Gauge Theories with Spontaneous Symmetry Breaking

Transcription

1 Gauge Theories with Spontaneous Symmetry Breaking Zhiguang Xiao May 4, 07

2 The Higgs Mechanism Realization of Symmetries we met: Global symmetry conserved current. SSB of Global Symmetry:interactions are constraint, coupling constants related; for continuous symm, current is still Conserved, massless Goldstone. Gauge symmetry: local symmetry, require massless vector bosons, interactions are highly restricted. What happens if we combine SSB and local gauge symmetry? gauge boson acquire masses, and we can choose a gauge s.t. the goldstone degree of freedom becomes the third d.o.f of the massvie gauge boson.

3 The Higgs Mechanism An Abelian example: L = 4 (Fµν + D µϕ V(ϕ, D µ = µ + iea µ. gauge symmetry: ϕ(x e iα ϕ(x, A µ(x A µ e µα(x. Choose Potential: V(ϕ = µ ϕ ϕ + λ (ϕ ϕ, µ > 0 minimum of the potential : ϕ = ϕ 0 = µ λ. We choose the real vev: ϕ = ϕ 0 = µ λ. Expand about ϕ 0, ϕ = ϕ 0 + (ϕ (x + iϕ (x, potential, quadratic terms: V(ϕ (µ ϕ + O(ϕ 3 i, m = µ, m = 0. ϕ goldstone. kinetic terms: D µϕ = µ ϕ µϕ iea µ( µ ϕ ϕ + ϕ µ ϕ + e ϕ ϕa µa µ = ( ϕ + ( ϕ + eϕ 0 A µ µ ϕ + e ϕ 0A µa µ +... A µ acquire mass: m A = e ϕ 0, m AA µ A µ = m A(A 0 A 0 A i A i, physical space-like A i have the correct sign of mass.

4 The Higgs Mechanism There is an interaction term: L eϕ 0A µ µ ϕ, mixing term between A µ gauge boson and ϕ goldstone. No eϕ 0A µ µ ϕ term, cancelled. Recall in QED, photon does not acquire mass, photon selfenergy PI (k g µν k µ k ν Π(k, Π(k does not have /k pole. photon propagator: i(gµν k µ k ν /k i ξkµ k ν. If Π(k has a /k pole, photon k ( Π(k k 4 acquire mass. Goldstone provides the right pole to give mass to gauge boson and make the vacuum polarization amplitude properly transverse. il eϕ 0A µ( ik µ ϕ (k = m A k µ ϕ (k Also treat mass term as perturbation:

5 The Higgs Mechanism massless vector: two physical transverse polarizations. Massive vector boson, three physical d.o.f.: µf µν + m A ν = 0, µa µ µ νf µν = 0. k ϵ = 0. k = m. Three space-like polarization vectors. k µ = (E, 0, 0, k, ϵ T( = (0,, 0, 0, ϵ T( = (0, 0,, 0, ϵ L = (k, 0, 0, E, longitudinal polarization. Roughly m speaking, Goldstone provides the third d.o.f of the massive gauge boson. Unitary gauge: the Goldstone does not appear as an independent physical particle. Decompose ϕ(x = e iα(x ϕ (x, ϕ a real field. ϕ (x = σ(x + ϕ 0. ϕ 0 = µ λ is the v.e.v. α(x can be gauged away.a µ = A µ + i e µα. L = 4 (Fµν + D µϕ V(ϕ = 4 (F µν + (D µϕ D µ ϕ V(ϕ rename ϕ ϕ ========== A A 4 (Fµν + ( µ iea µϕ( µ + iea µ ϕ V(ϕ m A = e ϕ 0 = 4 (Fµν + ( µϕ + e ϕ A µa µ V(ϕ = 4 (Fµν + ( µσ + e ( ( σ + ϕ 0 A µa µ V σ + ϕ 0 = + e ϕ 0 AµAµ + e σ A µa µ + ( e ϕ 0 σa µa µ V σ + ϕ 0 = e µ λ, m σ = µ. No goldstone modes in the Lagangian.

6 The Higgs Mechanism SSB with Non-Abelian gauge symmetry: ϕ i real. ( complex ϕ (ϕ + iϕ. Symmetric transformation: ϕ ( + iα a t a ijϕ j = ( α a T a ijϕ j, t a : pure imaginary, hermitian. T a = it a real anti-symmetric. Covariant derivative D µϕ = ( µ iga a µt a ϕ = ( µ + ga a µt a ϕ Kinetic term: D µϕ id µ ϕ i = µϕ µ ϕ + ga a µ µ ϕ it a ijϕ j + g A a µa bµ (T a ϕ i(t b ϕ i i SSB: ϕ i = (ϕ 0 i + ϕ i(x, Ω ϕ i Ω = (ϕ 0 i mass matrix: (m ab = g i (Ta ϕ 0 i(t b ϕ 0 i. (m ab positive semi-definite: v a m ab v b = g i [va (T a ϕ 0 i ] 0. Positive eigenvalues: massive gauge bosons. Zero-eigenvalues: massless gauge bosons. Unbroken generators: T a ij ϕ 0,j = 0, (m ab = 0, for b. Broken generators: T a ij ϕ 0,j 0, also T b (m ab 0. Linearly independent broken generators, (m ab nondegenerate, nonvanishing eigenvalues, massive gauge boson. Broken generators massive gauge bosons, also correspond to goldstones. Unbroken generators : massless gauge bosons.

7 The Higgs Mechanism Interaction vertex: ga a µ µ ϕ i(t a ijϕ 0,j; (T a ijϕ 0,j 0 broken generator T a, (T a ijϕ 0,j Goldstone excitation direction, µϕ i is projected onto the goldstone direction. Gauge boson propagator receives a contribution from the Goldstone boson. PI+ goldstone: Propagator = i k δ abg µν + ( i = i(g µν kµkν k k δ aa g µµ (gµ ν kµ k ν k [k m ] ab i k µk ν k k δ ab im a b ( i k δ b bg ν ν +...

8 The Higgs Mechanism A Non-Abelian example: ϕ = ( ϕ ϕ, ϕ (, doublet complex fields in SU( fundamental rep.. Covariant derivative : D µϕ = ( µ iga a µτ a ϕ. τ a = σa. SSB: ϕ = ( 0, we can always choose unitary transformation s.t. ϕ = 0 ( = ϕ v 0. Expand ϕ = ϕ 0 + π +iπ π 3 +iπ 4 Gauge boson mass from: D µϕ g ϕ 0 AµAµ ϕ 0 = g ( 0 v τ a τ b ( 0 v A a µa b,µ = g v 8 Aa µa a,µ A,,3 all are massive. m A = gv. All the SU( symmetry are broken: ( iτ 0 = ( ( iv π v 0, iτ 0 = ( ( v π v 0, iτ 3 0 = ( 0 π v iv 4 Goldstone : π, π, π 4.

9 The Higgs Mechanism SU( adjoint rep.: ϕ a, real fields, a =, Covariant derivative: D µϕ a = ϕ a + gϵ abc A b µϕ c. L = Dµϕa D µ ϕ a + µ ϕ λ 4 (ϕ vev: ϕ 0 = 3 i= (ϕi 0 = µ = λ v, -sphere, choose ϕ 0 = (0, 0, v. ( T = it = ϵ ab 0 = ( 00 (, T = it = 0 ( T 0 ( 0v ϕ 0 = =, v 0 ( ( T 00 ( v0 ϕ 0 = =, T 3 ϕ 0 = 0 v 0 (, T 3 = it 3 = 0 ( 0 ( 00 = 0 T, T generators are broken generators. T 3 is not broken. We would expect the A µ, A µ are massive, and A 3 µ is massless. mass terms from kinetic energy part:(using ϵ abc ϵ ab c = δ bb δ cc δ bc δ cb. Dµϕa D µ ϕ a g (ϵ abc A b µ ϕ 0,c(ϵ ab c Aµ,b ϕ 0,c a = g (A a µ Aa,µ ϕ 0 Ab µ Ac,µ ϕ b 0 ϕc 0 = g v ((A µ + (A µ Only A µ, A µ acquire masses: m = m = gv. A 3 remains massless. v

10 The Higgs Mechanism ( Unitarity gauge: ϕ = e i(t φ +t φ 00 φ φ, φ = Kinetic part: (Dµϕ (Dµφ = µ σ µσ + g ((A µ + (A µ (v + σ. v+σ. V[ϕ] V[φ] = µ (σ + v + λ (v + σ4 4. Two of the gauge bosons acquire the same mass and one remain massless.

11 The Higgs Mechanism Another example: SU(3 gauge theory: ϕ a in adjoint rep., a =,... 8 t a, 3 3 matrix in fundamental rep. tr(t a t b = δab, Define Φ = ϕ a t a, A µ = A a µt a. Φ transforms as: Φ e iαa t a Φe iαta = Φ + iα a [t a, Φ] +..., Infinitesimal: δφ = iα[t a, Φ]. If i[t a, Φ 0] = 0, t a unbroken generators. If i[t a, Φ 0] 0, t a broken generators. Covariant derivative:(d µϕ a = µϕ a + gf abc A b µϕ c, in Φ, D µφ = µφ ig[a µ, Φ] A µ mass term from kinetic part: [(Dµϕa ] = tr[(d µφ ] g tr ( [A µ, Φ][A µ, Φ] = g tr ( [t a, Φ][t b, Φ] A a µa b,µ. m ab = g tr ( [t a, Φ 0][t b, Φ 0]. Vev of Φ, Φ = Φ 0, hermitian and traceless, can be diagonalized by unitary transformaion, eigenvalues are real. Different vev can have different breaking pattern. Diagonal traceless real matrix: independent, ( and SU(3 generators: ( t a τ a 0 =, for a =,, 3; t 8 = t 4 = ( 0 ; t 5 = ( i 0 i ( 0. ( ; t 6 = ( 0 ; t 7 = ( 0 i i

12 The Higgs Mechanism ( Φ 0 = ϕ 0 = 3 ϕ 0 t 8 Φ 0 commutes with t, t, t 3, t 8 ; t,,3,8, unbroken generators, SU( U( unbroken. t 4,5,6,7 broken generators. A 4,5,6,7 massive gauge bosons. Mass matrix m ab = g tr ( [t a, Φ 0][t b, Φ 0] [t 4, Φ] = 3[t 4, t 8 ] ϕ 0 = 3it 5 ϕ 0, [t 5, Φ] = 3[t 5, t 8 ] ϕ 0 = 3it 4 ϕ 0, [t 6, Φ] = 3[t 6, t 8 ] = 3it 7 ϕ 0, [t 7, Φ] = 3[t 7, t 8 ] = 3it 6 ϕ 0, mass term: m ab = g tr ( [t a, Φ 0][t b, Φ 0] A a µa µ,b = 9g ϕ 0 a=4,5,6,7 Aa µa µ,a, m 4,5,6,7 = 9g ϕ 0 = (3g ϕ 0. m,,3,8 = 0.

13 The Higgs Mechanism ( Another: Φ 0 = ϕ 0 = ϕ 0 t 3. 0 Only t 3, t 8 commute with Φ 0, unbroken, U( U( unbroken. t,,4,5,6,7 are broken. m ab = g tr ( [t a, Φ 0][t b, Φ 0] [t, Φ] = [t, t 3 ] ϕ 0 = it ϕ 0, [t, Φ] = [t, t 3 ] ϕ 0 = it ϕ 0, [t 4, Φ] = [t 4, t 3 ] ϕ 0 = it 5 ϕ 0 [t 5, Φ] = [t 5, t 3 ] ϕ 0 = it 4 ϕ 0, [t 6, Φ] = [t 6, t 3 ] ϕ 0 = it 7 ϕ 0, [t 7, Φ] = [t 7, t 3 ] ϕ 0 = it 6 ϕ 0 mass term: m aba a µa b,µ = a=4,5,6,7 g ϕ 0 A a µa a,µ + 4g ϕ 0 A a µa a,µ a=, m, = 4g ϕ 0, m 4,5,6,7 = g ϕ 0, m 3,8 = 0. Gauge bosons acquire masses: gauge bosons corresponding to broken generators split to different masses.

14 Formal description of the Higgs Mechanism: Global symmetry of L 0 conserved current J µ. If we require the global symmetry parameter to depend on x, continuous and differentiable and nonvanishing in a small region: δl = ( µα a J µ,a. For on-shell fields d 4 xδl on shell fields = 0, d 4 x µαj µ,a = α µj µ,a = 0 µj µ,a = 0. J µ,a is the conserved Noether current. If we require the local invariance: introducing A µ L = L 0 ga a µj µ,a + O(A ; A a µ A a µ + µα + O(A g The gauge field is always coupled to the global current at O(g. Global SSB: for broken generators, J µ,a destroy or create the Goldstone π k, Ω J µ,a π k (p = ip µ F a ke ip x, in general F a k depends only on Lorentz invariant p of on-shell π k. F a k constants. F a k 0 for broken generators t a, and F a k = 0 for unbroken generators. J µ,a conserved: 0 = µ Ω J µ,a π k (p = p F a ke ip x, so p = 0. example: For weakly coupled scalar theory, Noether current J µ,a = µϕ it a ijϕ j, Ω µϕ it a ijϕ j ϕ i(p = ip µ(t a ϕ 0 ie ip x, so F a i = T a ijϕ 0j, nonzero only for broken generator direction of Goldstone exitation.

15 Formal description of the Higgs Mechanism: Local symmetry: A µ couples to the Neother current. The corresponding global symmetry is sponteneous broken. Vertex ia a µj µ,a, Amplitude for a gauge boson to convert to a Goldstone: d 4 x k; a ia b µ(xj µ,b (x π j(p = ϵ a µ ( gk µ F a jδ (4 (k p The vacuum polarization diagram of gauge boson: still transverse, (current conservation, ward id. The /k pole term will give the mass to gauge bosons in the propagator contribution from Goldstone. so m ab = g F a jf b j This can be used in any SSB theory, not only in broken by a scalar field v.e.v, but also by nonperturbative effects. Gauge bosons coupled to the SSB currents massive.

16 Gauge theory with SSB An expeimentally correct unified description of electro-weak Interactions, GWS theory. ( Gauge and Scalar sector: Gauge Symmetry: SU( U( Y. Gauge boson A a µ for SU(, and B µ for U( Y. Scalar field: ϕ = (ϕ, ϕ T in fund. rep of SU(. Charge / under U(. Transformation of ϕ: ϕ e iαa τ a e iβ/ ϕ. (τ a = σa, τ 0 = I. If the scalar field get vev: ϕ = ( 0 v : Unbroken generator: τ + τ 0 = ( 0 0 0, Symmetry para: α3 = β, α, = 0. Broken generator (can be a little arbitrary: τ, τ, τ 3 τ 0 = ( We expect that there are three massive gauge boson and one massless gauge boson. Define: ϕ = ( φ +iφ v+φ 3 +iφ 4, Goldstone: iτ ( 0 v = ( iv 0 φ, iτ ( 0 v = ( v 0 φ, i(τ τ 0 ( 0 v = ( 0 iv φ4

17 ( Covariant derivative for ϕ: D µϕ = ( µ iga a τ a ig B µτ 0 ϕ ϕ For direct product group, each simple group factor can have a separate gauge coupling constant. In general, G G, transformation in matrix rep U U = e ita α a e itb β b, we always use t a to represent t a I, T b represents I T b, U U = e ita α a e itb β b iga a µt a ig B b µt b U µu igua a µt a U ig UB b µ T b U =U µu + U µu iguaa µt a U ig U B b µ T b U U A a µt a U only transforms to combinations of ta, not to mix with T a, and similar for T b. So the two kinds of gauge fields transform separately, A a µt a i g U µu + UAa µt a U, Ba µt a i g U µu + UAa µt a U. g and g can be different.

18 Masses of gauge bosons: from kinematic term (D µϕ D µ ϕ: (0, v(iga a µτ a + i g Bµ( igaµ,b τ b i g Bµ = ( g v 4 Aa µa µa + g 4 BµBµ gg B µ A 3 µ = We have defined ( g v 4 (A µa µ + A µa µ + v 4 (ga3 µ g B µ = 4 g v W + W + ( 4 v (g + g Z 0 µz 0,µ W ± = (A µ ia µ, m W = gv, Z 0 µ = g + g (ga3 µ g B µ, m Z = v g + g ( 0 v We can also define another gauge boson orthogonal to Z 0 µ, which is massless A µ = g + g (g A 3 µ + gb µ, m A = 0 A µ massless gauge boson electromagnetic vector potential.

19 Covariant derivative in terms of W ±,Z 0 and A µ. SU( Generator T a in some rep; Y: U( generator. Define T ± = T ± it, D µ = µ iga a µt a ig YB µ = µ i g [W + T + + W T i ] g + g (g T 3 g YZ 0 µ g g i g + g (T3 + YA µ A µ massless, correspond to the unbroken generator T 3 + Y. Electric charge e = g gg, electric charge quantum number Q = +g T3 + Y. We will see later for electron: Q =. Define θ w weak mixing angle, sin θ w = ( Then, e = g sin θ w = g cos θ Z 0 w, = A g g +g, cos θw = g g +g. ( ( cos θw sin θ w A 3 sin θ w cos θ w B i g + g (g T 3 g i Y = g + g ((g +g T 3 g g Q = i (T 3 sin θ wq. cos θ w Covariant derivative: D µ = µ i g [W + T + + W T ] ig (T 3 Q sin θ wz µ ieqa µ, g = e. cos θ w sin θ w Origial gauge coupling: g, g e, θ w; vev: v m W = gv, m W = m Z cos θ w. e, θ w, m W independent paras.

20 Scalar Lagraigian: L ϕ = D µϕ + ( µ ϕ ϕ λ(ϕ ϕ (, In Unitarity gauge: ϕ = U(x 0 0 v + h(x. v + h(x h(x:higgs field, chargeless. (( For ϕ, Q = T 3 + Y = ( 0 0 = ( 0 0 0, T+ = ( 0 0 0, T = ( D µϕ =U(x ( µ i g [W + T + + W T ] ig ( (T 3 Q sin 0 θ wz µ ieqa µ cos θ w v + h(x ( =U(x i g W + (v + h(x µh(x + ig Z cos θ w µ(v + h(x D µϕ = g ( µh(x + 8 cos Z θ µ(v + h(x + g v w 4 W+ µ W µ( + h(x v = ( ( µh(x + m W W+ µ W µ + m Z ZµZµ( + h(x v Potential term in Lagrangian: (v = µ /λ, m h = µ. µ ϕ λ ϕ 4 = µ (v + h(x λ(v + h(x4 4 = m h h λmh h 3 4 λh4 + const.

21 D µϕ = ( ( µh(x + m WW + µ W µ + m ZZ µz µ( + h(x v µ ϕ λ ϕ 4 = m hh λmh h 3 4 λh4 + const. m h = µ = λv, proportional to v, λ. Interaction with gauge boson: 3-point coupling mass of the gauge boson. Self- 3-point interaction: m h.,

22 Gauge boson kinetic term:l 4 Faµν F a µν 4 Bµν B µν F µν = µa ν νa µ + g(a µ A3 ν A3 µ A ν, F µν = µa ν νa µ + g(a 3 µa ν A µa 3 ν, F 3 µν = µa 3 ν νa 3 µ + g(a µa ν A µa ν, B µν = µb ν νb µ. (F µν if µν = D µw + ν D νw + µ, (F µν + if µν = D µw ν D νw µ where D µ = µ iga 3 µ = µ ig(sin θ wa µ + cos θ wz µ = µ ie(a µ + cot θ wz µ W + : charge +, W : charge. F 3 µν = µa 3 ν νa 3 µ ig(w + µ W ν W µ W + ν = sin θ wf µν + cos θz µν ig(w + µ W ν W µ W + ν B µν = cos θ wf µν sin θ wz µν where F µν = µa ν νa µ, and Z µν = µz ν νz µ L 4 Fµν F µν 4 Zµν Z µν (D µ W ν D ν W µ (D µw + ν DνW+ µ + ie(f µν + cot θ wz µν W + µ W ν e sin θ w (W +µ W µ W+ν W ν W+µ W + µ W ν W ν

23 Coupling to fermions: If the quantum number of the fermion fields are fixed, the covariant derivatives are uniquely determined fixing the coupling between gauge bosons and fermions. Massless fermions: left-handed and right-handed fermions decouple in kinetic terms: ψi /ψ = ψ Li /ψ L + ψ Ri /ψ R, ψ L,R = ( γ5 ( ± γ5 ψ, ψl,r = ψ ( massive term m ψψ = m( ψ Lψ R + ψ Rψ L So for massless fermions, ψ L and ψ R could be in different representations of the gauge group. In early days, in experiment only left-handed neutrino and no right-handed neutrino were found. Left-handed neutrino presents in the beta decay, coming with lepton, e, or µ. W ± only couples with left-handed states of quark and lepton. We assign the left-handed fermion fields eg. (ν L, e L to SU( doublets, and right-handed fermion fields to singlet. (We still suppose there is no right-handed neutrinoes. We have then T 3 for fermions, and from experiment we have their Q charge quantum numbers. From Q = T 3 + Y, we can find out Y quantum numbers.

24 Consider only one family of fermions: left-handed ( Q T 3 Y right-handed Q T 3 Y ul Q L = 3 d L u R d 3 R 3 ( νl 0 E L = (ν Rnot consider e R 0 e L fermion fields: fermion and ( antifermion particle. ( u νe The first generation: Q L =, E d L =, u e R, d R, e R. ( L ( L c νµ The second generation: Q L =, E s L = µ, c R, s R, µ R. ( L ( L t ντ The third generation: Q L =, E b L = τ, t R, b R, τ R. L L

25 Fermion mass terms: ψ L: SU( fundamental rep; ψ R SU( singlet. Mass term: m ψψ = m( ψ Lψ R + ψ Rψ L not singlet, forbidden by SU( symmetry and U( Y. Scalar field ϕ also in SU( fundamental rep. We can construct interaction term: ( ϕ L e = λ e Ē L ϕe R + h.c. = λ e( ν e, ē L ϕ + v e R + h.c. λ e: Yukawa coupling constant. Y charges: Ē, ; ϕ, ; er :. Lepton mass term from scalar vev: ( 0 L e λ e( ν e, ē L v e R + h.c. = λev (ē Le R + ē Re L = λev ēe. m e = λev. mass Yukawa coupling λ e, and vev. m w = gv, mz = gv/ cos θw, me = 0.5MeV, mw = 80GeV. me m w = λ e λ g e is put by hand very small.

26 Quack mass: We have right-handed down quark Y charge: Q L, Y = 6 ; ur, Y = 3 ; dr, Y = 3 ; ϕ =. L Q = λ d QL ϕ d R λ uϵ ab ( Q L aϕ bu R + h.c. λ u,d Yukawa couplings. Y charge: QL ϕd R, ( = 0; ( Q L aϕ bu R, ( = 0 SU( fundamental rep. is pseudoreal: QLa M a c QLc, ϕ b M b b ϕ b, ϵ ab ( Q L aϕ bu R ϵ ab M a c M b b QLcϕ b = ϵ cb det M QLcϕ b = ϵcb QLcϕ b. Mass term L Q λ d v( d Ld R + d Rd L λ uv(ū Lu R + ū Ru L = λ d v d d λ uvū u m u = λ uv; m d = λ d v ( 0 Interaction with higgs: Unitary gauge, ϕ (v + h L Q + L e = m e( + h ēe m v d( + h d d m v u( + h ū u v Coupling mass

27 Fermions and Gauge boson Interaction: one generation of fermion Covariant derivative: id µ =i µ + ( g doublet i µ + L = Ē L id/e L + ē R id/e R + Q L id/q L + ū R id/u R + d R id/d R g [W + T + + W T ] + g (T 3 Q sin θ wz µ + eqa µ cos θ w ( cos θ w Q sin θ wz µ + eqa µ g W µ singlet i µ gq sin θ w cos θ w Z µ + eqa µ g W + µ g ( cos θ w Q sin θ wz µ + eqa µ L =Ē L i /E L + ē R i /e R + Q L i /Q L + ū R i /u R + d R i /d R + g(w + µ J+µ W + W µ J µ W + Z0 µ Jµ Z + eaµjµ EM

28 The second and third generation: e µ, τ, ν e ν µ, ν τ, u c, t, d s, b, Interaction with W ± µ, only left-handed fermions, charged current weak interaction. Interaction with Z µ, neutral current weak interaction, both left-handed and right-handed fermions, coupling different. No flavor changing neutral current at tree level. Interaction with A µ, electromagnetic current. vector coupling left right-handed couplings are the same.

29 We have three generation of quarks and leptons ( (( ( ( Q i u i u c t L = =, u i R = (u d s b R, c R, t R, d i R = (d R, s R, b R. d i L, L, L Gauge coupling:quarks of the same quantum number have the same gauge coupling. However, coupling with higgs will in general mix flavors. The most general renormalizable gauge invariant coupling L L Q = λ ij d Q i L ϕ d j R λij uϵ ab ( Q i L aϕ bu j R + h.c. λ ij d, λij u: most general, not necessarily symmetric or Hermitian, complex-valued matrices. We could diagonalize the matrices by unitary transformations different on leftand right- handed fermion fields. λ ij u = (U ud uw u ij ; D u = ( D u D u D 3 u λ ij d = (U dd d W d ij ; D d = D d D d, D 3 d, D i u : real D i d : real change variables: u i R W ij uu j R, di R W ij d dj R, ui L U ij uu j L, di L U ij d dj L.

30 change variables to mass eigenstates: u i R W ij uu j R, di R W ij d dj R, ui L U ij uu j L, d i L U ij d dj L. L Q = D i d d i L(v + h(xd i R D i uū i L(v + h(xu i R + h.c. masses: m i d = D i dv, m i u = D i uv Interaction with higgs: coupling m i d d i L(h(xd i R, proportional to fermion masses ū i Lγ µ u i L (ū LU u i γ µ (U uu L i = ū i Lγ µ u i L invariant. ψl,rγ µ ψ L,R invariant under the transformations. The J µ EM, Jµ Z are of this form. QCD coupling are also invariant like J µ EM vector coupling. J ± W has interaction between UL and dl J ± W ū i Lγ µ d i L (ū LU u i γ µ (U d d L i = ū Lγ µ (U uu d d L = ū i Lγ µ V ij d j L V ij : Cabibbo-Kobayashi-Maskawa matrix.

31 We can still have transformation freedom: q i L e iαi q i L, q i R e iαi q i R, does not change mass and kinetic terms, and other diagonal interaction terms. Mass term:d i d d i L(v + hd i R D i de iβ i di L (v + he iβ i d i R, invariant; interaction term ū i Lγ µ u i L ū i Le iα i γ µ e iα i u i L, invariant. CKM: ū i LV ijd j L ūi Le iα i V ije iβ j d j L example: for two generations: V ij U(, i, j =,, four parameters, one real, three phases: (ū L, ū L V ( d L d L ( (ū L e iα, ū L e iα e iβ d V L e iβ d L ( =(ū L, ū L e i(α α e i(β α V d L e i(β α d L Eliminate three independent ( phases of V. V has only one real freedom, cos θc sin θ c Cabbibo angle. V = sin θ c cos θ c V ij a unitary 3 3 matrix, 9 free para: 3 parameterize real othogonal matrix, 6 phases; redefining the 6 fermion phases, five independent phase freedom can be used to eliminate the CKM phases. So, we can obsorb 5 phases into fermions. Only three real parameters and one phase parameter is left in V. One matrix element is complex. Complex coupling constant CP violation. CP violation only presents in the charged current weak interaction with W ±.

32 CKM matrix, parameterization: 0 0 e iδ/ 0 0 c 3 0 s 3 V = 0 c 3 s s 3 c e iδ/ s 3 0 c 3 e iδ/ 0 0 c s s c e iδ/ 0 0 c c 3 s c 3 s 3e iδ = s c 3 c s 3s 3e iδ c c 3 s s 3s 3e iδ s 3c 3 s s 3 c c 3s 3e iδ c s 3 s c 3s 3e iδ c 3c 3 c ij = cos θ ij, s ij = sin θ ij

33 Leptonic sector: if there is no right neutrino, no neutrino mass, there will be no CKM like matrix. L l = λ ij l Ēi L ϕe j R + h.c. just need to diagonalize λ ij l = U l D l W, transform e i L U l e L, ν L U l ν L, e R W l e R. in J + W : νi Lγ µ e i L invariant.