0 Ψ d γ α. Ψ u π + 0 Ψ d γ 5 γ α. Ψ u π + ). 2

Transcription

1 PHY 396 K/L. Solutions for homework set #25. Problem 1a: For the sake of definiteness, let s focus on the decay of the positive pion, π + µ + ν µ. In the Fermi s current-current theory 1 of this weak decay, the J current annihilates the positive pion while the J + current creates the positive muon and the neutrino. In light of eqs. 2 for the currents, µ + ν µ Ĵ +α L 0 = µ + ν µ 1 2 Ψ νµ 1 γ 5 γ α Ψ µ 0 = 1 2ūν µ1 γ 5 γ α vµ +, S.1 while 0 Ĵ α L π + = cosθ c 0 Ψ d γ α Ψ u π + 0 Ψ d γ 5 γ α Ψ u π +. 2 S.2 Inside the on the RHS, the axial isospin current annihilates the pion according to eq. 3 while the vector isospin current has a zero matrix element between the pion and the vacuum states since the vector isospin symmetry is unbroken. Consequently, and therefore 0 Ĵ α L π + = f π cosθ c 2 p α S.3 µ + ν µ LFermi π + = G F f π cosθ c p α π ūν µ 1 γ 5 γ α vµ +. S.4 At the tree level, this immediately translates into the decay amplitude 4. Problem 1b: Let s sum the amplitude 4 or rather its mod-square over the spin states of the outgoing muon and neutrino. As usual, we get a Dirac trace M 2 = G 2 F f2 π cos2 θ c spins spins where ūν 1 γ 5 p π v µ 2 S.5 = G 2 F f2 π cos 2 θ c tr p ν +m ν 1 γ 5 p π p µ M µ 1 γ 5 p π 1 γ 5 p π = p π 1 γ 5 = p π 1+γ 5 = 1 γ 5 p π. S.6 Since the neutrino s mass m ν is so much smaller than any other mass and momentum in the 1

2 process, we may safely neglect it, hence tr = tr p ν 1 γ 5 p π p µ M µ 1 γ 5 p π = tr 1 γ 5 p π p ν 1 γ 5 p π p µ M µ using a bγ 5 = +γ 5 a b = tr 1 γ 5 2 p π p ν p π p µ M µ using a bγ 5 = +γ 5 a b = 2tr 1 γ 5 p π p ν p π p µ M µ using 1 γ 5 2 = 21 γ 5 = 2tr 1 γ 5 p π p ν p π p µ using tr a b c = trγ 5 a b c = 0 S.7 = 16p π p ν p π p µ 8p 2 π p νp µ 8iǫ αβγδ p α π pβ ν pγ π pδ µ. Moreover, the last term on the bottom line vanishes by the antisymmetry of the ǫ tensor = ǫ αβγδ p α π pγ π = 0, so we are left with M 2 = G 2 F f2 π cos2 θ c spins 16p π p ν p π p µ 8p 2 π p νp µ. S.8 Now consider the kinematics of the pion decay. For the 3 on-shell momenta satisfying p π = p µ +p ν we have 2p π p µ = M 2 π +M 2 µ, 2p π p ν = 2p µ p ν = M 2 π M 2 µ, S.9 hence 16p π p ν p π p µ 8p 2 πp ν p µ = 4M 2 π M 2 µm 2 π+m 2 µ 4M 2 πm 2 π M 2 µ = 4M 2 π M 2 µ M 2 µ, and therefore S.10 Mπ + µ + ν µ 2 = 4G 2 F f2 π cosθ c Mπ 2 Mµ M 2 µ. 2 spins S.11 Also, the phase-space factor for the two-body decay of a spinless particle is dphase space = p c.m. 32πM 2 π d 2 Ω c.m. = net phase space = p c.m. 8πM 2 π S.12 2

3 where p c.m. = E νpionframe = 2p πp ν 2M π = M2 π M2 µ 2M π. S.13 Thus, ombining all these formulae, we obtain the net tree-level pion decay rate as Γπ + µ + ν µ = spins Mπ + µ + ν µ 2 net phase space = 4G 2 F f2 π cosθ c Mπ 2 MµM 2 µ 2 M2 π Mµ 2 16πMπ 3 2 = G2 F f2 π cos 2 θ c M π Mµ 2 1 M2 µ 4π Mπ 2. S.14 Numerically, for M π 140 MeV, M µ 106 MeV, f π 93 MeV, G F = GeV 1, and θ c 13, this formula gives us Γπ + µ + ν µ ev = 1 Γ seconds. S.15 This is quite close to the experimental mean lifetime of a charged pion τ = s. Problem 1c: The charged weak current J L +α can create a positron and an electron-type neutrino just as well as it can create a positive muon and a muon-type neutrino. The amplitude for both π + e + ν e and π + µ + ν µ processes is given by exactly the same equation 4, hence proceeding exactly as in part b we find Γπ + e + ν e = G2 F f2 π cos 2 θ c M π 4π M 2 e 2 1 M2 e Mπ 2. S.16 But comparing this formula to eq. S.14, we find that M e M µ leads to a much smaller decay rate into electrons rather than muons, Γπ + e + ν e Γπ + µ + ν µ = G2 F f2 π cos2 θ c M π 8π M 2 e 2 / 1 M2 e G 2 F fπ 2cos2 θ c M π Mπ 2 8π M 2 µ 1 M2 µ M 2 π 2 = M2 e M 2 µ 1 M e/m π M µ /M π

4 To explain this very small ratio of decay rates, consider the chirality structure of the amplitude 4. The 1 γ 5 factor projects the Dirac factors vµ + and uν µ onto their left-handed components. Indeed, in the frame of the initial pion M u ν µ γ 0 1 γ 5 γ 0 vµ = u ν µ vµ S.17 = u L ν µv L µ +, so both fermions must have left-handed chirality. At the same time, the angular momentum conservation requires the two fermions to have similar helicities. Indeed, since the pion has no spin, it its rest frame J net = 0 and hence J ν = J µ. But the muon and the neutrino travel in opposite directions, p ν = p µ, so opposite angular momenta mean similar helicities: λ = p J p = λ ν = +λ µ. S.18 Finally, for massless particles, the helicity follows from the chirality and vice verse, but the relation is different for the particles and the antiparticles: For a particle such as e, or µ, or ν helicity and chirality have the same sign they are both left or both right. But for an antiparticle such as e + or µ + helicity and chirality have opposite signs. So if both the neutrino and the charged muon were massless, similar helicities according to eq. S.18 would require them to have opposite chiralities, one left and the other right. At the same time, the weak interactions S.17 require left chiralities of both fermions. Since the two requirements contradict each other, the weak decay of a pion into two massless leptons is impossible. To make the decay happen, one of the two leptons must have mis-matched helicity and chirality, and this requires mass. Since the neutrino is much lighter than the charged lepton, the mismatch happens for the µ + : it has left chirality but also left helicity because the neutrino has 4

5 λ = 1 2, which is a mis-match for an anti-lepton. The cost of this mis-match is a suppression factor 1 β M/E in the decay amplitude, which comes from the muon Dirac wave + E pηl vµ + L = E +pη L = v L µ + L = 1 β 2Eη L. S.19 This factor is not too small for the mildly-relativistic muon, but it would be much smaller for an ultra-relativistic positron in the π + e + ν e decay. And that s why spinless mesons do not like to decay into light leptons: the smaller the lepton mass, the smaller the Dirac wave v or u becomes for the mis-matched chirality and helicity. Problem 2a: In the matrix notation for the non-abelian gauge fields, F µν = µ A ν ν A µ + ig[a µ,a ν ], S.20 hence g 2 16π 2 ǫαβγδ tr F αβ F γδ = g 2 4π 2 ǫαβγδ tr α A β γ A δ + ig3 4π 2 ǫαβγδ tr [A α,a β ] γ A δ g4 16π 2 ǫαβγδ tr [A α,a β ][A γ,a δ ] [two gluons] [three gluons] [four gluons]. S.21 Thanks to the cyclic symmetry of the trace, for any matrices X, Y, and Z, tr[x,y]z = trx[y,z]. S.22 Applying this rule to the 4 gluon term in the decomposition S.21, we have [4 gluon anomaly] ǫ αβγδ tr [A α,a β ][A γ,a δ ] = ǫ αβγδ tr A α [A β,[a γ,a δ ]]. S.23 In this double-commutator formula, we may use the Jacobi identity [A β,[a γ,a δ ]] + [A γ,[a δ,a β ]] + [A δ,[a β,a γ ]] = 0. S.24 Since the ǫ αβγδ is symmetric with respect to cyclic permutations of the last three indices β 5

6 γ δ β, it follows that ǫ αβγδ [A β,[a γ,a δ ]] = 0 S.25 and hence [4 gluon anomaly] = 0. S.26 Quod erat demonstrandum. Problem 2b, first part: Reducing the 3-gluon anomaly calculation to the loops of massive Pauli Villars compensators only. As explained in class, for a massless fermion µ ν µ ν µ ν S.27 iq α γ 5 γ α = γ 5 igγ ν γ 5 igγ µ but for a massive fermion such as the Pauli Villars compensator χ PV, µ ν µ ν µ ν µ ν = + 2iMγ 5 iq α J 5α γ 5 igγ ν γ 5 igγ µ S.28 6

7 Now, let s apply these rules to the quadrangle anomaly diagram 7 regulated à la Pauli Villars, regulated = massless quark + heavy PV regulator J 5α J 5α J 5α S.29 Multiplying the axial current J 5α q by iq α, we get regulated = regulated regulated + PV regulator only iq α J 5α 2iMγ 5 S.30 Without the regulation, the two triangular diagrams on the right hand side would diverge as UV cutoff scale +1, but subtracting similar loops of the heavy Pauli Villars fermions makes them converge. Consequently, we may shift the momentum integration variables p µ p µ +const separately for each diagram, and this would lead to cancellation of the triangle diagrams once we sum over gluon permutations. 7

8 Actually, it is enough to sum over just the cyclic permutations of the three gluons which have the same group factor trt 3 T 2 T 1. Note the cyclic symmetry of the trace. Summing the triangular diagrams on the right hand side of eq. S.30, we have diagram-by-diagram cancellation: regulated + regulated + regulated = regulated regulated regulated S.31 Strictly speaking, this cancellation involves separate shifting of the integration momentum for each diagram, but that s OK since the regulated diagrams are finite. But the quadrangle diagrams involving only the massive PV regulator loops cf. the last term in eq. S.30 do not cancel after we sum over gluon permutations, and that s what leads to the quadrangle anomaly: 8

9 regulated = PV regulator only + gluon permutations. iq α J 5α 2iMγ 5 S.32 Problem 2b, second part: Evaluating the loops S.32 of the Pauli Villars compensators. For any particular order of the 3 gluons for example, for the order explicitly shown on the RHS of eq. S.32, we have 2 b µ 1 a λ 3 c ν p 1 p 2 p 3 = +tr colort c T b T a p 4 d 4 p 1 2π 4 tr Dirac S.33 2iMγ 5 where 2iMγ 5 i i p tr Dirac = tr 4 M +i0 igγν p 3 M +i0 igγ µ i i p 2 M +i0 igγλ p 3 M +i0 S.34 = 2Mg 3 N D 9

10 for 4 D = p 2 1 M2 +i0, S.35 i=1 N = tr γ 5 p 4 +Mγ ν p 3 +Mγ µ p 2 +Mγ λ p 1 +M. S.36 As usual, we may express the denominator in terms of the Feynman parameters x,y,z,w, thus 1 D = 24 d 4 x,y,z,wδx+y +z +w 1 1 [l 2 Ok 2 M 2 +i0] 4 S.37 where the details of the Ok 2 expression are not important because the Pauli Villars mass M is much bigger than all the external momenta. On the other hand, the relations between the shifted loop momentum l µ and the propagator momenta p µ 1,pµ 2,pµ 3,pµ 4 are rather important for the numerator S.36, so here they are: p 1 = l + q 1 for q 1 = y +z +wk 1 z +wk 2 wk 3, p 2 = l + q 2 for q 2 = + xk 1 z +wk 2 wk 3, p 3 = l + q 3 for q 3 = + xk 1 + x+yk 2 wk 3, S.38 p 4 = l + q 4 for q 4 = + xk 1 + x+yk 2 + x+y +zk 3, thus N = tr γ 5 l +M+ q 4 γ ν l +M+ q 3 γ µ l +M+ q 2 γ λ l +M+ q 1. S.39 Before we evaluate this trace, let s put it in the context of the momentum integral d 4 l I = M 2π 4 N [l M O + ] S.40 in the M limit. This integral is UV-convergent, so it s dominated by the loop momenta l M, hence by dimensional analysis we expect N = OM 4 and I = OM. The q 1,...,q 4 10

11 momenta in the numerator S.39 and in the Oq 2 term in the denominator make for small corrections, so let s expand the interal S.40 in powers of q i /M: Iq i = M C 0 + i q qi α 2 C1 i,α + O. S.41 M Note that all the terms involving second or higher powers of q i carry negative powers of the Pauli Villars mass M, so in the eventual M limit they may be neglected. Thus, all we need to calculate are the q-independent term and the linear-in-q i terms. Consequently, we may expand both the numerator N and the denominator D = [l 2 m 2 = Oq 2 +i0] 4 in powers of q µ i and stop the expansion after the linear terms. For the denominator, this means simply 1 D 1 [l 2 M 2 +i0] 4, S.42 while for the numerator S.39 we have N N i=1 q i,α N α i S.43 where N N i=1 q i,α N α i, N 0 = tr γ 5 l +Mγ ν l +Mγ µ l +Mγ λ l +M, N 1 = tr γ 5 l +Mγ ν l +Mγ µ l +Mγ λ γ α, N 2 = tr γ 5 l +Mγ ν l +Mγ µ γ α γ λ l +M, N 3 = tr γ 5 l +Mγ ν γ α γ µ l +Mγ λ l +M, N 4 = tr γ 5 γ α γ ν l +Mγ µ l +Mγ λ l +M. S.44 11

12 Now let s evaluate these traces, starting with N 1 = M 3 tr γ 5 γ ν γ µ γ λ γ α + M tr γ 5 lγ ν lγ µ + lγ ν γ µ l + γ ν lγ µ l γ λ γ α using lγ ν lγ µ + lγ ν γ µ l + γ ν lγ µ l = 4l ν l µ l 2 γ ν γ µ = MM 2 l 2 tr γ 5 γ ν γ µ γ λ γ α + 4Ml ν l µ tr γ 5 γ λ γ α S.45 = MM 2 l 2 4iǫ νµλα + 4Ml ν l µ 0 = 4iMM 2 l 2 ǫ νµλα. Similarly, N 4 = 4iMM 2 l 2 ǫ ανµλ. S.46 In the remaining three traces we move the last l +M factor forward and use l +Mγ 5 l +M = l +MM lγ 5 = M 2 l 2 γ 5. S.47 Consequently, N 2 = M 2 l 2 tr γ 5 γ ν l +Mγ µ γ α γ λ = 4iMM 2 l 2 ǫ νµαλ, N 3 = M 2 l 2 tr γ 5 γ ν γ α γ µ l +Mγ λ = 4iMM 2 l 2 ǫ ναµλ, N 0 = M 2 l 2 tr γ 5 γ ν l +Mγ µ l +Mγ λ = M 2 l 2 M tr γ 5 γ ν { l,γ µ }γ λ = MM 2 l 2 2l µ tr γ 5 γ ν γ λ = 0. Thus, the q-independent term in the numerator vanishes while the linear-in-q terms all have the same form apart from the order of indices in the ǫ tensor. Reordering the indices and changing 12

13 the sign of a i as necessary, we arrive at N = 4iMM 2 l 2 ǫ νµλα q 1 q 2 +q 3 q 4 α + OM 2 q 2. S.48 Plugging this numerator into the momentum integral S.40, we obtain d 4 l N q 2 I = M 2π 4 D = ǫνµλα q 1 q 2 +q 3 q 4 α J + O M S.49 where d J = 4iM 2 4 l M 2 l 2 d 4 2π 4 [l 2 M 2 +i0] 4 = l 1 4iM2 2π 4 [l 2 M 2 +i0] 3 = 4M 2 d 4 l E 2π 4 1 [l 2 E +M2 ] 3 = 4M2 16π 2 = 1 8π 2. 0 l 2 E dl2 E [l 2 E +M2 ] 3 = 1 4π 2 0 xdx x+1 3 S.50 Moreover, the gluon momenta enter eq. S.49 in combination q 1 q 2 + q 3 q 4 = k 1 k 3 S.51 which does not depend on any of the Feynman parameters. Hence, integrating over those parameters is completely trivial, 1 24 d 4 x,y,z,wδx+y +z +w 1 = 1, 0 S.52 so the bottom line for the diagram S.33 is A1,2,3 = g3 4π 2 trtc T b T a ǫ νµλα k 1 +k 3 α. S.53 13

14 As we saw in eqs. S.30 and S.31, [ ] quadrangle anomaly = gluon permutations regulated = gluon permutations PV regulator only S.54 J 5α 2iMγ 5 so all we need to do now is to sum eq. S.53 over 3! = 6 permutations of the three gluons. For the three cyclic permutations, the group factors is the same and the ǫ tensor is the same, hence A1,2,3 + A2,3,1 + A3,1,2 = g3 4π 2 trtc T b T a ǫ νµλα 2k 1 +2k 2 +2k 3 = 2q α. S.55 For the other three permutations in the opposite cyclic order, the ǫ tensor changes sign while the group factor becomes trt a T b T c instead of trt c T b T a = trt a T c T b. Thus altogether, [quadrangle anomaly] = g3 4π 2 ǫ νµλα2q α trt a T c T b T a T b T c = + g3 2π 2 q αǫ αλµν trt a [T b,t c ] S.56 or in terms of the local gluons fields A a λ, Ab µ and A c ν, α J 5α 3g = 1 3! g3 2π 2 i α ǫ αλµν tr A λ [A µ,a ν ] = ig3 4π 2 ǫαλµν tr α A λ [A µ,a ν ]. S.57 Comparing this formula to the second line of eq. S.21 part a, we see that the quadrangle diagrams 6 generate precisely the three gluon part of the non-abelian anomaly 1. 14

15 Problem 4: In d = 2n Euclidean dimension, a gauge theory with massless fermions has Lagrangian L E = F a µν 2 + Ψγµ D µ Ψ S.58 where {γ µ,γ ν } = +2δ µν, while the axial symmetry acts according to Ψx exp +iθxγ Ψx, Ψx Ψxexp +iθxγ. S.59 For x-independent θ this is a symmetry of the Lagrangian S.58, while for an x dependent θx L E = Ψγ µ i µ θγψ = i µ θ Ψγ µ ΓΨ = i µ θ J µ A S.60 for the axial current J µ A = ΨΓγµ Ψ, hence SE classical = +i d d xθx µ J µ A x. S.61 At the same time, in a non-trivial gauge field background the axial symmetry of the fermionic path integral D[Ψx] D[Ψx] carries a non-trivial Jacobian Detexp2iθxΓ, which is equivalent to S quantum E = logdet exp2iθxγ = 2iTr θxγ S.62 where the functional trace Tr involves both summing over Dirac and gauge indices and integration over d d x. In terms of the ordinary matrix trace over the Dirac, color, and flavor indices, S quantum = 2i d d x tr x Γ x reg where the matrix element x Γ x = Γ δ d x x must be regulated to smooth out the delta-function. The S classical and S quantum should cancel each other, hence µ J µ A x = 2tr x Γ x reg x ΓĜ x = 2tr S.63 for some regulating operator Ĝ. 15

16 As explained in of Weinbergs book which was your reading assignment for this homeworks, the regulating operator Ĝ must commute with the Dirac operator D in the background gauge field and also with the Γ matrix, so Ĝ should be a smooth function of D2. Also, Ĝ should suppress the high-momentum modes p M for some UV cutoff scale M while affecting the low-momentum modes p M as little as possible, hence Ĝ = G D 2 /M 2. S.64 for some smooth function Gt which goes to one for t 0 and to zero for t, Gt 1 t = D 2 /M 2 S.65 Altogether, µ J µ A x ΓG D x = 2tr 2 /M 2 x. S.66 Since the covariant derivatives do not commute with each other, [D µ,d ν ] = igf µν, we have D 2 = D 2 + g 2 F µνσ µν S.67 and consequently G D 2 /M 2 = G D 2 /M 2 g 2M 2 G D 2 /M 2 F µν σ µν + g2 8M 4 G D 2 /M 2 F µν F αβ σ µν σ αβ +. S.68 In d = 4, the Dirac trace trγ 5 G comes from the second-derivative term in this expansion because we need four γ µ matrices or equivalently two σ µν matrices to accompany the γ 5. In other even dimensions d = 2n, we need d γ µ matrices or n σ µν matrices to accompany the Γ 16

17 inside the Dirac trace. Specifically, in 2n Euclidean dimensions {γ µ,γ ν } = 2δ µν, σ µν = i 2 [γµ,γ ν ], Γ = i n γ 1 γ 2 γ 2n, S.69 hence tr Γγ α1 γ α k = 0 for k < 2n, tr Γγ α1 γ α 2n = 2i n ǫ α1 α2n, S.70 and trγσ α1β1 σ α kβ k = 0 for k < n, trγσ α1β1 σ αnβn = +2 n ǫ α1β1 αnβn. S.71 Therefore, the leading term in the Dirac trace trγg comes from the n th derivative term in the expansion S.68, tr Dirac ΓG D 2 /M 2 = 1 n g n! 2M 2 G n D 2 /M 2 tr DiracΓF αβ σ αβ n = 1 n g n! M 2 G n D 2 /M 2 ǫ α1β1 αnβn F F α1β1 αnβn + subleading terms, S.72 wherethesubleadingtermscarryhigherpowersof1/m 2,sotheymaybeneglectedintheM limit. Now let s calculate the matrix element x G n D 2 /M 2 x of the n th derivative of the regulator. Going from the coordinate basis to the momentum basis, we have d x G n D 2 /M 2 d p x = 2π d e ipx G n D 2 /M 2 e +ipx d d p p µ id µ 2 S.73 = 2π d Gn E M 2 where on the last line the derivative D µ acts on the Fx Fx in eq. S.72 andalso on A ν x in other D ν in the expansion of G n. In the integral S.73, the overall scale of the momentum p follows from the regulator Gp 2 /M 2, hence p µ = OM while the derivatives D µ are effectively 17

18 Oexternal momenta of the background photon or gluon fields. For the regulation purposes, we assume that M is much larger than all such external momenta, hence 1 d d p p µ id µ 2 M d 2π d Gn M 2 = 1 M d d d p 2π d Gn p 2 /M 2 + O k 2 ext M 2 S.74 while the leading term on the right hand side is a O1 constant. To calculate this constant, we use spherical coordinates in 2n Euclidean dimensions, Consequently, d 2n p = 1 d 2n p M 2n 2π 2n Gn p 2 /M 2 = 2π n n 1! p2n 1 dp = = 1 4π n n 1! 1 4π n n 1! π n n 1! p2 n 1 dp 2. 1 M 2n 0 0 dp 2 p 2n 2 G n p 2 /M 2 dtt n 1 G n t integrating n 1 times by parts = 1n 1 4π n 0 dtg t = 1n 4π n [ G0 G ] = 1n 4π n S.75 regardless ofthespecific formofgtfunctions, aslongasg0 = 1andG = 0, cf.fig.s.65. Therefore, x tr Dirac ΓG D 2 /M 2 x = 1 n! n +g ǫ α1β1 αnβn F α1β1 4π x F αnβn x, S.76 and the axial anomaly follows from the trace of the RHS here over the species indices color and flavor of the fermionic fields, µ J µ A x = 2 n! +g 4π n ǫ α1β1 αnβn tr species Fα1β1 x F αnβ n x. S.77 This completes our analysis of the axial anomaly in d = 2n Euclidean dimensions. In Minkowski spacetime of 2n 1 space dimensions plus 1 time, there is an overall minus sign 18

19 hiding in the ǫ tensor, hence µ J µ A x = 2 n! n +g ǫ α1β1 αnβn trf α1β1 4π x F αnβn x. 5 Quod erat demonstrandum. 19