Physics 217 Solution Set #5 Fall 2016

Transcription

1 Physics 217 Solution Set #5 Fall Repeat the computation of problem 3 of Problem Set 4, but this time use the full relativistic expression for the matrix element. Show that the resulting spin-averaged differential cross section is given by ( dσ dω = α 2 4 p 2 β 2 sin 4 1 β 2 sin 2 θ ), (θ/2) 2 where p is the three-momentum of the electron and β v/c is the electron velocity. This is the famous Mott formula for the Coulomb scattering of relativistic electrons. We begin with eq. (45) of Solution Set 4, It follows that 1 T 2 fi 2 = λ,λ = = = Z 2 e 4 2 p p 4 Z 2 e 4 2 p p 4 T fi = u(p )γ 0 Ze 2 u(p) p (1) p 2 λ,λ u (λ ) (p )γ 0 u(p) (λ) 2 λ,λ u (λ ) α (p )γαβ 0 u(p)(λ) β u(p)(λ) ρ γ0 ρσ u(λ ) (p ) σ ( ) Z 2 e 4 2 p p 4γ0 αβ u (λ) β (p)u(λ) ρ (p) λ Z 2 e 4 2 p p Tr γ 0 (/p+m)γ 0 (/p +m) ] 4 γ 0 ρσ ( u σ (λ ) (p )u (λ ) α (p ) λ = 2Z2 e 4 p p 4 2p 0 p 0 p p +m 2]. (2) We shall evaluate this expression in the laboratory frame. Assume that the initial electron moves in the z-direction, and the final electron moves in the x z plane at and angle θ with respect t the z-axis. That is, the electron four-vectors are given by p = ( E; 0, 0, p ), p = ( E; p sinθ, 0, p cosθ ), where we have used the fact that energy conservation requires that E = E, which is enforced by the delta function in eq. (34) of Solution Set 4. Moreover, E = E implies that p = p = E 2 m 2. (3) 1 )

2 Thus, we obtain 2p 0 p 0 p p +m 2 = 2E 2 E 2 + p 2 cosθ+m 2 = 2E 2 p 2 (1 cosθ) = 2 E 2 p 2 sin 2 (θ/2) ], (4) after using m 2 = E 2 p 2. Finally, we need to evaluate, p p 2 = 2 p 2( 1 cosθ ) = 4 p 2 sin 2 (θ/2), (5) Inserting eqs. (4) and (5) back int eq. (2) yields 1 T 2 fi 2 = λ,λ Z 2 e 4 E 2 4 p 4 sin 4 p 2 sin 2 (θ/2) ]. (θ/2) The above result can be written in terms of the relativistic velocity. Recall that the relativistic energy and momentum are given respectively by E = γm and p = γm v (in units of c = 1). Thus, β v = p /E. hence, 1 T 2 fi 2 = λ,λ Z 2 e 4 1 β 2 4 p 2 β 2 sin 4 sin 2 (θ/2) ]. (6) (θ/2) Using eq. (41) of Solution Set 4, we end up with the relativistic generalization of the Rutherford scattering formula, dσ dω = Z 2 α 2 1 β 2 4 p 2 β 2 sin 4 sin 2 (θ/2) ], (7) (θ/2) where α e 2 /(4π) is the fine structure constant. Note that the non-relativistic Rutherford formula obtained in eq. (47) of Solution Set 4, is obtained from eq. (7) in the limit of β v 1 by using the non-relativistic formula p m v and setting β = 0 in the numerator of eq. (7). 2. In class, the total cross section for e + e µ + µ at very high center-of-mass energies ( s m µ ) was found to be where α 1/137. σ = 4πα2 3s, (8) (a) Restore the appropriate factors of and c in the formula for σ, so that you can evaluate σ in units of area. Consider the formulae for the relativistic energy and momentum, E = γmc 2, p = γm v, 2

3 where γ (1 v 2 /c 2 ) 1/2. It the follows that E = pc2 v, (9) where p p and v v. In addition, recall the Heisenberg uncertainty relation, What eqs. (9) and (10) teach us is that x p 1. (10) 2 x] = /p], E] = cp], where x] means the dimensions of x (in this case, length), etc. Combining these two relations yiekds x] = c/e]. Since s in eq. (8) has units of E 2 ], it follows that the correct form of eq. (8) with the units of and c restored is σ = 4π 2 c 2 α 2, (11) 3s in order to obtain a cross-section in units of squared length. (b) Find the value of σ in nanobarns (1 nb= cm 2 ), for s = GeV (this is the center-of-mass energy of the B-factory at the KEKB collider in Tsukuba, Japan). Using the table of physical constants given in C. Patrignani et al. (Particle Data Group), 2016 Review of Particle Physics, Chin. Phys. C, 40, (2016), 1 we find the following numerical values, 2 c = GeV fm, α 1 = , where 1 fm = cm. The same table of physical constants also gives where 1 mb = 10 3 barns = cm 2. For s = GeV, eq. (8) yields ( c) 2 = GeV 2 mb, σ = 4π( GeV2 mb) 3(10.58 GeV) 2 )( ) 2 = mb = nb, where 1 nb = 10 9 barns = cm 2. 1 Thistableisavailableathttp://www-pdg.lbl.gov/2016/reviews/rpp2016-rev-phys-constants.pdf. 2 The table of physical constants cited above gives more accurate versions of these constants. I have truncated them to five or six significant figures, which is still far more than we really need for this problem. 3

4 3. Compute the differential cross section, dσ/dω, in the center-of-momentum frame, for Bhabha scattering, e + e e + e. You may assume that the total energy of the initial e + e system in the center-of-momentum frame is much larger than the electron mass, m e, in which case (to good approximation) you may set m e = 0 in your calculation. There are two Feynman diagrams that contribute at second order in the perturbation series for the S-matrix. These two contributions must be added in the invariant matrix element before squaring. Be sure that you have the correct relative sign between these diagrams. Consider the scattering process e + e e + e. The incoming momenta of the electron and positron are denoted by p 1 and p 2, respectively, and the outgoing momenta by p 1 and p 2, respectively. We introduce the three kinematic invariants, s = (p 1 +p 2 ) 2 = (p 1 +p 2) 2 = 2p 1 p 2 = 2p 1 p 2, (12) t = (p 1 p 1 )2 = (p 2 p 2 )2 = 2p 1 p 1 = 2p 2 p 2, (13) u = (p 1 p 2 )2 = (p 2 p 1 )2 = 2p 1 p 2 = 2p 2 p 1, (14) where we have the electron and positron masses to zero. In the massless limit, s+t+u = 0. Two Feynman diagrams contribute at tree-level to e + e e + e, e p 1 p 1 e e p 1 p 1 e p 1 +p 2 p 1 p 1 e + p 2 p 2 e + e + p 2 p 2 e + (a) (b) and the corresponding invariant amplitudes will be denoted by M a and M b. Note that the arrows indicate the flow of fermion number (or equivalently the flow of negative charge). To avoid too many arrows, we also label the electron and positron lines with the corresponding four-momenta, which point in the direction of the arrow as shown. For example, the fourmomentum of the incoming positron is p 2. This is equivalent to a four-momentum of p 2 that is pointing in the opposite direction as the incoming positron. Employing the QED Feynman rules, im a = u (λ 1 ) (p 1)ieγ µ u (λ 1) (p 1 )v (λ 2) (p 2 )ieγ ν v (λ 2 ) (p 2) im b = u (λ 1 ) (p 1)ieγ µ v (λ 2 ) (p 2)v (λ 2) (p 2 )ieγ ν v (λ 1) (p 1 ) ig µν (p 1 p 1 )2, ig µν (p 1 +p 2 ) 2, wherethehelicities, λ 1, λ 2, λ 1, andλ 2 ofthecorresponding incoming andoutgoingelectrons and positrons, respectively, are specified. 4

5 Using eqs. (12) and (13), The total invariant amplitude is M a = e2 ) t u(λ 1 (p 1 )γµ u (λ1) (p 1 )v (λ2) (p 2 )γ µ v (λ ) 2 (p 2 ), (15) M b = e2 s u(λ 1 ) (p 1)γ µ v (λ 2 ) (p 2)v (λ 2) (p 2 )γ µ u (λ 1) (p 1 ), (16) M = M a M b. (17) Note the relative minus sign. This is due to the fact that the ordering of spinors in M a and M b are indicated schematically by the corresponding four-momenta of the external particles, (p 1p 1 p 2 p 2) and (p 1p 2p 2 p 1 ), respectively. Note that the two orderings of the external four-momenta are related by three transpositions. The fact that this is an odd number implies that there is a relative minus sign between the two amplitudes, as exhibited in eq. (17). (a) After averaging over the initial state helicities and summing over the final state helicities, express the result for the squared invariant matrix element for Bhabha scattering in terms of the Mandelstam variables, s, t and u. Show that the differential cross-section can be cast into the following form, dσ dω = α2 2s u 2 ( 1 s + 1 t) 2 + ( ) ] 2 t ( s ) 2 +, s t where α e 2 /(4π). NOTE: s+t+u = 0 in the approximation where m e = 0.] We must now square the total invariant amplitude, eq. (17), and sum over final helicities and average over initial helicities. We introduce the following notation, M 2 ave 1 4 λ 1,λ 2,λ 1,λ 2 M 2, wherethefactorof 1 accountsfortheaveraging ofthetwohelicities oftheincoming electron 4 and the two helicities of the incoming positron. Then, eq. (17) yields λ M 2 ave = M a 2 ave + M b 2 ave 2Re(M am b ) ave. (18) The squared matrix elements above can be evaluated by employing the projection operaotrs, u α (λ) (p)u(λ) β (p) = (/p+m e) αβ, v α (λ) (p)v(λ) β (p) = (/p m e) αβ,, where the four-component spinor indices are explicitly exhibited. In the present calculation, we are setting m e = 0. 5 λ

6 Thus the squared matrix elements summed over helicities can be converted into traces, 3 M a 2 ave = e4 4t Tr( ) ( γ µ /p 2 1 γ ν /p 1 Tr γµ /p 2 γ ) ν/p 2, (19) M b 2 ave = e4 4s Tr( ) ( ) γ µ /p 2 2 γν /p 1 Tr γµ /p 1 γ ν /p 2, (20) 2Re(M a M b ) ave = e4 2ts Tr( ) γ µ /p 1 γ ν /p 2 γ µ /p 2 γν /p 1. (21) I will illustrate in detail the calculations that produced the above results. To make the steps of the computations clear, I shall eplicity exhibit the four-component spinor indices. First, we compute, M a 2 ave = e2 4ts = e2 4ts = e2 4ts λ 1,λ 2,λ 1,λ 2 λ 1,λ 2,λ 1,λ 2 λ 1,λ 2,λ 1,λ 2 u (λ 1 ) α (p 1)γ µ αβ u(λ 1) β (p 1 )v (λ 2) γ (p 2 )(γ µ ) γδ v (λ 2 ) δ (p 2) ] u (λ 1 ) ρ (p 1 )γν ρσ u(λ 1) σ (p 1 )v (λ 2) τ (p 2 )(γ ν ) τκ v (λ 2 ) κ (p 2 )] u (λ 1 ) α (p 1 )γµ αβ u(λ 1) β (p 1 )v (λ 2) γ (p 2 )(γ µ ) γδ v (λ 2 ) δ (p 2 )] v (λ 2 ) κ (p 2)(γ ν ) κτ v (λ 2) τ (p 2 )u (λ 1) σ (p 1 )γ ν σρu (λ 1 ) ρ (p 1) ] γ µ αβ u(λ 1) β (p 1 )u (λ 1) σ (p 1 )γ ν σρu (λ 1 ) ρ (p 1)u (λ 1 ) α (p 1) (γ µ ) γδ v (λ 2 ) δ (p 2)v (λ 2 ) κ (p 2)(γ ν ) κτ v (λ 2) τ (p 2 )v (λ 2) γ (p 2 ) = e2 4ts γµ αβ (/p 1) βσ γ ν σρ(/p 1) ρα (γ µ ) γδ (/p 2) δκ (γ ν ) κτ (/p 2 ) τγ = Tr ( γ µ /p 1 γ ν /p 1) Tr ( γµ /p 2γ ν /p 2 ), where I have used u u γ 0, v v γ 0 and the identities (γ 0 ) 2 = I and γ 0 γ µγ 0 = γ µ. Following the spinor indices (explicitly written out above) allows one to express the final result as a product of two traces. The computation of M b 2 ave is performed in a similar manner. The end result can be obtained from the expression for M a 2 ave by the interchange of p 1 p 2. The calculation of 2Re(M am b ) ave proceeds similarly, and the details will be presented below. Note that after a little practice, one should be able to directly obtain eqs. (19) (21) without laboriously writing out all the spinor indices. 3 In the derivation of eq. (21), the evaluation of the trace yields a result that is manifestly real, so we can drop the Re symbol. 6

7 Finally, we compute, 2Re(M a M b) ave = e4 2ts = e4 2ts = e4 2ts λ 1,λ 2,λ 1,λ 2 λ 1,λ 2,λ 1,λ 2 λ 1,λ 2,λ 1,λ 2 u (λ 1 ) α (p 1)γ µ αβ u(λ 1) β (p 1 )v (λ 2) γ (p 2 )(γ µ ) γδ v (λ 2 ) δ (p 2) ] u (λ 1 ) ρ (p 1 )γν ρσ v(λ 2 ) σ (p 2 )v(λ 2) τ (p 2 )(γ ν ) τκ u (λ 1) κ (p 1 ) ] u (λ 1 ) α (p 1 )γµ αβ u(λ 1) β (p 1 )v (λ 2) γ (p 2 )(γ µ ) γδ v (λ 2 ) δ (p 2 )] u (λ 1) κ (p 1 )(γ ν ) κτ v (λ 2) τ (p 2 )v (λ 2 ) σ (p 2)γ ν σρu (λ 1 ) ρ (p 1) ] γ µ αβ u(λ 1) β (p 1 )u (λ 1) κ (p 1 )(γ ν ) κτ v (λ 2) τ (p 2 )v (λ 2) γ (p 2 ) (γ µ ) γδ v (λ 2 ) δ (p 2)v (λ 2 ) σ (p 2)γ ν σρu (λ 1 ) ρ (p 1)u (λ 1 ) α (p 1) = e4 2ts γµ αβ (/p 1) βκ (γ ν ) κτ (/p 2 ) τγ (γ µ ) γδ (/p 2) δσ γ ν σρ(/p 1) ρα = e4 2ts Tr( ) γ µ /p 1 γ ν /p 2 γ µ /p 2 γν /p 1. Following the spinor indices (explicitly written out above) allows one to express the final result as a single trace. It is straightforward to evaluate the traces of eqs. (19) (21) using the identities TrI = 4, Tr(γ µ γ ν ) = 4g µν, Tr(γ µ γ ν γ α γ β ) = 4 ( g µν g αβ +g µβ g να g µα g νβ), where I is the 4 4 identity matrix. We then obtain, M a 2 ave = e4 4t Tr( ) ( γ µ /p 2 1 γ ν /p 1 Tr (γµ /p 2 γ ) ν/p 2 = 4e4 ( p1µ p t 2 1ν +p 1νp 1µ g )( ) µνp 1 p 1 p µ 2p ν 2 +pν 2 p µ 2 g µν p 2 p 2 = 8e4 t 2 ( p1 p 2 p 1 p 2 +p 1 p 2p 1 p 2 ). (22) Using eqs. (12) (14) to write the dot products in eq. (22) in terms of s, t and u, we obtain M a 2 ave = 2e4 t 2 (s2 +u 2 ). (23) The calculation of M b 2 ave is nearly identical. Indeed, a careful comparison of eqs. (19) and (20) show that they differ by the interchange of p 1 p 2 as previously noted, which in light 7

8 of eqs. (12) and (14) corresponds to s t, with u unchanged. Performing this substitution of variables in eq. (23) then yields M b 2 ave = 2e4 s 2 (t2 +u 2 ). (24) To evaluate the trace in eq. (21), we first derive the following gamma matrix identities, γ µ γ µ = 4, (25) γ µ γ α γ µ = 2γ α, (26) γ µ γ α γ β γ µ = 4g αβ, (27) γ µ γ α γ β γ ν γ µ = 2γ ν γ β γ α. (28) These identities are proved by repeated use of the anticommutation relation, { γ µ, γ ν} = 2g µν. (29) For example, eq. (25) is derived by multiplying eq. (29) by 1 2 g µν and summing over µ and ν. To derive eq. (26), we write (2g µα γ α γ µ )γ µ = 2γ α 4γ α = 2γ α, after using eq. (25). To derive eq. (27), we write γ µ γ α γ β γ µ = (2g µα γ α γ µ )γ β γ µ = 2γ β γ α +2γ α γ β = 4g αβ, after using eqs. (26) and (29), Finally, to derive eq. (28), we write γ µ γ α γ β γ ν γ µ = (2g µα γ α γ µ )γ β γ ν γ µ = 2γ β γ ν γ α 4g βν γ α = 2(2g βν γ ν γ β )γ α 4g βν γ α = 2γ ν γ β γ α, after using eqs. (27) and (29). Using eqs. (27) and (28) to perform the sums over µ and ν, it follows that 2Re(M a M b) ave = e4 2ts Tr( ) γ µ /p 1 γ ν /p 2 γ µ /p 2γ ν /p 1 = e4 ts Tr( ) /p 2 γ ν /p 1 /p 2 γν /p 1 Using eq. (14) to evaluate the dot products, we end up with = 4e4 ts p 1 p 2 Tr( ) /p 2 /p 1 16e 4 = ts p 1 p 2 p 2 p 1. (30) 2Re(M a M b ) ave = 4e4 u 2. (31) ts 8

9 Using eqs. (??, (24) and (31), the spin-averaged squared matrix element, eq. (18) is given by, ] s M 2 2 ave = +u 2 2e4 + t2 +u 2 + 2u2, t 2 s 2 st where the extra minus sign that appears on the interference term in eq. (18) has been accounted for. We can rewrite the above result in the following more convenient form, ( 1 M 2 ave = 2e4 u 2 s ( ) ] 2 t ( s ) 2 + +, (32) t) s t The differential cross section is given by eq. (25) of the class handout entitled, Twoparticle Lorentz invariant phase space. Since all particles involved in the scattering process have the same mass (which is actually being set to zero in this calculation), it follows that using eq. (32), we arrive at dσ dω = α2 2s dσ dω = 1 64π 2 s M 2 ave. u 2 ( 1 s + 1 t) 2 + where α e 2 /(4π) is the fine structure constant. ( ) ] 2 t ( s ) 2 +, (33) s t (b) Integrate over the azimuthal angle, and express dσ/d cos θ as a function of the center-of-momentum scattering angle θ. Sketch the shape of the angular distribution of the scattered electron. What feature of the Feynman diagram is responsible for the divergence of dσ/dcosθ as θ 0? Consider a 2 2 scattering process in which none of the helicities are measured. We are free to choose the center-of-momentum frame and choose the incoming particle threemomenta to lie along the z-direction. We are also free to choose the plane of scattering to be the x-z plane. Thus, there is no azimuthal angle dependence in the differential scattering cross section, and we are free to integrate over the angle φ to obtain 2π. Hence, writing dω = dcosθdφ, it follows that In particular, eq. (33) yields dσ dcosθ = πα2 s dσ 2π dcosθ = 0 dσ dφ = 2πdσ dω dω. u 2 ( 1 s + 1 t) 2 + ( ) ] 2 t ( s ) 2 +, (34) s t 9

10 In the center-of-momentum frame, the four-momenta of the incoming and outgoing electrons and positrons, in the limit where all masses are set to zero, are: p 1 = 1 2 s ( 1; 0, 0, 1), p 2 = 1 2 s ( 1; 0, 0, 1), p 1 = 1 2 s ( 1; sinθ, 0, cosθ), p 1 = 1 2 s ( 1; sinθ, 0, cosθ), where θ is the scattering angle. It follows from eqs. (13) and (14) that t = 2p 1 p 1 = 1 2 s(1 cosθ) = s sin2 (θ/2), u = 2p 1 p 2 = 1 2 s(1+cosθ) = s cos2 (θ/2). Inserting the above expressions for t and u into eq. (34), and simplifying the resulting expression, the end result is given by dσ dcosθ = πα2 1+sin 8 ] (θ/2)+cos 8 (θ/2) s sin 4. (35) (θ/2) A sketch of the angular distribution of Bhabha scattering in units of πα 2 /s, is exhibited in the figure below, which shows the behavior of the angular function in eq. (35). dσ/dcosθ θ 10

11 The divergence as θ 0 corresponds to the kinematic region in which the exchanged photon in the t-channel Feynman diagram shown below is nearly on-shell. e p 1 p 1 e p 1 p 1 e + p 2 p 2 e + Thus, the t-channel propagator is diverging as t 0. This is a well-known feature that we have seen before in Rutherford scattering, whose angular distribution also contains a factor of sin 4 (θ/2) in the denominator. Indeed, if one integrates dσ/dcosθ to get the total cross section, one finds that the total cross section is infinite. This simply reflects the fact that the Coulomb interaction has infinite range one can never escape the influence of a 1/r potential. 4. The interaction Lagrangian of scalar electrodynamics is: L I = iea µ H µ H + ( µ H )H +] +e 2 A µ A µ H + H, (36) where e is the usual coupling strength of quantum electrodynamics. This Lagrangian describes the interaction of photons and charged scalars, H ± (where H = H + ] ). The Feynman rules for scalar electrodynamics are exhibited below. Note that four-momenta specified in the first Feynman rule point in the direction of the arrows as indicated. γ µ p H+ ie(p+p ) µ p H + γ µ H + ie 2 Cg µν γ ν H + 11

12 (a) By explicitly computing the first-order S-matrix element for the scattering process γγ H + H, determine the number C that appears in the Feynman rule given below for the γγh + H vertex. Only the term e 2 A µ A µ H + H from L I cf. eq. (36)] contributes to the scattering process γγ H + H at first order. Hence, S (1) fi = ie 2 d 4 x p 1 p 2 :A µ (x)a µ (x)h + (x)h (x): k 1 k 2, where p 1 is the four-momentum of the incoming H +, p 2 is the four-momentum of the incomingh,andk 1 andk 2 arethefour-momentaofthetwooutgoingphotons, respectively. We shall also write λ 1 and λ 2 for the corresponding helicities of the two outgoing photons. To evaluate this S-matrix element, we contract in all possible ways the operators with the corresponding external states. There are two possible contractions, S (1) fi = ie 2 = 2ie 2 d 4 x { p 1 p 2 A µ (x)a µ (x)h + (x)h (x) k 1 k 2 + p 1 p 2 A µ (x)a µ (x)h + (x)h (x) k 1 k 2 d 4 xe ix (k 1+k 2 p 1 p 2 ) ǫ(k 1,λ 1 ) ǫ(k 2,λ 2 ) = 2ie 2 (2π) 4 δ(p 1 +p 2 k 1 k 2 )g µν ǫ µ (k 1,λ 1 )ǫ ν (k 2,λ 2 ), where we have used the following rules for contractions of the photon fields with the initial states and the scalar fields with the final states, A µ (x) k, λ = ǫ µ (k,λ)e ik x, p 1 H (x) = e ip 1 x, p 2 H + (x) = e ip 2 x. where ǫ(k 1,λ 1 ) and ǫ(k 2,λ 2 ) are the polarization vectors of the two incoming photons, respectively. The invariant matrix element is defined by Hence, S fi = δ fi +i(2π) 4 δ(p 1 +p 2 k 1 k 2 )M. im = 2ie 2 g µν ǫ µ (k 1,λ 1 )ǫ ν (k 2,λ 2 ). (37) The Feynman rules provide a diagrammatic method for evaluating im (note the factor of i here, which is conventional). The Feynman rule for an incoming photon of fourmomentum k and helicity λ is } µ,λ k ǫ µ (k,λ) where the solid dot indicates the vertex where the incoming photon will interact. 12

13 Thus, in order to reproduce eq. (37), we conclude that the Feynman rule for the γγh + H vertex must be γ µ H + 2ie 2 g µν γ ν H + That is, C = 2. (b) At O(e 2 ), there are also contributions at second order to the S-matrix element. Using the above Feynman rules, write down the complete O(e 2 ) invariant matrix element for the scattering process γγ H + H. Simplify as much as possible each term that contributes to the matrix element by using the kinematical constraints of the problem. There are three Feynman diagrams that contribute at O(e 2 ) to the scattering process, γγ H + H. γ k 1 p 1 H + γ k 1 p 1 H + k 1 γ H + p 1 p 1 k 1 k 1 p 2 γ k 2 p 2 H γ k 2 p 2 H γ k 2 p 2 H (a) (b) (c) Note that the arrows indicate the direction of flow of positive charge. In order to reduce the number of arrows, we also employ the arrows to denote the direction of the four-momenta of the charged scalars. Thus, the outgoing H with four-momentum p 2 is indicated in the the three Feynman diagrams above as the flow of four-momentum p 2 into the vertex. Using the Feynman rules given in part (a), we can immediately write down the contributions of the three Feynman diagrams above to the invariant matrix element, im a = ie(2p 1 k 1 ) ǫ(k 1,λ 1 ) ] ie(p 1 k 1 p 2 ) ǫ(k 2,λ 2 ) ] i (p 1 k 1 ) 2 M, (38) 2 im b = ie(k 1 2p 2 ) ǫ(k 1,λ 1 ) ] ie(k 1 p 2 +p 1 ) ǫ(k 2,λ 2 ) ] i (k 1 p 2 ) 2 M, (39) 2 im c = 2ie 2 ǫ(k 1,λ 1 ) ǫ(k 2,λ 2 ), (40) where M is the mass of the charged scalar. 13

14 Note that (p 1 k 1 ) 2 M 2 = 2p 1 k 1, (k 1 p 2 ) 2 M 2 = 2k 1 p 2, after using k 2 1 = 0 (since the photon is massless) and p 2 1 = p 2 2 = M 2. Next, we use four momentum conservation, k 1 +k 2 = p 1 +p 2, to write p 1 k 1 p 2 = k 2 2p 2, k 1 p 2 +p 1 = 2p 1 k 2, which allow us to write M a and M b is a slightly more symmetrical form, M a = e2 (2p 1 k 1 ) ǫ 1 (k 2 2p 2 ) ǫ 2 2p 1 k 1, (41) M b = e2 (k 1 2p 2 ) ǫ 1 (2p 1 k 2 ) ǫ 2 2p 2 k 1, (42) where ǫ 1 ǫ(k 1,λ 1 ) and ǫ 2 ǫ 2 ǫ(k 2,λ 2 ). The total invariant matrix element at O(e 2 ) is given by M = M a +M b +M c. (43) Note the relative plus signs, since we are dealing with a scattering process involving only bosons. Using eqs. (40), (41) and (42), the end result is M = e 2 (2p1 k 1 ) ǫ 1 (k 2 2p 2 ) ǫ 2 + (k ] 1 2p 2 ) ǫ 1 (2p 1 k 2 ) ǫ 2 +2ǫ 1 ǫ 2. (44) 2p 1 k 1 2p 2 k 1 We can simplify eq. (44) further using k 1 ǫ 1 = k 2 ǫ 2 = 0 cf. eq. (50) of Solution Set 4], which then yields p1 ǫ M = 2e 2 1 p 2 ǫ 2 + p ] 1 ǫ 2 p 2 ǫ 1 ǫ 1 ǫ 2. (45) p 1 k 1 p 2 k 1 (c) The matrix element obtained in part (b) takes the form M = M µν ǫ µ 1 (k 1)ǫ ν 2 (k 2), where k 1 and k 2 are the initial photon four-momenta and ǫ 1 and ǫ 2 are the corresponding photon polarization four-vectors. Verify that k µ 1 M µνǫ ν 2 (k 2) = k ν 2 M µνǫ µ (k 1 ) = 0. That is, the replacement of either ǫ 1 k 1 and/or ǫ 2 k 2 in M yields an expression that vanishes. These relations are consequences of the gauge invariance of electrodynamics and provide an important check that all the contributing Feynman diagrams of O(e 2 ) have been correctly included. It also serves as a check of the value of C that you obtained in part (a). Let us rewrite eq. (44) equation as follows, M = 2e 2 p1µ p 2ν + p ] 1νp 2µ g µν ǫ µ 1 p 1 k 1 p 2 k ǫν 2. (46) 1 14

15 First, we replace ǫ 1 with k 1, p1µ p 2ν + p ] 1νp 2µ g µν k µ 1 p 1 k 1 p 2 k ǫν 2 = (p 2 +p 1 k 1 ) ǫ 2 = k 2 ǫ 2 = 0, 1 after using momentum conservation and k 2 ǫ 2 = 0. Second, we replace ǫ 2 with k 2, p1µ p 2ν + p ] ( ) ( ] 1νp 2µ p2 k g µν ǫ µ p 1 k 1 p 1k2 ν 2 p1 k 2 = p 1µ + )p 2µ k 2µ ǫ µ 1. (47) 2 k 1 p 1 k 1 p 2 k 1 To make further progress, note that, (p 1 k 1 ) 2 = (k 2 p 2 ) 2 = p 1 k 1 = p 2 k 2, (48) (p 1 k 2 ) 2 = (k 1 p 2 ) 2 = p 1 k 2 = p 2 k 1, (49) due to four-momentum conservation, k1 2 = k2 2 = 0 and p2 1 = p2 2 = M2. It therefore follows that the two ratios of dot products in eq. (47) are both equal to 1. Hence, eq. (47) yields, p1µ p 2ν + p ] 1νp 2µ g µν ǫ µ p 1 k 1 p 1k2 ν = (p 1 +p 2 k 2 ) ǫ 1 = k 1 ǫ 1 = 0, 2 k 1 After using momentum conservation and k 1 ǫ 1 = 0. Finally, if we replace both ǫ 1 k 1 and ǫ 2 k 2 in eq. (46), we obtain, p1µ p 2ν + p ] ( ) ( ] 1νp 2µ p2 k g µν k µ 1 p 1 k 1 p 2 k kν 2 = 2 p1 k 2 p 1µ + )p 2µ k 2µ k µ 1 1 p 1 k 1 p = 2 k k2 1 = 0. 1 Note that the gauge invariance checks that we have just performed also confirm the relative signs in eq. (43) as well as the value of C = 2 obtained in part (a). REMARKS: It is interesting to note that if we write M = M µν ǫ µ 1 (k 1)ǫ ν 2 (k 2) prior to invoking the transversality of the polarization vectors, k 1 ǫ 1 = k 2 ǫ 2 = 0, then we would use eq. (44) to identify M µν = e 2 (2p1 k 1 ) µ (k 2 2p 2 ) ν 2p 1 k 1 + (k ] 1 2p 2 ) µ (2p 1 k 2 ) ν +2g µν. 2p 2 k 1 In this case, one can show that gauge invariance in electrodynamics implies that k µ 1 M µν = k ν 2 M µν = 0. (50) Let us verify this form of the gauge invariance check. First, k µ 1 M µν = e 2 k1 (2p 1 k 1 )(k 2 2p 2 ) ν 2p 1 k 1 + k 1 (k 1 2p 2 )(2p 1 k 2 ) ν 2p 2 k 1 +2k 1ν = e 2 (k 2 2p 2 ) ν (2p 1 k 2 ) ν +2k 1ν ] = 2e 2 (k 1 +k 2 p 1 p 2 ) ν = 0, after using k 2 1 = 0 and four-momentum conservation. ] 15

16 Second, k 2M µ µν = e 2 (2p1 k 1 ) µ k 2 (k 2 2p 2 ) + (k ] 1 2p 2 ) µ k 2 (2p 1 k 2 ) +2k 2µ 2p 1 k 1 2p 2 k 1 = e 2 (2p1 k 1 ) µ k 2 (k 2 2p 2 ) 2p 2 k 2 + (k 1 2p 2 ) µ k 2 (2p 1 k 2 ) 2p 1 k 2 +2k 2µ = e 2 (k 1 2p 1 ) µ +(k 1 2p 2 ) µ +2k 2µ ] = 2e 2 (k 1 +k 2 p 1 p 2 ) µ = 0, after using k2 2 = 0 and four-momentum conservation. Note that in the penultimate step above, we used the the results of eqs. (48) and (49). Thus, we have confirmed that eq. (50) is indeed satisfied. ] 16