1 Weak Interaction. Author: Ling fong Li. 1.1 Selection Rules in Weak Interaction

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1 Author: ing fong i 1 Weak Interaction Classification Because the strong interaction effects are diffi cult to compute reliably we distinguish processe involving hadrons from leptons in weak interactions. a eptonic weak interactions Here all particles are leptons and there is no complication from strong interactions. Once the interaction is known we can calculate these processes accurately since the interaction is weak and perturbation theory is applicable. Examples: µ e + ν µ + ν _ e ν µ + e ν µ + e τ µ + ν τ + _ ν µ τ e + ν τ + _ ν e b Semi-leptonic interactions These processes involve both leptons and hadrons. Since we can calculate the leptonic parts quite reliabley these processes can be used to study the properties of hadrons very similar to the case of ep scattering. Examples : π µ + ν µ K + π 0 + e + + ν e n p + e + ν e νµ + p µ + + n c Non-leptonic interactions Here all particles are hadrons and they are the most diffi cult reactions to study because of the strong interaction effects. This class of reactions differ from the normal strong interaction in the slower decay rates and smaller crossections. Example : K + π + + π 0 K 0 π + + π + π 0 Σ + P + π 0 Λ P + π 1.1 Selection Rules in Weak Interaction a eptonic Interaction Two neutrino experiments: ν from β-decay and ν from π decay are different If they were the same then the following chain of reactions should be possible n p + e + ν ν + p µ + + n However only e + is observed in the final product and no µ + has been seen. A simple explanation that is the neutrino from β-decay called ν e is different from neutrino from π-decay accompanied by µ called ν µ and there is also muon number and electron number conservation. In these conservation laws we assign the electron number e as Similarly for the muon number µ e ν e e = 1 e + ν e e = 1 µ ν µ µ = 1 µ + ν µ µ = 1

2 As a consequenc of these conservation laws the reaction µ ± e ± + γ are forbidden and experimentally this is indeed the case. epton number conservations seem to hold up very well for many years until neutrino oscillations have been observed recently. b Semi-leptonic decays a The strangeness changing reactions S 0 seem to be about a factor of 10 or so smaller than those which conserve strangeness S = 0. b It has been observed that hadrons in the strangeness changing decays satisfy the selection rule S = Q For examples K + π 0 µ + ν µ but K + π + π + e _ ν e c Absence of S = 1 neutral currents For example Γ K µ + µ Γ K + µ + ν 10 9 d No S = transistion been obsverbed For example Ξ ne _ ν e When quark model was developed all these properties can be accormodated by writing the hadronic weak current as J had µ = [_ uγ µ 1 γ 5 d cos θ c + _ uγ µ 1 γ 5 s sin θ c ] θ c 0.5 is the Cabbibo angle. c Non-leptonic interactions Here we havethe I = 1/ rule Γ K + π + π 0 Γ K s π + π This rule is very diffi cult to explain because the strong interaction effect is hard to study accurately. 1. Milestones of Weak Interaction a Neutrino and Nuclear β decay The emission of electron e from many different nuclei A Z A Z e was observed at early study of radioactivity. The unusual feature is that the energy spectrum of e is continuous rather than a sharp line as expected in a body decay. If the basic mechanism for e emission is n p + e the energy momentum conservation will require e to hav a single energy while experimentally a continuous distribution of energy was observed. Pauli 1930 postulated the presence of neutrino which carries away energy and momentum in nuclear β -decay n p + e + _ ν e so that energy momentum conservation can be saved.

3 b Fermi Theory Fermi 1934 proposed to explain the β decay by making analogy with QED to write down the weak interaction agrangian in the form F = G F [ pxγ µ nx][ēxγ µ ν e x] + h.c. G F : Fermi coupling constant Fitting nuclear β decay reates give G F 10 5 Mp M p is the proton mass This theory works very well for J = 0 β decays of many nuclei. interaction was added ater Gamow-Teller to account for J = 1 nuclear β decays. GT = G F [ pxγ µ γ 5 nx][ēxγ µ γ 5 ν e x] + h.c. c Parity violation and V - A theory θ τ puzzle In 1950 s it was observed that there are two decays θ π + + π even parity τ π + + π + π 0 odd parity while θ and τ have the same mass charge and spin. It is very diffi cult to understand these features if the parity is a good symmetry : ee and Yang proposed that parity is not conserved in weak decays and suggested many experiments to test this hyposethsis : C. S. Wu showed that e in 60 Co decay has the property σ p 0 σ p spin and momentum of e This implies that the parity symmetry is violated in this decay. d V-A theory 1958 Feynman and Gell-Mann Sudarshan and Marshak Sakurai As a result of parity violation the effective weak interaction was casted in the form with V A currents eff = G F J µj µ + h.c.

4 and J λ x = J l λ x + J h λ x J λ l x = _ ν e γ λ 1 γ 5 e + _ ν µ γ λ 1 γ 5 µ leptonic current 1 J λ h x = _ uγ λ 1 γ 5 cos θ c d + sin θ c s θ c : Cabibbo angle Note that in V-A form the fermion fields are all left-handed. Define ψ 1 1 γ 5ψ Then we can simplify the form of weak currents J λ l x = ν eγ λ e + ν µ γ λ µ +... hadronic current Phenomenologically the V-A theory has been quite successful in most of the weak interactions phenomena. Diffi culties: 1 Not renormalizable In the Fermi theory 4 fermions interaction operator has dimension 6 and is not renormalizable. In other words the higher order graphs are more and more divergent. For example in µ decay we have Violate unitarity The tree amplitude for ν µ + e µ + ν e has only J = 1 partial wave at high energies and cross section has the form σν µ e G F S S = m e E On the other hand unitarity for J=1 cross section is σj = 1 < 1 S Thus σν µ e violates unitarity for E 300GeV. Since unitarity is a consequence of conservation of probablity this violation is unacceptable. Intermidate Boson TheoryIVB In analogy with QED we can introduce vector boson W to couple to the V-A current to mediate the weak interaction W = gj µ W µ + h.c. For example the µ decay is now mediated by W-exchange. Since weak interaction is short range we need W-boson to be massive M W 0. Use the massive W-boson

5 propagator in the form g µν + kµ k ν M W k M W gµν M W when k µ M W We see that this reproduces 4-fermion interaction with MW = G F In this theory the scattering ν µ + e µ + ν e no longer violates unitarity. But the violation of unitarity shows up in other processes like ν + ν W + + W and the theory is still non-renormalizable. g 1.3 Construction of SU U1 model Choice of group In IVB theory the interaction is described by W = gj µ W µ + h.c For simplicity we neglect all other fermions excepts ν e and write the current as J µ = νγ µ 1 γ 5 e Recall that in the electromagnetic interaction we have em = ej em µ A µ J em µ = ēγ µ e Define the eletromagnetic and weak charges as the intergals of the time-component of the currents T + = 1 d 3 xj 0 x = 1 d 3 xν 1 γ 5 e T = T + Q = d 3 xj0 em x = d 3 xe e We can compute the commutator [T + T ] = T 3 T 3 = 1 d 3 x[ν 1 γ 4 5 ν e 1 γ 5 e] Q This means that these 3 charges T + T and Q don t form a SU algebra. The reason is that in order for the eletric charge operator Q to be a generator of SU it has to be traceless. In our case here it is not the case. Futhre more the weak charges T ± have the V A form while the em charge Q is pure vector. At this point there are alternatives:

6 a Introduce another guage boson coupled to T 3. The generatros corresponding to these 4 gauge bosons can then form the group SU U 1. This will be the choice we will adapt eventually. b We can add new fermions to modify the currents such that T + T and Q do form a SU algebra Georgi and Glashow 197. Here we introduce new fermions to extend the multiplet into triplets E γ 5 ν e cos α + N sin α e and a singlet The weak charge is then T + = 1 It is then straightforward to verify that γ 5 E + N e γ 5 N cos α ν e sin α d 3 x [ E + 1 γ 5 ν e cos α + N sin α ] + ν e cos α + N sin α 1 γ 5 e + E γ 5 N + N 1 + γ 5 e [T + T ] = Q with Q = [ ] d 3 x E E e e Clearly in this model only electromagnetic current is neutral and others all carry charges. When neutral weak current reactions were discovered in 1973 this model is ruled out. Unitarity argument Equivalently we can argue from unitarity that it is necessary to introduce either new leptons or a new guage boson. Consider in IVB theory the reaction ν + _ ν W + + W both W s are longitudinally polarized. The lowest order amplitude corresponding to the following graph is T ν _ ν W + W = i _ v p ig /ε 1 γ 5 The polarization vectors = g _ v p /ε /p /k /ε 1 γ 5 p k u p m e ε i µ k with ε i ε j = δ ij and k ε i = 0 i /p /k m e ig /ε 1 γ 5 u p

7 maybe chosen in the rest frame of W boson as ε i 0 = 0 ε i j = δ ij For a moving W boson with k µ = E 0 0 k with k = E Mw we can make a orentz transformation along the z axis. The transverse polarizations do not change while the longitudinal polarization becomes ε 3 µ = 1 k 0 0 E. In the high energy limit with k = E M W + we see that M W E ε 3 µ = k µ MW + O M W E Then for longitudinally polarized W bosons the scattering amplitude in Eq becomes g _ T v p /k /k k /p /k 1 γ p k M W M 5 u p 3 W g _ v p /k 1 γ 5 u p M W To show more explicitly this amplitude is a pure J = 1 partial wave we take k µ = E k e p µ = E 0 0 E k µ = E k e p µ = E 0 0 E with e = sin θ 0 cos θ Since ν and _ ν have opposite helicities we have u p = E 1 σ p χ 1/ = 1 E χ σ 1/ z E _ v p = σ Eχ p 1/ E 1 = Eχ 1/ σ z 1 χ 1/ = Then the combination in Eq3 becomes We have then χ 1/ = 1 _ v p /k 1 γ 5 u p = Eχ 1/ = 4Eχ 1/ 1 1 E k σ e E k σ e χ 1/ k σ e E χ 1/ = 4Ek sin θ T G F E sin θ as E 4 The partial wave expansin for the helicity amplitude is of the form T λ3 λ 4 λ 1 λ E θ = J + 1 Tλ J 3 λ 4 λ 1 λ E d J µλ θ J=M λ 1 = λ = 1/ and λ 3 = λ 4 = 0 are the helicities of the initial and final particles with λ = λ 1 λ = 1 µ = λ 3 λ 4 and M = max λ µ = 1. d J µλ θ is the usual rotation matrix with d1 10 θ = sin θ. It is clear that T in Eq4 corresponds to a pure J = 1 partial wave and violates the unitarity bound of T J=1 E constant at high energies. To cancel this bad high energy behavior we need other diagrams for this reaction. There are possibilities: s-channel or u-channel exchange diagrams.

8 a The heavy lepton alternative; the u-channel exchange in the following diagrama yields the amplitude _ T u ν ν W + W = g _ v p /ε /p /k /ε 1 γ 5 p k u p m E = g _ v p MW /k 1 γ 5 u p If g = g this will cancel the bad behavior given in Eq3 b The neutral vector boson alternative; the s-channel exchange in the diagram b given above gives the amplitude _ T s ν ν W + W = i _ v p ifγ β 1 γ5 u p αµν ε µ k ε ν k [ ] [ ] i g αβ + k + kα k + k β 1 MZ k + k MZ Choose the ZW W coupling to have Yang-Mills structure αµν = if [ k k α g µν k + k ν g µα + k + k µ g αν ] we get αµν ε µ k ε ν k = if [ k k α ε ε k ε ε α + k ε ε α ] if M W [ k k α k k ] and T s ff _ v p MW /k 1 γ 5 u p Thus if we choose ff = g this will also canel the amplitude in Eq3. This corresponds to the case of adding an additional U 1 symmetry discussed before. In fact if one demands that all the amplitudes which violate unitarity be cancelled out one ends up with a renormalizable agrangian which is the same as the one derived from the algebraic approach. We now choose the gauge group to be SU U 1. The agrangian for the gauge fields is then = 1 4 F iµν F i µν 1 4 Gµν G µν

9 F i µν = µ A i ν ν A i ν + gɛ ijk A j µa k ν SU gauge fields G µν = µ B ν µ B µ U 1 gauge field Fermions Clearly from the stuctrue of the weak charge current given in Eq1 ν e form a doublet under SU l = 1 ν 1 γ 5 e For convenience we introduced left-handed and right-handed fields ψ 1 1 γ 5ψ ψ R γ 5ψ ψ = ψ + ψ R Then T + = ν + e d 3 x T = e + ν d 3 x Q = e + e + e + R e R Note that Q T 3 = [ 1 ν+ ν + e + e e + R e R]d 3 x It is straightforward to show that [Q T 3 T i ] = 0 i = 1 3 Thus we can take Q T 3 to be U1 charge Y Q T 3 sometime called weak hypercharge. The Y charges for fermions are ν l = Y = 1 e R Y = With these quantum numbers the agrangian for gauge coupling is For example e = l iγ ν D ν l l + l R iγ ν D ν l R 5 D ν ψ = ν ig τ A ν ig Y B νψ D ν l = ν ig τ A ν ig Y B νl Spontaneous Symmetry Breaking The symmetry braking pattern we want is SU U1 U1 em. Choose scalar fields in SU doublet with hypercharge Y = 1 φ φ = Y = 1 The agrangian contaning φ is of the form φ 0 3 = D µ φ D µ φ V φ and D µ φ = µ ig τ A µ ig B µφ V φ = µ φ φ + λφ φ

10 In addition there is a coupling between leptons and scalar field φ 4 = f _ φe R + h.c. As we have seen before the spontaneous symmetry breaking is generated by the vaccum expectation value < φ > 0 = 0 φ 0 = 1 0 µ with v = v λ Write the scalar field in the form φx = U 1 ξ 0 v+ηx U ξ = exp[ i ξx τ ] 6 v Gauge Transformation The scalar field in Eq6 is in the form of gauge transformation. We can then simplify the form by a gauge transformation φ = U ξφ = 1 0 v + ηx τ A µ = U ξ τ A µ U 1 ξ i g µuu 1 Now the field ξx disappears from the agrangian as a consequence of gauge invariance. From 4 Yukawa coupling VEV of the scalar field gives 4 = f 1 l < φ > e R + h.c. + f ηx _ e e R + h.c. as consequence the electron is now massive with electron mass m e = f v Mass spectrum We now list the mass spectrum of the theory after the spontaneous symmetry breaking: a Fermion mass b Scalar masshiggs c Gauge boson masses From the covariant derivative in 3 3 = v χ g m e = fv V φ = µ η + λvη 3 + λ 4 η4 m η = µ τ A µ + g B µ µ A g τ we get the mass terms for the gauge bosons + g B µ 0 χ + χ = 1 3 = v 8 {g [A 1 µ + A µ ] + ga 3 µ g B µ } + = M W W +µ W µ + 1 M ZZ µ Z µ +

11 W + µ = 1 A 1 µ ia µ M W = g v 4 1 Z µ = g + g g A 3 gb µ MZ = g + g 4 v The fielda µ = 1 g +g g A 3 µ + gb µ massless photon does not appear in 3 and is massless. This clearly corresponds to photon field. For convience we define tan θ W = g g θ W : Weinberg angle or weak mixing angle Then we can write Z µ = cos θ W A 3 µ sin θ W B µ M Z = g v A µ = sin θ W A 3 µ cos θ W B µ Note that there is a relation between M W M Z and θ W of the form ρ = M W M Z cos θ W = 1 which is a consequence of the doublet nature of the scalar fields. 4 sec θ W The weak interactions mediated by W and Z bosons can be read out from Eq5 a Charged current cc = g J µw µ + h.c. J µ = J 1 µ + ij µ = 1 νγ µ1 γ 5 e Again to get 4-fermion interaction as low energy limit we require which implies that g 8M W = G F v = This is usually referred to as the weak scale. b Neutral Current The argrangian for the neutral currents is G F 46Gev NC = gj 3 µa 3µ + g J Y µ B µ = ej em µ A µ + g cos θ W J Z µ Z µ and e = g sin θ W J Z µ = J 3 µ sin θ W J em µ

12 is the weak neutral current. We can define the weak neutral charge as Q Z = J0 Z d 3 x = T 3 sin θ W Q This means that the coupling strength of fermions to Z-boson is proportional to the quantum number T 3 sin θ W Q. In particluar Z boson can contribute to the scattering ν e + e ν e + e The measurement of this corss section in the 1970 s give sin θ W 0.. This yields M W 80 GeV and M Z 90 GeV Generalization to more than one family. From 4-fermion and IVB theory the form of weak currents of leptons and hadrons gives the following multiplets structure νe νµ u e e µ R µ R u d R d R s R θ d θ = cos θ C d + sin θ C s The neutral current in the down quark sector is of the form NC = [ d θ γ µ sin θ C 3 d θ sin 1 θ W 3 d R γ µ d R + s R γ µ s R ]Z µ = sin θ W 3 [ d γ µ d + s γ µ s + sin θ W cos θ W d γ µ s + s γ µ d +... The term d γ µ s + s γ µ d gives rise to S = 1 neutral current processes e.g. K µ + + µ with same order of magnitude as charged current interaction. But experimentally R = ΓK µ + + µ ΓK + µ + ν 10 8 Thus we can not have S = 1 neutral curent process at the same order of magnitude as the charged current process. GIM mechanism Glashow Iliopoulos and Maiani 1970 suggested that there is a 4-th quark the charm quark c which couples to the orthogonal combination s θ = sin θ c d + cos θ c s so that the multiplets look like u c d θ s θ

13 As a result the S = 1 neutral current is canceled out. The new current is of the form d θ sin θ W γ µ d θ + s θ sin θ W γ µ s θ = sin θ W dγ µ d + sγ µ s which conserves the strangeness. Quark mixing Before the spontaneous symmetry braking fermions are all massless becauseψ and ψ R have different quantum numbers under SU U 1. So the mass term ψ ψ R + h.c. is not invariant under SU U1 groups and can not appear in the agrangian. When we have more than one doublets ψ ir ψ i all have the same quantum numbers with respect to SU U1 group we call them "weak eigenstates". Here the index i labels the different families of weak eigenstates. When spontaneous symmetry breaking takes place fermions obtain their masses through Yukawa coupling. Y = f ij ḡ i u Rj + f ijḡ i d Rj φ + h.c. Note that from the requirement of renormalizability we need to write down all possible terms consistent with SU U 1 symmetry. Since Yukawa coupling constants f ij f ij are arbitrary the fermion mass matrices are in general not diagonal. When mass matrices are diagonalized the mass matrix we obtain the mass eigenstates which are not the same as the weak eigenstates. The mass matrices in the up and down sectors are given by m u v ij = f ij m d ij = f v ij These matrices which are sandwiched between left and right handed fields can be diagonalized by bi-unitary transformations i.e. given a mass matrix m ij there exits unitary matrices S and T such that S mt = m d is diagonal. Basically S is the unitary matrix which diagnoalizes the hermitian combination mm + i. e. Biunitary transformation Write Define and Define a matrix V by Then So V is unitary and we have S + mm + S = m d m d = m 1 m m 1 m d = m H = Sm d S V H 1 m m 3 m 3 hermitian V V = H 1 mm H 1 = H 1 Sm d S H 1 = H 1 H H 1 = 1 S HS = m d = S mv S = m d Or S mt = m d with T = V S

14 If we write the left-handed doublets weak eigenstates as u q 1 = d c q = These weak eigenstates are related to mass eigenstates by unitary transformations u u d d = S u = S c d s c Note that in the coupling to charged gauge boson W ± we have s s W = g W µ [ _ q 1 γ µ τ q 1 + q γ µ τ q ] + h.c. and is invariant under unitary transformation in q 1 q space i.e. q 1 q1 q = V V V = 1 = V V q We can use this feature to put all mixing in the down quark sector q u c d i = d s s = U Here U is a unitary matrix. Clearly we can extend this to 3 generations with result q i : u d c s t b d s b = U Now U is a 3 3 unitary matrix usually called the Cabibbo-Kobayahsi-Maskawa CKM matrix. CP violation Phase CP violetion can come form complex coupling to gauge bosons. The gauge coupling of W ± to quarks is governed by the 3 3 unitary matrix U discussed above. This unitary matrix U can have many complex entries. However in diagonalzing the mass matrices S mm S = m d There is ambiguity in the matrix S in the form of diagonal phases. In other words if S diagonalizes the mass matrix so does S e iα S = S e αn We can use this property to redefine the quark fields to reduce the phase in U. It turns out that for n n unitary matrix number of independent physical phases left over is n 1n Thus to get CP violetion we need to go to 3 generations or more Kobayashi Mskawa. Here we give a constructive proof of this statement. et us start with a first doublet written in the form q 1 = u U 11 d + U 1 s + U 13 b d s b d s If U 11 has phase δ U 11 = R 11 e iδ R 11 real

15 then this phase δ can be absorbed in the redefiniton of the u quark field and we can write q 1 = e iδ u u = ue iδ u R 11 d + U 1 s + U 13 b Similarly we can factor out the complex phases of U 1 and U 31 by redefinition of c and t quark fields. These overall phases are immaterial because there are no gauge couplings between doublets with different family indices. Finally we can absorb two more phases of U 1 and U 13 by a redefinition of the s and b fields. The doublets now take the form u R 11 d + R 1 s + R 13 b t R 31 d + R 3 e iδ 3 s + R 33 e iδ 4 b c R 1 d + R e iδ 1 s + R 3 e iδ b Now we have reduced the number of parameters to 13. The normalization conditons of each down-like state gives 3 real conditions aen orthogonality conditions among different states give 6 real conditions on the parameters Now we are down to 4 parameters. Since we need 3 parameters for the real orthogonal matrix we end up with one independent phase. Flavor consevation in neutral current interaction It turns out that the coupling of neutral Z boson to the fermions conserve flavors. This can be illustrated as follows. We first write the neutral currents in terms of quark fields which are weak eigenste J Z µ = i _ ψ i γ µ [ T3 ψ i sin θ W Q ψ i ] ψ i Separate into left- and right-handed fields and distingush the up and down components [ ] [ _ u 1 iγ µ sin θ W u _ i + d 3 iγ µ sin θ W 3 J Z µ = i [ ] [ ] + _ u Riγ µ sin θ W u _ Ri + d 1 3 Riγ µ sin θ W d Ri 3 ] d i Since weak eigen states q i and mass eigen states q i are related by unitary matries u i = U u ij u j We see that these unitary matrices cancel out in the combination _ u iu i so that the neutral current in terms of mass eigenstates has the same form as the one in terms of weak eigenstates. Thus it conserves all quark flavor. Note this feature is due to the fact that all quarks with same helicity and electric charge have the same quantum number with repect to SU U 1 gauge group. In other words if there are quarks in representations other than SU doublets then there will be flavor changing neutral current if the new quarks have the same eletric charge as either u or d quark.