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Electroweak Unification and Properties of W and Z Bosons Electroweak unification Decays of the W boson Decays of the Z boson Electroweak unification and properties of W and Z bosons Harri Waltari University of Helsinki & Helsinki Institute of Physics University of Southampton & Rutherford Appleton Laboratory Autumn 2018 H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson Contents In this lecture we shall Construct a common theory for weak and electromagnetic interactions Compute the predictions of the electroweak theory for gauge boson decays Discuss the experimental discovery of W and Z bosons This lecture corresponds to chapters 15.1, 15.3 and 15.4 of Thomson's book. H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson The simple SU(2) model has problems With the SU(2) model we are on the right track, since the interactions of the W -bosons with fermions come out right. However, there are a number of problems: The W -boson does not have electromagnetic interactions even though it is charged and you cannot introduce them in the same way as for other particles The gauge boson masses are not compatible with the gauge symmetry and SU(2) is not confining so they are not due to the potential energy of weak interactions (leptons have a SU(2) charge and are free particles) Even the fermion mass terms mf ( L R + R L) are not compatible with SU(2) since left- and right-handed fields have different transformation rules We shall solve the first one now and postpone the two latter ones later. H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson Gauging hypercharge and allowing gauge boson mixing gives the correct vertices The solution lies in the Gell-Mann{Nishijima formula for charge: (W ) Q = I3 + Y =2 ) Gauge weak isospin and hypercharge and then you will have a gauge theory for electric charge, too Hypercharge is an additive quantum number, so the first Ansatz is to associate a U(1) symmetry to it 0 We associate a gauge field Bµ and a gauge coupling g to the symmetry ) The covariant derivative becomes i i 0 Y Dµ = @µ − igτ Wµ − ig 2 Bµ for left-handed fermions and 0 Y Dµ = @µ − ig 2 Bµ for right-handed fermions We then assume that since there are two neutral spin-1 fields that they can mix, we have the combinations 3 3 Aµ = Bµ cos θW + Wµ sin θW and Zµ = Wµ cos θW − Bµ sin θW , where θW is the weak mixing angle (or Weinberg angle) H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson Gauging hypercharge and allowing gauge boson mixing gives the correct vertices We associate Aµ with the photon, whereas Zµ is a neutral gauge boson, which will couple to neutrinos The charges for the left-handed doublet come out right if it has Y = −1 From the covariant derivative we may read that the neutrino couples 3 0 to a combination proportional to gWµ − g Bµ g g 0 Hence cos θW = p and sin θW = p g 2+g 02 g 2+g 02 The left-handed electron then couples to the photon by 1 0 gg 0 (g sin θW + g cos θW ) ) e = p , known as the unification 2 g 2+g 02 condition Notice that Y = −2 for the right-handed electron leads to the same coupling with the photon H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson Gauging hypercharge and allowing gauge boson mixing gives the correct vertices 2 The current value for sin θW = 0:231 (which is the parameter that can be measured directly in a number of ways) The kinetic terms for the gauge field have a term 2 ijk ilm jµ kν l m g W W WµWν Now setting e.g. k = m = 3 we get photon components in the four-boson vertex ) a coupling W +W −γγ with the strength e2 as one would expect Hence the SU(2)L×U(1)Y theory provides at least qualitatively the known features of electromagnetic and weak interactions and provides a chance to predict the outcomes of a large number of processes, to which we turn next H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson Massive gauge bosons have three polarization states The W -boson has a mass of roughly 80 GeV, hence it has a rest frame Since in a rest frame you cannot say, which directions are transverse, you have to have three polarization states (instead of two for massless ones, as required by the Maxwell equations) The basis vectors for the polarization states can be chosen as 1 µ = −p (0; 1; i; 0); + 2 1 µ = p (0; 1; −i; 0); − 2 µ 1 L = (p; 0; 0; E); mW where ± are circularly polarized states, L is the longitudinal state and the z-axis is chosen in the direction of the momentum H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson We compute the W-boson decay width in its rest frame Using Feynman rules we get the matrix element: Matrix element for the decay W − ! e−ν M = pg λ(p )u(p )γµ 1 (1 − γ5)v(p ) 2 µ 1 3 2 4 It is easiest to compute this in the rest frame of the W -boson, where p1 = (mW ; 0; 0; 0) Neglecting the fermion masses (this is a good approximation to all 2 2 −3 fermions as even mb=mW ' 3 × 10 ) leads to p3 = E(1; sin θ; 0; cos θ) and p4 = E(1; − sin θ; 0; − cos θ) H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson Orthogonality of polarization vectors gives us three terms We shall sum over polarizations λ, but since the polarization vectors are orthogonal MMy just reduces to the sum of squares for each y y y polarization (M+M+ + M−M− + MLML) The factor 1 − γ5 reduces the leptonic part of the matrix element to just a product of the left-chiral electron and the right-chiral antineutrino giving T 0− sin θ=21 0 cos θ=2 1 µ 1 5 B cos θ=2 C µ B sin θ=2 C u(p3)γ (1 − γ )v(p4) = E B C γ B C 2 @− sin θ=2A @− cos θ=2A cos θ=2 − sin θ=2 = 2E(0; − cos θ; −i; sin θ) = mW (0; − cos θ; −i; sin θ) noting that E = mW =2 H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson Orthogonality of polarization vectors gives us three terms We get gm gm M = W (0; 1; −i; 0) · (0; − cos θ; −i; sin θ) = W (1 + cos θ) − 2 2 gm gm M = − W (0; 1; i; 0) · (0; − cos θ; −i; sin θ) = W (1 − cos θ) + 2 2 gmW gmW ML = (0; 0; 0; 1) · (0; − cos θ; −i; sin θ) = − p sin θ 2 2 2 2 2 2 2 g mW 2 2 2 g mW Hence hjMj i = 12 ((1 + cos θ) + (1 − cos θ) + 2 sin θ) = 3 This decay width does not depend on the angles so we predict an isotropic distribution in the W -boson rest frame Plugging this in with the kinematical factors gives 2 − − g mW Γ(W ! e νe ) = 48π H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson SU(2) invariance allows us to compute the total decay width from one of the decays Since all fermions are in SU(2) doublets, we have − − − − − − Γ(W ! e νe ) = Γ(W ! µ νµ) = Γ(W ! τ ντ ) For quarks we need to take into account color and CKM-mixing, the 2 former giving a factor of 3 and the latter a factor of jVqq0 j Since five quarks are lighter than mW , there are six possible hadronic decay modes at the quark level The unitarity conditions of the CKM matrix tell us that 2 2 2 2 2 2 jVud j + jVus j + jVubj = 1 and jVcd j + jVcs j + jVcbj = 1 Overall the quark modes are equivalent to six lepton generations, 2 3g mW giving a tree-level prediction of Γtot = 16π ' 2:03 GeV and BR(W ! hadrons) = 2=3, BR(W ! `ν) = 1=3 H. Waltari Electroweak unification and properties of W and Z bosons Electroweak unification Decays of the W boson Decays of the Z boson Loop corrections explain the discrepancy Experimentally we have Γtot = 2:085 GeV and BR(W ! hadrons) = 67:4% Most of the difference is explained by taking into account the NLO correction to the quark decays, which gives effectively a factor 0 1 + αs /π to the Wqq vertices 2 2 Since αs (q = mW ) ' 0:12, we get 1 + αs /π ' 1:038, which then gives the correct branching ratio and also a total width close to the experimental value Altogether we have a fair understanding of the W -boson decays based on the electroweak theory H.
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