 lines of eq. S.60), ϕ 1 x) = ξ 1x) N 1 N N + ϕ 1 x), ϕ x) = ξ x) N 1 N N + ϕ x), tr ϕ 1 ) tr ϕ ) 0. S.65) With this decomposition, the ξ 1 and the ξ are both invariant under the SUN) which puts each of them into its own singlet multiplet while the each of the traceless parts ϕ 1 and ϕ makes each own adjoint multiplet. This multiplet structure agrees with the masses we obtained in part d). Indeed, all N 1 members of the adjoint multiplet ϕ 1 have the same mass αc, while the singlet ξ 1 has a different mass α+nβ)c. On the other hand, both the adjoint multiplet ϕ and the singlet ξ are massless. The reason for this degeneracy goes beyond the un-broken SUN) symmetry; instead, both the ϕ and the ξ are Goldstone bosons of the spontaneously broken symmetries in G/H = ) / SUN) L SUN) R U1) SUN). S.66) Specifically, the singlet ξ is the Goldstone boson of the broken U1) symmetry. Indeed, the U1) s generator commutes with all the other generators, so it belongs in its own singlet of the symmetry, and the corresponding Goldstone particle should also be a singlet. Now consider the non-abelian generators. Generators TL a of the SUN) L form an adjoint multiplet of the SUN) L, but are invariant under the SUN) R. Likewise, generators TR a of the SUN) R form an adjoint multiplet of the SUN) R, but are invariant under the SUN) L. In other words, under an U L,U R ) SUN) L SUN) R they transform as T a L U LT a L U L, Ta R U RT a R U R. S.67) When the SUN) L SUN) R is broken down to a single SUN) spanning U L = U R = U, both TL a and Ta R transform as T a L UTa L U, T a R UTa R U, S.68) which puts them into two adjoint multiplets of the unbroken SUN). Equivalently, we may 15

16 form two adjoint multiplets out of T a V = Ta L + Ta R and T a A = Ta L Ta R, S.69) which act on the scalar fields according to T a V Φ = i [λa,φ], T a A Φ = i {λa,φ}. S.70) The TV a generate the unbroken SUN) symmetry, cf. eq. S.47). The Ta A generators are spontaneously broken, hence there should be an adjoint multiplet of massless Goldstone bosons. And indeed there is the ϕ. 16