iδ jk q 2 m 2 +i0. φ j φ j) 2 φ φ = j

Transcription

1 PHY 396 K. Solutions for problem set #8. Problem : The Feynman propagators of a theory follow from the free part of its Lagrangian. For the problem at hand, we have N scalar fields φ i (x of similar mass m, hence in momentum space φ j φ k iδ jk q m +i. (S. Note the δ jk factor the two fields connected by a propagator must be of the same species. Graphically, this means that both ends of the propagator carry the same species label j k. Likewise, the external lines should also carry a species label of the incoming or outgoing particle in question. For the external lines, these labels are fixed (for a particular process, while for the internal lines we sum over j,,...,n. The Feynman vertices follow from the interactions between the fields; for the theory in question, the interactions comprise the quartic potential V 4 λ 8 ( φ φ j φ j φ j. (S. so the vertices have valence 4. To extract their specific form, we should rewrite the potential (S. in the form V 4 4 j,k,l,m λ jklm φ j φ k φ l φ m (S.3 where the coefficients λ jklm are totally symmetric in all 4 indices. Specifically, V 4 λ ( φ j φ j λ φ j φ j φ l φ l λ 3δ jk δ lm φ j φ k φ l φ m j j,l j,kl,m λ ( δjk δ lm +δ il δ km +δ jm δ kl φ j φ k φ l φ m, 4 j,kl,m (S.4

2 hence λ jklm λ ( δ jk δ lm +δ il δ km +δ jm δ kl, (S.5 and therefore the Feynman vertices φ j φ l iλ ( δ jk δ lm +δ il δ km +δ jm δ kl. (S.6 φ k φ m Now consider the scattering process φ j +φ k φ l +φ m. At the tree level i.e., at the lowest order of the perturbation theory there is only one diagram φ j φ l (S.7 φ k φ m hence M tree (φ j +φ k φ l +φ m λ ( δ jk δ lm +δ jl δ km +δ jm δ kl (S.8 independent of the particles momenta. Specifically, M(φ +φ φ +φ λ, M(φ +φ φ +φ λ, (S.9 M(φ +φ φ +φ 3λ. The partial cross sections in the CM frame follow from these amplitudes via eq. (4.85 of the

3 textbook or eq. (7 of my notes on phase space: For elastic scattering, dσ M 64π Ec.m., (S. hence dσ(φ +φ φ +φ λ 64π Ec.m., dσ(φ +φ φ +φ dσ(φ +φ φ +φ λ 64π Ec.m., 9λ 64π Ec.m.. (S. To calculate the total cross sections, we integrate over dω, which gives the factor of 4π when the two final particles areof distinct species, but for the same species, we only get π because of Bose statistics. Thus, σ tot (φ +φ φ +φ σ tot (φ +φ φ +φ σ tot (φ +φ φ +φ λ 6πEc.m. λ 3πEc.m. 9λ 3πE c.m.,,. (S. Problem (a: In perturbation theory, the Feynman propagators follow from the quadratic part of the Lagrangian (and hence free Hamiltonian, while the vertices follow from the cubic, quartic, etc., terms treated as perturbation. For the linear sigma model s Lagrangian (, L L free V pert, L free i ( µφ i + ( µσ M σ λf (S.3 σ, (S.4 3

4 V pert 3λf 6 σ3 + λf i σφ i (S.5 + 3λ 4 σ4 + λ 4 i σ φ i + λ (. 8 i φ i The free Lagrangian (S.4 describes one massive field σ plus N massless fields π i, hence two types of scalar propagators, σ σ i q M σ +i and π j π k iδjk q +i, (S.6 and the ππ propagator carries a label j k,...,n specifying a particular species of the pion field. As to the perturbation (S.5, it has two cubic terms and 3 quartic terms, hence two types of valence 3 vertices and three types of valence 4 vertices: the σσσ and σππ vertices σ σ 3iλf and π j σ iλfδ jk, (S.7 the σσσσ and σσππ vertices σ π k σ σ π j σ 3iλ and iλδ ik (S.8 σ σ and finally the ππππ vertex similar to what we had in problem, π k σ π j π l iλ ( δ jk δ lm +δ jl δ km +δ jm δ kl. (S.9 π k π m This completes the Feynman rules of the linear sigma model. 4

5 Problem (b: As explained in class, a tree diagram (L with E 4 external lines has either (A one valence 4 vertex and no propagators, or else (B two valence 3 vertices and one propagator. Topologically, there are three diagrams of type (B with different arrangements of incoming versus outgoing external lines, so altogether there are 4 tree diagrams. Specifically for the π j +π k π l +π m scattering, the diagrams are π j (p π l (p iλ ( δ jk δ lm +δ jl δ km +δ jm δ kl, (S. π k (p π m (p... π j (p π l (p (iλfδ jk i s (iλfδ lm, (S. π k (p π m (p... π j (p π l (p (iλfδ jl i t (iλfδ km, (S. π k (p π m (p... π j (p π l (p (iλfδ jm i u (iλfδ kl, (S.3 π k (p π m (p 5

6 where s,t,u are the Mandelstam variables s def (p +p (p +p, t def (p p (p p, u def (p p (p p. (S.4 Note that in the diagrams (S., (S., and (S.3, the internal line belongs to the σ field rather than to any π i fields since there are no πππ vertices but only ππσ. Also, each of the diagrams (S., (S., and (S.3 yields a different combination of Kronecker δδ forthej,k,l,mindicesofthefourpions, whilethefirstdiagram(s.yieldsall three combinations. So when we total up the four tree diagrams amplitudes, it s convenient to reorganize the net tree amplitude by the j,k,l,m indexology, thus M ( π j (p +π k (p π l (p +π m (p ( δ jk δ lm λ + λ f s ( δ jl δ km λ + λ f t ( δ jm δ kl λ + λ f u. (S.5 Problem (c: The Lagrangian ( of the linear sigma models has a very important relation between the quartic coupling λ, the cubic coupling κ λf, and the σ particle s mass M σ λf, thus ( M σ λf λ ( κ λf. (S.6 Thanks to this relation, λ + (λf sm σ λsλm σ +(λf sm σ λs sm σ (S.7 and likewise λ + (λf tm σ λt tm σ and λ + (λf um σ λu um σ. (S.8 6

7 Thanks to these formulae, the scattering amplitude (S.5 simplifies to ( M λ δ jk δ lm s sm σ + δ jl δ km t t + δ jm δ kl u u. (S.9 Now consider the low-energy limit of this amplitude. In the CM frame, all 4 pions have the same energy E, hence s (E tot cm 4E, t 4E sin (θ/, u 4E cos (θ/, (S.3 and therefore s,t,u O(E. (S.3 Consequently, when the pion s energies are much smaller than the σ particle s mass, the denominators in the amplitude (S.9 may be approximated as sm σ tm σ um σ m σ λf. (S.3 Consequently, the scattering amplitude (S.9 simplifies to M ( +λ M σ + ( ( E f δ jk δ lm s + δ jl δ km t + δ jm δ kl 4 u + O. (S.33 Themagnitudeofthisamplitude isgenerallyo(e /f, sointhelow-energy limititbecomes quite small. Now consider the ππ ππ scattering in a completely general frame of reference. Since the pions are massless, Mandelstam s s,t,u variables may be written as s def (p +p (p +p +(p p +(p p, t def (p p (p p (p p (p p, u def (p p (p p (p p (p p, (S.34 so whenever any one of the four momenta becomes small, all 3 of the s,t,u become small. In particular, when 3 of the momenta are O(M σ or smaller while the fourth momentum 7

8 becomes much smaller, we have s,t,u O(M σ p smallest M σ. (S.35 Consequently, the scattering amplitude becomes as in eq. (S.33, and its magnitude is generally M (s,t,u λ M σ < P smallest λ M σ. (S.36 The physical 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 O(p 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.9; 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. Problem (d: In the low-energy limit E M σ, the tree-level ππ ππ amplitudes may be approximated as in eq. (S.33. In particular, M(π +π π +π λt ( λe 4 + O 4 M(π +π π +π λs ( λe 4 + O 4 M(π +π π +π λ(s+t+u + O t f, s f, ( λe 4 M 4 σ ( λe 4 O M 4 σ since s+t+u 4m π (S.37 8

9 Translating these amplitudes into the partial and total scattering cross-sections, we obtain dσ(π +π π +π 64π s t f 4 E c.m. 64π f 4 sin4 θ c.m., σ tot (π +π π +π E c.m. 48πf 4, dσ(π +π π +π σ tot (π +π π +π E c.m. 3πf 4, 64π s s f 4 E c.m. 64π f 4, dσ(π +π π +π ( λ 64π s O E 8 8 ( E σ(π +π π +π 6 O c.m. f 4 4 E c.m. f 4. ( E 6 O c.m. f 4 4, (S.38 Eqs. (S.38 complete the required part of the homework problem. But for completeness sake, let me derive a more accurate low-energy approximation for the same-species scattering, for example π +π π +π. Let s go back to the general-energy amplitude (S.9 and expand it to the second power in s,t,u. Thus, λs s λt t λu u λs λt λu + λs 4, + λt 4, + λu M 4 σ, (S.39 and therefore M(π +π π +π λs sm σ λ M σ λt tm σ (s+t+u + λ M 4 σ λu um σ (s +t +u. (S.4 In the center of mass frame, s + t + u 6E 4 + 6E 4 sin 4 (θ/ + 6E 4 cos 4 (θ/ 8E 4 (3+cos θ ( Ecm tot 4 3+cos θ, (S.4 9

10 hence and therefore dσ(π +π π +π λ E 6 cm 56π M 8 σ (3+cos θ, (S.4 σ(π +π π +π 7λ E 6 cm 8πM 8 σ. (S.43 Problem 4(a: For the ultra-relativistic electrons and positrons, the spinors u(p, s and v(p, s become chiral. Indeed, by inspection of eqs. (4, u has chirality matching the electron s helicity left for λ and right for λ + while v has chirality opposite to the positron s helicity left for λ + and right for λ. At the same time, the amplitude (3 depends on the electron s and positron s spin states via the Dirac sandwich v(e + γ ν u(e which does not mix helicities. Indeed, ( σν vγ ν u v γ γ ν u v u v σ L σ νu L + v R σ νu R, ν (S.44 so if u and v are chiral, then they should have the same chirality both left or both right or else vγ ν u. In terms of helicities, u(e and v(e + being both left-handed means λ(e while λ(e+ + ; likewise, u(e and v(e + being both right-handed means λ(e + while λ(e+. Thus, a non-zero amplitude (3 requires the electron and the positron to have opposite helicities, λ(e + λ(e ; for similar helicities of the two initial particles, the amplitude vanishes. Q.E.D. For future reference, let me calculate the explicit Dirac sandwiches (S.44 for the ultrarelativistic electron and positron spinors (4: ( ( ( σν v(e + L γ νu(e L E ξl η L σ ν ( ( ( σν v(e + L γ νu(e R E σ ν η L ξ R, (5 E η L σ νξ R, (S.45

11 ( ( ηr ( σν v(e + R γ νu(e L +E ξl σ ν ( ( ηr ( σν v(e + R γ νu(e R +E σ ν ξ R +E η R σ νξ L, (S.46. (5 As promised, the sandwich and hence the amplitude (3 vanishes for the same helicities. This fact is of practical importance for the electron-positron colliders. Any kind of fermion pair production µ µ +, or τ τ +, or q q which proceeds through a virtual vector particle a photon, or Z, or even something not yet discovered would have the v(e + γ ν u(e factor in the amplitude, so the electron and the positron must have have opposite helicities, or they would not annihilate each other and make pairs. Nowsupposewehavealongitudinallypolarizedelectronbeam sayλ + only but the positron beam is un-polarized. Because of eq. (5, only the left-handed positrons would collide with the right-handed electrons and produce pairs, while the left-handed positrons would do something else. Likewise, if the electron beam has the λ polarization, then only the right-handed positrons would collide with our left-handed electrons and make pairs, while the left-handed positrons would do something else. Thus, as far as the pairproduction is concerned, the positron beam could just as well be longitudinally polarized with λ(e + λ(e. In other words, if we want to study polarization effects in fermion pair production, it s enough to longitudinally polarize just the electron beam. We do not need to polarize the positron beam which is much harder to do because the electrons of a definite helicity would automatically select positrons of the opposite helicity. Problem 4(b: For the ultra-relativistic muons, the u(µ and v(µ + are chiral, and the chiralities behave exactlysimilartotheelectronandthepositroninpart(a: thediracsandwichū(µ γ ν v(µ + vanishes unless u and v have the same chirality, which requires the µ and the µ + to have

12 opposite helicities. Specifically, ( ξl ū(µ L γ νv(µ + L E ( ξl ū(µ L γ νv(µ + R E ( ū(µ R γ νv(µ + L E ξ L ( ( σν σ ν ( σν η L ( ηr σ ν ( ( σν σ ν ( ( σν ū(µ R γ νv(µ + L E ξ L η L ( ηr σ ν, (6 +E ξ L σ νη R, (S.47 E ξ R σ νη L, (S.48. (6 Eqs. (6 and similar formulae for the other fermion-antifermion pairs produced with ultra-relativistic speeds in electron-positron collisions assure that the fermion and the antifermion always have opposite helicities. Experimentally, this means that if for some event we are able to determine the helicity of one final particle, then we may infer the second final particle s helicity without any further experimental effort. Problem 4(c: The electron moves in the positive z direction, so its helicity λ is the same as its S z the z component of its spin. Hence, the ξ spinors corresponding to the helicities are ξ(e L ( and ξ(e R (. (S.49 The positron moves in the negative z direction, so its helicity is opposite from S z, hence ξ(e + L ( and ξ(e + R (. (S.5 However, the η spinors in eqs. (. for the positrons have opposite spins from these ξ

13 spinors, specifically η σ ξ, thus η(e + L ( +i ( i and η(e + R. (S.5 Substituting these component spinors into eqs. (S.45 and (S.46, we obtain ( v(e + L γ νu(e R E ( iσ ν +ie (σ ν E (,i,+, ν, ( (7 v(e + R γ νu(e L +E (+i σ ν +ie ( σ ν E (,+i,+, ν. When taking the and matrix elements of the σ ν and σ ν matrices, please remember that σ ν (,σ x,σ y,σ z while σ ν (,σ x,σ y,σ z. Problem 4(d: Suppose for a moment θ and the µ move in the same directions as e. Then the muons have exactly the same spinors u and v as the e of the same charge and helicity, and similarly to eqs. (7 we have v(µ + L γ νu(µ R E (,i,+,ν and v(µ + R γ νu(µ L E (,+i,+,ν. (S.5 For the muons, the amplitude (3 involves the ū(µ γ ν v(µ + Dirac sandwich rather than the v(µ + γ ν u(µ, but these two sandwiches are related by complex conjugation ( vγ ν u ūγ ν v ūγ ν v, (S.53 as well as raising the index ν, hence ū(µ R γν v(µ + L E [ (,+i,, ν] E (,i,, ν, ū(µ L γν v(µ + R E [ (,i,, ν] E (,+i,, ν. (S.54 Eqs. (S.54 apply for θ. For other muon directions, we may simply rotate the 3

14 4 vectors (S.54 through angle θ in the xz plane, thus ū(µ R γν v(µ + L E (,icosθ,,+isinθν, ū(µ L γν v(µ + R E (,+icosθ,,isinθν. (9 Problem 4(e: Substituting the Dirac sandwiches (7 and (9 into the pair production amplitude (3, we obtain µ L,µ + R Me L,e + R µ R,µ + L Me R,e + L e (+cosθ, µ R,µ + L Me L,e + R µ L,µ + R Me R,e + L e (cosθ, (S.55 while all the other polarized amplitudes vanish by eqs. (5 and (6: µ any,µ + any Me L,e + L µ any,µ + any Me R,e + R, µ L,µ + L Me any,e + any µ R,µ + R Me any,e + (S.56 any. The partial cross-sections ( follow from these amplitudes according to ( dσ M p 64π s p. (S.57 Problem 4(f: Summing the polarized cross-sections ( over the muons helicities, we get dσ(e L +e+ R µ any +µ+ any dσ(e R +e+ L µ any +µ+ any α 4s (+cosθ + α 4s (cosθ + + α s (+cos θ (S.58 while dσ(e L +e+ L µ any +µ + any dσ(e R +e+ R µ any +µ + any. (S.59 4

15 Averaging these cross-sections over the electron s and positron s helicities gives dσ(e avg +e+ avg µ any +µ+ any 4 ( α s (+cos θ + α s (+cos θ + + α 4s (+cos θ, which is exactly what we found in class for the un-polarized cross-section when E M µ. (S.6 5