Dirac Trace Techniques

Transcript

1 Dirac Trace Techniques Consider a QED amplitude involving one incoming electron with momentum p and spin s, one outgoing electron with momentum p and spin s, and some photons. There may be several Feynman diagrams contributing to this amplitude, but they all have the same external legs and the corresponding factors up, s and ūp, s. Consequently, we may write the amplitude as e,... M e,... = ūp, s Γup, s 1 where Γ is comprises all the other factors of the QED Feynman rules; for the moment, we don t want to be specific, so Γ is just some kind of a 4 4 matrix. In many experiments, the initial electrons come in un-polarized beam, 50% having spin s = and 50% having s = 1 2. At the same time, the detector of the final electrons measures their momenta p but is blind to their spins s. The cross-section σ measured by such an experiment would be the average of the polarized cross-sections σs, s with respect to initial spins s and the sum over the final spins s, thus σ = 1 2 s s σs, s. 2 Similar averaging / summing rules apply to the un-polarized partial cross-sections, dσ dω = 1 2 s s dσs, s dω, 3 etc., etc. Since all total or partial cross-sections are proportional to mod-squares M 2 of amplitudes M, we need to know how to calculate M 2 def = 1 2 Ms, s 2 4 s s for amplitudes such as 1. 1

2 First, To do such a calculation efficiently, we need to recall two things about Dirac spinors. If M = ūp, s Γup, s then M = ūp, sγup, s 5 where Γ = γ 0 Γ γ 0 is the Dirac conjugate of the matrix Γ; for a product γ λ γ ν of Dirac matrices, γ λ γ ν = γ ν γ λ. Second, and likewise u α p, s ū β p, s = p + m αβ 6 s Combining these two facts, we obtain s u γ p, s ū δ p, s = p + m γδ. 7 M = ūp, s Γup, s 2 = ūp, s Γup, s ūp, sγup, s s,s s,s = ū δ p, s Γ δα u α p, s ū β p, s Γ βγ u γ p, s s,s δ,α β,γ = Γ δα Γ βγ u α p, s ū β p, s and hence α,β,γ,δ = α,β,γ,δ s s u γ p, s ū δ p, s Γ δα Γ βγ p + m αβ p + m γδ = matrix product p + m Γ p + m Γ γ = tr p + m Γ p + m Γ e,... M e,... = ūp, s Γup, s = 1 M 2 = tr p +m Γ p+m Γ. s,s A similar trace formula exists for un-polarized scattering of positrons. In this case, the γγ 8 9 2

3 amplitude is e +,... M e +,... = vp, sγvp, s 10 note that vp, s belongs to the outgoing positron while vp, s belongs to the incoming s +, and we need to average M 2 over s and sum over s. Using v α p, s v β p, s = p m αβ 11 s and working through algebra similar to eq. 8, we arrive at e +,... M e +,... = vp, sγvp, s = 1 M 2 = tr p m Γ p m Γ. s,s 12 Now suppose an electron with momentum p 1 and spin s 1 and a positron with momentum p 2 and spin s 2 come in and annihilate each other. In this case, the amplitude has form... M e 1, e + 2,... = vp 2, s 2 Γup 1, s 1 13 for some Γ, and if both electron and positron beams are un-polarized, we need to average the M 2 over both spins s 1 and s 2. Again, there is a trace formula for such averaging, namely... M e 1, e + 2,... = vp 2, s 2 Γup 1, s 1 = 1 M 2 = tr p 2 m Γ p 1 +m Γ. s 1,s 2 form Finally, for a process in which an electron-positron pair is created, the amplitude has 14 e 1, e+ 2,... M... = ūp 1, s 1Γvp 2, s 2, 15 and if we do not detect the spins of the outgoing electron and positron but only their momenta, then we should sum the M 2 over both spins s 1 and s 2. Again, there is a trace 3

4 formula for this sum, namely e 1, e+ 2,... M... = ūp 1, s 1 Γvp 2, s 2 = M 2 = tr p 1 +m Γ p 2 m Γ. s 1,s 2 Processes involving 4 or more un-polarized fermions also have trace formulae for the spin sums. For example, consider an e + e + collision in which a muon pair µ + µ + is created. There is one tree diagram for this process, µ µ + p 1 p 2 16 q 17 and it evaluates to p 1 p 2 e e + i µ, µ + M e, e + = igλν q 2 ūµ ieγ λ vµ + ve + ieγ ν ue = ie2 s ūµ γ ν vµ + ve + γ ν ue 18 where s = q 2 = p 1 + p 2 2 = p 1 + p 2 2 = E 2 c.m. 19 is the square of the total energy in the center-of-mass frame. When the initial electron and positron are both un-polarized, and we do not measure the spins of the muons but only their moments, we need to average the M 2 over both initial spins s 1, s 2 and some over both final spins s 1, s 2, thus dσ dω c.m. = M 2 64π 2 s where M 2 1 M s 1,s 2,s 1,s 2 4

5 For the amplitude 18 at hand, M M = e4 s 2 ūµ γ ν vµ + ve + γ ν ue note sums over two separate Lorentz indices λ and ν = e4 s 2 ūµ γ ν vµ + vµ + γ λ uµ and consequently vµ + γ λ uµ ū e γ λ ve + ve + γ ν ue ū e γ λ ve + M 2 = e4 4s 2 ūµ γ ν vµ + vµ + γ λ uµ s 1,s 2 ve + γ ν ue ūe γ λ ve + s 1,s 2 = e4 4s 2 tr p 1 + M µ γ ν p 2 M µ γ λ tr p 2 m e γ ν p 1 + m e γ λ Calculating Dirac Traces Thus far, we have learned how to express un-polarized cross-sections in terms of Dirac traces i.e., traces of products of the Dirac γ λ matrices. In this section, we shall learn how to calculate such traces. Dirac traces do not depend on the specific form of the γ 0, γ 1, γ 2, γ 4 matrices but are completely determined by the Clifford algebra {γ µ, γ ν } γ µ γ ν + γ ν γ µ = 2g µν. 23 To see how this works, please recall the key property of the trace of any matrix product: trab = trba for any two matrices A and B. Proof: trab = α AB αα = α A αβ B βα = β β BA ββ = trba. 24 This symmetry has two important corollaries: 5

6 All commutators have zero traces, tr[a, B] = 0 for any A and B. Traces of products of several matrices have cyclic symmetry trabc Y Z = trbc Y ZA = trc Y ZAB = = trzabc Y. Using these properties it is easy to show that 25 trγ µ γ ν = 4g µν = tr a b = 4ab 4a µ b µ. 26 Indeed, trγ µ γ ν = trγ ν γ µ = tr 1 2 {γ µ, γ ν } = trg µν = g µν tr1 = g µν 4, 27 where the last equality follows from Dirac matrices being 4 4 and hence tr1 = Next, all products of any odd number of the γ µ matrices have zero traces, trγ µ = 0, trγ λ γ µ γ ν = 0, trγ λ γ µ γ ν γ ρ γ σ = 0, etc., 29 and hence tr a = 0, tr a b c = 0, tr a b c d e = 0, etc. 30 To see how this works, we can use the γ 5 matrix which anticommutes with all the γ µ and hence with any product Γ of an odd number of the Dirac γ s, γ 5 Γ = Γγ 5. Consequently, Γ = γ 5 γ 5 Γ = γ 5 Γγ 5 = 1 2 [γ5 Γ, γ 5 ] 31 and hence trγ = 1 2 tr[γ5 Γ, γ 5 ] = Products of even numbers n = 2m of γ matrices have non-trivial traces, and we may calculate them recursively in n. We already know the traces for n = 0 and n = 2, so consider 6

7 a product γ κ γ λ γ µ γ ν of n = 4 matrices. Thanks to the cyclic symmetry of the trace, trγ κ γ λ γ µ γ ν = trγ λ γ µ γ ν γ κ = tr 12 {γ κ, γ λ γ µ γ ν } 33 where the anticommutator follows from the Clifford algebra 23: 1 2 {γ κ, γ λ γ µ γ ν } = g κλ γ µ γ ν g κµ γ λ γ ν + g κν γ λ γ µ, 34 cf.homework set#8, problem 2c. Therefore, and hence trγ κ γ λ γ µ γ ν = g κλ trγ µ γ ν g κµ trγ λ γ ν + g κν trγ λ γ µ = 4g κλ g µν 4g κµ g λν + 4g κν g λµ 35 tr a b c d = 4abcd 4acbd + 4adbc. 36 Note that in eq. 35 we have expressed the trace of a 4 γ product to traces of 2 γ products. Similar recursive formulae exist for all even numbers of γ matrices, trγ ν1 γ ν2 γ νn = tr 1 2 {γ ν1, γ ν2 γ νn } For example, for n = 6 for even n only n = 1 k g ν1ν k tr γ ν2 γ νk γ νn. k=2 37 trγ κ γ λ γ µ γ ν γ ρ γ σ = g κλ trγ µ γ ν γ ρ γ σ g κµ trγ λ γ ν γ ρ γ σ + g κν trγ λ γ µ γ ρ γ σ g κρ trγ λ γ µ γ ν γ σ + g κσ trγ λ γ µ γ ν γ ρ = 4g κλ g µν g ρσ g µρ g νσ + g µσ g νρ 4g κµ g λν g ρσ g λρ g νσ + g λσ g νρ + 4g κν g λµ g ρσ g λρ g µσ + g λσ g µρ 4g κρ g λµ g νσ g λν g µσ + g λσ g µν + 4g κσ g λµ g νρ g λν g µρ + g λρ g µν. For products of more γ matrices, the recursive formulae 37 for traces produce even more terms 105 terms for n 8, 945 terms for n = 10, etc., etc., so it helps to reduce 38 7

8 n whenever possible. For example, if the matrix product inside the trace contains two a matrices for the same 4 vector a µ next to each other, you can simplify the product using a a = a 2, thus tr a a b c = a 2 tr b c. 39 Also, when a product contains γ α and γ α with the same Lorentz index α which should be summed over, we may simplify the trace using γ α γ α = 4, γ α aγ α = 2 a, γ α a bγ α = +4ab, γ α a b cγ α = 2 c b a, 40 etc., cf. homework set#8, problem 2d. In the electroweak theory, one often needs to calculate traces of products containing the γ 5 matrix. If the γ 5 appears more than once, we may simplify the product using γ 5 γ 5 = 1 and γ 5 γ ν = γ ν γ 5. For example, tr 1 γ 5 γ µ p 1 γ 5 γ ν q = tr 1 γ 5 γ µ 1 + γ 5 p γ ν q = tr 1 γ 5 1 γ 5 γ µ p γ ν q = 2 tr 1 γ 5 γ µ p γ ν q because 1 γ 5 2 = 1 2γ 5 + γ 5 γ 5 = 21 γ 5 = 2 tr γ µ p γ ν q 2 tr γ 5 γ µ p γ ν q. 41 When the γ 5 appears just one time, we may use γ 5 = iγ 0 γ 1 γ 2 γ 3 to show that trγ 5 = 0, trγ 5 γ ν = 0, trγ 5 γ µ γ ν = 0, trγ 5 γ λ γ µ γ ν = 0, 42 while trγ 5 γ κ γ λ γ µ γ ν = 4iǫ κλµν. 43 For more γ ν matrices accompanying the γ 5 we have trγ 5 γ ν1 γ νn = 0 odd n, 44 8

9 while for even n = 6, 8,... there are recursive formulae based on the identity γ 5 γ λ γ µ γ ν = g λµ γ 5 γ ν g λν γ 5 γ µ + g µν γ 5 γ λ iǫ λµνρ γ ρ. 45 Muon Pair Production As an example of trace technology, let us calculate the traces 22 for the muon pair production. Let s start with the trace due to summing over muons spins, tr p 1 + M µγ λ p 2 M µγ ν = tr p 1 γλ p 2 γν + M µ trγ λ p 2γ ν M µ tr p 1γ λ γ ν 46 M 2 µ trγ λ γ ν. In the second line here, we have three γ matrices inside each trace, so those traces vanish. In the third line, trγ λ γ ν = 4g λν. Finally, the trace in the first line follows from eq. 35, tr p 1 γλ p 2 γν = p aα p 2β trγα γ λ γ β γ ν = p aα p 2β 4 g αλ g βν g αβ g λν + g αν g λβ = 4p λ 1 p ν 2 4p 1 p 2 gλν + 4p ν 1 p λ Altogether, tr p 1 + M µγ λ p 2 M µγ ν = 4p λ 1 p ν 2 + 4p ν 1 p λ 2 4Mµ 2 + p 1 p 2 gλν 48 = 4p λ 1 p ν 2 + 4p ν 1 p λ 2 2s g λν where on the last line I have used s p 1 + p 2 2 = p p p 1 p 2 = 2M2 µ + 2p 1 p

10 Similarly, for the second trace 22 due to averaging over electron s and positron s spins, we have tr p 2 m e γ ν p 1 + m e γ λ = tr p 2 γ ν p 1 γ λ + m e tr p 2 γ ν γ λ m e trγ ν p 1 γ λ m 2 e trγ νγ λ = 4p 2ν p 1λ 4p 2 p 1 g νλ + 4p 2λ p 1ν 50 + m e 0 m e 0 m 2 e 4g νλ = 4p 2ν p 1λ + 4p 2λ p 1ν 4p 2 p 1 + m 2 e g λν = 4p 2ν p 1λ + 4p 2λ p 1ν 2s g λν, where on the last line I have used s p 1 + p 2 2 = p p p 2p 1 = 2m 2 e + 2p 2p It remains to multiply the two traces and sum over the Lorentz indices λ and ν: tr p 1 + M µ γ ν p 2 M µ γ λ tr p 2 m e γ ν p 1 + m e γ λ = = 4p ν 1 p λ 2 + 4p λ 1 p ν 2 2s g νλ 4p 2ν p 1ν + 4p 2λ p 1ν 2s g νλ = 16p ν 1 p λ 2 + p λ 1 p ν 2 p 2ν p 1λ + p 2λ p 1ν 8s g νλ p 2ν p 1λ + p 2λ p 1ν 8s g νλ p λ 1 p ν 2 + p ν 1 p λ 2 + 4s 2 g λν g λν = 16 2 p 1 p 1p 2 p 2 + p 2 p 1p 1 p 2 8s 2p 1 p 2 8s 2p 1 p 2 + 4s2 4 = 32p 1p 1 p 2p p 2p 1 p 1p s M 2 µ + m 2 e where the last line follows from eqs. 49 and 51. Hence, in eq. 22 we have 52 M all spins M 2 = 4e4 s 2 2p 1 p 1p 2 p 2 + 2p 2 p 1p 1 p 2 + sm 2 µ + m2 e

11 p ν 1,2 Finally, let s work out the kinematics of pair production. In the center-of-mass frame, = E, ±p and p ν 1,2 = E,p, same E = 1 2 E cm but p p. Therefore, p 1p 1 = p 2p 2 = E 2 p p, p 2p 1 = p 1p 2 = E 2 + p p, s = 4E 2, 2p 1p 1 p 2p 2 + 2p 2p 1 p 1p 2 = 2E 2 p p 2 + 2E 2 + p p 2 = 4E 4 + 4p p 2 54 = 4E 4 + 4p 2 p 2 cos 2 θ, and consequently M 2 = e p 2 p 2 E 4 cos 2 θ + M2 µ + m 2 e E where p 2 = E 2 Mµ 2 and p 2 = E 2 m 2 e. We may simplify this expression a bit using the experimental fact that the muon is much heavier than the electron, M µ 207m e, so we need ultra-relativistic e to produce µ, E > M µ m e. This allows us to neglect the m 2 e term in eq. 55 and let p 2 = E 2, thus M 2 = e M2 µ E M2 µ E 2 cos 2 θ, 56 and consequently the partial cross-section is dσ dω c.m. = α2 4s 1 + M2 µ E M2 µ E 2 cos 2 θ 1 M2 µ E 2 57 where the root comes from the phase-space factor p / p for inelastic processes. Looking at the angular dependence of this partial cross-section, we see that just above the energy threshold, for E = M µ +small, the muons are produced isotropically in all directions. 11

12 On the other hand, for very high energies E M µ when all 4 particles are ultra-relativistic, dσ dω c.m. 1 + cos 2 θ. 58 In homework set#12 you will see that the polarized cross sections depend on the angle as 1 ± cosθ 2 where the sign ± depends on the helicities of initial and final particles; for the un-polarized particles, we average / sum over helicities, and that produces the averaged angular distribution 1 + cos 2. Finally, the total cross-section of muon pair production follows from eq. 57 using d 2 Ω = 4π, d 2 Ω cos 2 θ = 4π 3, 59 hence σ tot e e + µ µ + = 4π 3 α 2 s 1 + M2 µ 2E 2 1 M2 µ E Well above the threshold σ tot σ tot 0 4π 3 α 2 s. 61 At the threshold the cross-section is zero, but it rises very rapidly with energy and for E = 1.25M µ, σ tot reaches 85% of σ tot 0. σ tot /σ tot 0 1 M µ E 12