Electro-weak interactions Marcello Fanti Physics Dept. | University of Milan M. Fanti (Physics Dep., UniMi) Fundamental Interactions 1 / 36 The ElectroWeak model M. Fanti (Physics Dep., UniMi) Fundamental Interactions 2 / 36 Electromagnetic vs weak interaction Electromagnetic interactions mediated by a photon, treat left/right fermions in the same way g M = [¯u (eγµ)u ] − µν [¯u (eγν)u ] 3 1 q2 4 2 1 − γ5 Weak charged interactions only apply to left-handed component: = L 2 Fermi theory (effective low-energy theory): GF µ 5 ν 5 M = p u¯3γ (1 − γ )u1 gµν u¯4γ (1 − γ )u2 2 Complete theory with a vector boson W mediator: g 1 − γ5 g g 1 − γ5 p µ µν p ν M = u¯3 γ u1 − 2 2 u¯4 γ u2 2 2 q − MW 2 2 2 g µ 5 ν 5 −−−! u¯3γ (1 − γ )u1 gµν u¯4γ (1 − γ )u2 2 2 low q 8 MW p 2 2 g −5 −2 ) GF = | and from weak decays GF = (1:1663787 ± 0:0000006) · 10 GeV 8 MW M. Fanti (Physics Dep., UniMi) Fundamental Interactions 3 / 36 Experimental facts e e Electromagnetic interactions γ Conserves charge along fermion lines ¡ Perfectly left/right symmetric e e Long-range interaction electromagnetic µ ) neutral mass-less mediator field A (the photon, γ) currents eL νL Weak charged current interactions Produces charge variation in the fermions, ∆Q = ±1 W ± Acts only on left-handed component, !! ¡ L u Short-range interaction L dL ) charged massive mediator field (W ±)µ weak charged − − − currents E.g. weak decays, n ! pe ν¯e , µ ! e ν¯eνµ Weak neutral current interactions νµ νµ Conserves fermion charge, ∆Q = 0 Acts on both left-handed and right-handed components ¡Z (but with different strengths!) e e Short-range interaction weak neutral µ ) neutral massive mediator field Z currents − − E.g. neutrino scattering, νµe ! νµe M. Fanti (Physics Dep., UniMi) Fundamental Interactions 4 / 36 Seeking a symmetry group 4 gauge fields: 2 neutral (γ; Z) and 2 charged (W ±) ) symmetry group must have 4 parameters Cannot be just one group (e.g. U(2)), because electromagnetic interaction couples to uL; uR; vL; vR with same strength, while weak charged current only couples to uL; vR g0 P3 k Easiest way is a composition of two groups, SU(2) ⊗ U(1) : ! ei 2 θ0Y · eig k=1 θk T (note different couplings g; g 0 for groups SU(2); U(1)) Naively one could think at U(1) as the group for electromagnetism and at a SU(2) for weak (charged and neutral) interactions, but this again does not work, because weak charged currents couples only to uL; vR, while weak neutral currents also couple with uR; vL | despite with different strength ) The gauge boson of the U(1) group cannot be the photon! U(1) group: g0 i θ0Y µ e 2 ) one parameter θ0 ) one gauge field B Y is the weak hypercharge operator | each particle is an eigenstate of Y SU(2) group: P3 k ig k=1 θk T µ e ) 3 parameters θk ) 3 gauge fields Wk (k = 1; 2; 3) 2 2 2 2 2 T k are the weak isospin operators | each particle is an eigenstate of T and T 3 (T ≡ T 1 + T 2 + T 3) M. Fanti (Physics Dep., UniMi) Fundamental Interactions 5 / 36 Weak isospin and weak hypercharge for fermions T = weak isospin, Y = weak hypercharge : must commute with each other Weak charged currents couple to L, not to R ) treat L; R differently. uL L are isospin doublets: L ≡ with same hypercharge YL dL τ k ) T k ≡ ; L 2 0 1 0 −i 1 0 recall Pauli's matrices: τ = ; τ = ; τ = 1 1 0 2 i 0 3 0 −1 k R are isospin singlets: uR , dR ) T R = 0, no coupling to W -fields | and different hypercharge eigenvalues, Yu;R; Yd;R T 3 QY Here uL; dL stay for generic \up" or \down" states in isospin space, ν +1=2 0 u νe e can be e.g. up-/down-quarks, or leptons: ≡ − −1 d e− e −1=2 −1 L L L On the opposite, uR ; dR are their chiral counterparts, but they (νe)R 0 0 0 don't belong to doublets. However, uL; uR have same electric (e)R 0 −1 −2 charge, and likewise for dL; dR T3 eigenvalues are quantized (as from SU(2) algebra) u +1=2 +2=3 1=3 d −1=2 −1=3 Y eigenvalues are free ) chosen such to fulfill L Y = 2(Q − T 3) (u)R 0 +2=3 4=3 (d)R 0 −1=3 −2=3 M. Fanti (Physics Dep., UniMi) Fundamental Interactions 6 / 36 Weak hypercharge and weak isospin for gauge bosons 3 Extend the relation Y = 2(Q − T ) to any particle | i.e. also to vector bosons. U(1) and SU(2) are separate groups The B-boson has T = 0 and Y = 0 ) has charge Q = 0 ) no B-Wk interactions in L ) TB = 0 and YW = 0 (if B had Y 6= 0, it'd couple to itself: not foreseen since U(1)Y is abelian) The 3 W-bosons are assigned Y = 0( ) have charge Q = T3) 0 1 1 0 0 1 0 0 1 are in the adjoint representation of SU(2) W1 = @ 0 A ; W2 = @ 1 A ; W3 = @ 0 A ) dimension 3 0 0 1 0 0 0 0 1 T = −i 0 0 1 δW µ = −g θ W µ 1 @ A a abc b c 0 −1 0 0 0 0 −1 1 µ µ ) [Tb] = −ibac T = −i 0 0 0 δWa = igθb [TbW ]a ac 2 @ A µ 1 0 0 = igθb [Tb]ac Wc 0 0 1 0 1 T3 = −i @ −1 0 0 A ) T3 W3 = 0 ; T3 W1 = iW2 ; T3 W2 = −iW1 0 0 0 T 3 QY B 0 0 0 ± 1 Define W = p (W1 ± iW2) W 0 0 0 2 3 ± ± W + +1 +1 0 ) T3 W3 = 0 and T3 W = ± W W − −1 −1 0 M. Fanti (Physics Dep., UniMi) Fundamental Interactions 7 / 36 Interaction Lagrangian 0 0 g P3 k g X k µ ! ei 2 θ0Y · eig k=1 θk T )L EW = − ¯γ Y Bµ − g ¯γ T W k int 2 µ µ | {zY } k | {zk } Jµ Jµ Recall: k uL k τ L-fermions: L ≡ , with T L ≡ and one common YL dL 2 0 1 0 −i 1 0 recall Pauli's matrices: τ = ; τ = ; τ = 1 1 0 2 i 0 3 0 −1 k R-fermions: separate uR; dR , with T R ≡ 0 and different Yu;R; Yd;R g 0 g g 0 g 0 g LEW = − ¯γ BµY − ¯ γ W k,µτ k = − ¯ Y B= − ¯ Y B= − ¯ τ kW= k int 2 µ 2 L µ L 2 R R 2 L L 2 L L ! g 0 g 0 1 g 0Y B= + gW= 3 g(W= 1 − iW= 2) u = − u¯ Y B= u − d¯ Y B= d − u¯ d¯ L L R u;R R R d;R R L L 1 2 0 3 2 2 2 g(W= + iW= ) g Y B= − gW= dL | {z } L R-couplings | {z } L-couplings k def ¯ k Y def ¯ fermionic currents: Jµ = γµT and Jµ = γµY Y 1 Q = + T 3 ) electromagnetic current: Jem def= ¯γ Q = JY + J3 2 µ µ 2 µ µ M. Fanti (Physics Dep., UniMi) Fundamental Interactions 8 / 36 Physical electroweak fields g 0 Recall: LEW = − JY · B − g J1 · W 1 + J2 · W 2 + J3 · W 3 int 2 Weak charged currents: ± 1 2 + − ¯ ± 1 1 2 define J = J ± iJ (explicitely, J =u ¯LγµdL and J = dLγµuL) and W = p W ± iW µ µ 2 ! = − 1 1 2 2 1 + − − + 1 ¯ 0 W uL ) J · W + J · W = p W · J + W · J = p u¯L dL + 2 2 W= 0 dL NOTE: W= ∓ change isospin by ∆T 3 = ±1 and keep Y unchanged ) change fermion charge by ∆Q = ±1 Y 1 Electromagnetic currents: (recall: Q = + T 3 ) Jem = JY + J3 ) 2 2 1 1 we need a term −e Jem · A ≡ −e JY + J3 · A , can we extract it from − g 0JY · B + gJ3 · W 3 ? 2 2 Yes, if A is a \mixture" of B; W 3 ) introduce a rotation (B; W 3) ! (A; Z) B = A cos θW − Z sin θW 3 W = A sin θW + Z cos θW θW is the \Weinberg" electroweak mixing angle M. Fanti (Physics Dep., UniMi) Fundamental Interactions 9 / 36 Physical electroweak fields 1 g 0 g 0JY · B + gJ3 · W 3 = (A cos θ − Z sin θ ) JY + g (A sin θ + Z cos θ ) J3 2 2 W W W W g 0 g 0 = JY cos θ + gJ3 sin θ A + gJ3 cos θ − JY sin θ Z 2 W W W 2 W 0 Jem 1 z }| { 0 B1 Y 3C g cos θW = e ) can recover the e B J + J C · A term if: @2 A g sin θW = e e e Useful relations: B = 0 A − Z = A cos θW − Z sin θW 0 g g e = g sin θW = g cos θW e e 0 2 2 W 3 = A + Z = A sin θ + Z cos θ gg e e g g 0 W W e = | to enforce + = 1 pg 2 + g 02 g 0 g Then the part coupled to Z becomes: 0 3 g Y e cos θW 3 e sin θW em 3 e 3 2 em gJ cos θW − J sin θW = J − J − J = J − sin θW J 2 sin θW cos θW sin θW cos θW Electroweak interactions: 2 weak charged weak neutral 3 z }| { e.m.