<<

Gauge in QED ● The Lagrangian density for the free e.m. field is

= 1 F F µν Lem − 4 µν where F µν is the field strength tensor

0 E E E − x − y − z  Ex 0 Bz By  F µν = ∂µAν ∂ν Aµ = − −  E B 0 B   y z − x     Ez By Bx 0   −    Thus = 1 (E2 B2) Lem 2 − ● In A0 = 0 gauge the density is ∂A π = = E ∂t − hence = π A˙ = 1 (E2 + B2) Hem · − Lem 2 1 ● Gauge symmetry: F µν and hence are under transformation Lem µ µ µ µ A A! = A + ∂ χ → 1 2 µ N.B. A mass term 2 mγ A Aµ is forbidden by gauge symmetry. ● The full Lagrangian for charged Dirac interacting with the e.m. field is = + where L Lem LD = iψ¯γµ(∂ + ieA )ψ mψ¯ψ LD µ µ −

iφ(x) ● Under a phase transformation ψ ψ! = e ψ, we have → µ iψ¯!γ (∂ + ieA i∂ φ)ψ! mψ¯!ψ! LD → µ µ − µ −

● Invariance under phase transformations of the fermions thus requires a compensating gauge transformation of the e.m. field

A A! = A + ∂ χ = A ∂ φ/e µ → µ µ µ µ − µ 2 i.e. χ(x) = φ(x)/e. Conversely, gauge invariance of the whole Lagrangian will − be preserved provided we simultaneously change the phase of the field

ieχ(x) ψ ψ! = e− ψ →

● Gauge transformations of e.m. field phase transformation of (charged) Dirac ⇒ field. ● Conversely, if we demand symmetry of under local (xµ-dependent) phase L transformations of ψ, this requires the existence of a (massless) vector field to

cancel the term involving ∂µφ(x).

● The gauge symmetry occurs because the derivative ∂µ only appears in the combination called the covariant derivative

Dµ = ∂µ + ieAµ

Then Dµψ transforms in the same way as ψ itself:

ieχ ieχ Dµ! ψ! = [∂µ + ieAµ + ie(∂µχ)]e− ψ = e− Dµψ

3 ● Note that also involves only D since Lem µ F µν = ∂µAν ∂ν Aµ = DµAν Dν Aµ − −

● Recall that e.m. phase (gauge) symmetry conservation of electric current ⇒ and charge (Noether) ● Successive gauge transformations commute:

ieχ ieχ ieχ ieχ ie(χ +χ ) e− 1 e− 2 = e− 2 e− 1 = e− 1 2

This is an Abelian gauge symmetry, with gauge U(1). Thus is a U(1) . ● It is believed that all fundamental interactions are described by some form of gauge theory.

4 Non-Abelian Gauge Symmetry

● Suppose the Lagrangian involves two fermion fields (e.g. νe and e−) and we demand symmetry under transformations that mix them together while preserving normalization and orthogonality:

ψ ψ! = αψ + βψ 1 → 1 1 2 ψ ψ! = γψ + δψ 2 → 2 1 2 We require

αα∗ + ββ∗ = γγ∗ + δδ∗ = 1

αγ∗ + βδ∗ = γα∗ + δβ∗ = 0

Hence

ψ1 α β ψ1 Ψ = Ψ! = UΨ  ψ2  →  γ δ   ψ2  ≡      

5 where

α β α∗ γ∗ 1 0 =  γ δ   β∗ δ∗   0 1        i.e. UU † = I. Hence the of coefficients U is a unitary matrix. ● U has 4 complex elements satisfying 4 constraints 4 real parameters. It can ⇒ be written as

U = exp[iα0 + iα1τ1 + iα2τ2 + iα3τ3]

where τ1,2,3 are the Pauli matrices

0 1 0 i 1 0 τ1 = , τ2 = − , τ3 =  1 0   i 0   0 1  −       and α0,1,2,3 are the real parameters. ● The exponential of a matrix A can be defined as

1 2 exp[A] = I + A + 2 A + . . .

6 Note that in general

exp[A ] exp[A ] = exp[A ] exp[A ] = exp[A + A ] 1 2 % 2 1 % 1 2 the gauge [U(2)] is non-Abelian. ⇒ ● We can write U = eiα0 V where eiα0 U(1), the Abelian symmetry group, and iα τ ∈ V = e · SU(2), the non-Abelian group of 2 2 unitary matrices with unit ∈ × . Check:

det U = exp[iTr (α τ )] = exp[0] = 1 · 1 U † = exp[ iα∗ τ †] = exp[ iα τ ] = U − − · − ·

● The matrices τ1,2,3 are the generators of the group SU(2). An infinitesimal gauge transformation can be written as

iα τ U = e · I + iα τ ( ·

7 ● In fact we can define the exponential for matrices like this: A N eA = lim I + N N →∞ ' ( iα τ 1/& e · = lim (I + i + α τ ) & 0 → · ● For simplicity we shall usually consider infinitesimal gauge transformations. If necessary, we can then build up finite ones as above. ● Consider a small SU(2) gauge transformation of the form

ψ1 g Ψ = Ψ! = I + i ω(x) τ Ψ  ψ  → 2 · 2 ) * 1   i.e. α1,2,3 = 2 g ω1,2,3 small. By analogy with QED, we expect gauge invariance µ µ µ to require the presence of vector fields W1 , W2 , W3 , coupling to the fermions via the covariant derivative g Dµ = ∂µ + i W µ τ 2 · = iΨγµD Ψ LD µ 8 N.B. Mass term ΨΨ to be discussed later. ∝ ● What transformation law must W µ have to make gauge invariant? We 1,2,3 LD need DµΨ to transform just like Ψ itself. Then g D! Ψ! = 1 + i ω τ D Ψ µ 2 · µ ) g * Ψ! = Ψ 1 i ω τ − 2 · ) µ * µ Ψ!γ D! Ψ! = Ψγ D Ψ ⇒ µ µ (up to terms of order ω2, which we are neglecting). µ ● This implies a more complicated gauge transformation law for W1,2,3. With

µ µ µ g µ D D! = ∂ + i W ! τ → 2 · we have µ µ g µ g D! Ψ! = ∂ + i W ! τ 1 + i ω τ Ψ 2 · 2 · As we have seen, this shou)ld equal * ) * g g g 1 + i ω τ DµΨ = 1 + i ω τ ∂µ + i W µ τ Ψ 2 · 2 · 2 · ) * ) * ) * 9 Thus 1 µ g µ g µ g µ g − ∂ + i W ! τ = 1 + i ω τ ∂ + i W τ 1 + i ω τ 2 · 2 · 2 · 2 · ) g * ) g * ) * = ∂µ + i W µ τ i (∂µω) τ 2 · − 2 · g2 (ω τ W µτ W µτ ω τ ) + (ω2) − 4 j j k k − k k j j O Now τ τ τ τ = 2i + τ j k − k j jkl l Hence µ µ µ µ W ! = W ∂ ω g + ω W l l − l − jkl j k

µ ● To preserve gauge invariance we have introduced gauge fields W1,2,3 via the covariant derivative. Clearly we also have to add a gauge field part to the LG Lagrangian, to allow propagation of the gauge fields. ❖ We might expect to be 1 F µν F where LG − 4 j jµν F µν = ∂µW ν ∂ν W µ, by analogy with QED. j j − j But this is not gauge invariant! 10 µ ❖ From transformation law for Wj we have

µν µν µν µ ν ν µ F ! = F g+ ω F g + [(∂ ω )W (∂ ω )W ] j j − jkl k l − jkl k l − k l and so

1 µν 1 µν g µν F ! F ! = F F + + F [(∂ ω )W (∂ ω )W ] − 4 j jµν − 4 j jµν 2 jkl j µ k lν − ν k µl 2 using antisymmetry of +jkl and neglecting terms of order ω . ● To get rid of the extra term we must define

= 1 Gµν G LG − 4 j jµν where Gµν = ∂µW ν ∂ν W µ g + W µW ν j j − j − jkl k l

● Notice that now contains terms that represent self-interactions of the LG gauge fields:

11 1 g+ ∂µW ν ∂ν W µ W W W W W vertex − 2 jkl j − j kµ lν ⇒ 1 2 3 + , w2

w1

w3

1 g2+ + W µW ν W W W W W W vertices − 4 jkl jmn k l mµ nν ⇒ k l k l wk wl

wk wl

12 Weak Interactions ● We can use a two-component notation to represent the two charge states of a given species of lepton or :

ψνe ψu Ψe = , Ψq =  ψe   ψd      ● Then the is described by an SU(2) gauge theory: the gauge

invariance w.r.t. Ψ Ψ! = UΨ is weak symmetry. → ● The interaction term is g ΨγµW τ Ψ where 2 µ · µ µ µ ψν W W iW Ψ = e , Ψ = ψ¯ , ψ¯ , W µ τ = 3 1 − 2 e   e νe e ·  µ µ µ  ψe W1 + iW2 W3 + , −     Defining 1 W ± = (W1 iW2) √2 ∓

13 w+ g g ¯ µ + -i γ µ we get a term ψν γ W ψe √2 e µ ⇒ 2 ν e - e

● We know from experiment that the W± in fact only interact with the left-handed fermion states ψ = 1 (1 γ5)ψ and correspondingly with L 2 − right-handed antifermions:

¯ 1 ¯ 5 (ψ)R = ψL = 2 ψ(1 + γ )

● Thus by Ψ we really mean ΨL:

Ψ γµW τ Ψ = 1 Ψ(1 + γ5)γµW τ(1 γ5)Ψ L µ · L 4 µ · − = 1 Ψγµ(1 γ5)2W τ Ψ 4 − µ · = 1 Ψγµ(1 γ5)W τΨ 2 − µ · using (1 γ5)2 = 1 + (γ5)2 2γ5 = 2(1 γ5). − − − 14 w+ w

νe + νe eL eR g -i γµ γ 5 2 2 (1− )

● We say that the left-handed fermions have I = 1 ( 2I + 1 = 2 w 2 ⇒ w states, e.g. νe, eL−), transforming under weak isospin gauge transformations as

ψL1 ig 1 ω τ ΨL = e 2 · ΨL  ψL2  →  

● The right-handed fermions have Iw = 0, i.e. only 1 state (eR−), invariant under weak isospin transformations:

ψ e0ψ = ψ R → R R

15 ● This is all fine except that it implies that fermions cannot have mass! The mass term in the Lagrangian is

mψ¯ψ = 1 mψ¯(1 γ5)(1 γ5)ψ + 1 mψ¯(1 + γ5)(1 + γ5)ψ 4 − − 4 = mψRψL + mψLψR

● Since ψR and ψR do not transform while ψL and ψL do, this is clearly not gauge invariant (unless m = 0). We postpone this problem for the moment. . .

16 Electroweak Interactions S. Glashow, S. Weinberg, A. Salam, 1961–68

● We interpreted (W iW )/√2 as W± fields, but what about W ? 1 ∓ 2 3 Could it represent the Z0 boson? ❖ It has the right kind of vertices: w3 w3

- eL νe - eL νe

❖ However, the Z0 also interacts with right-handed electrons. Also is has a

different mass from W±, so they cannot belong in an exact symmetry

multiplet (a weak isospin triplet, Iw = 1). ● But there is another neutral gauge boson, the photon. Therefore we suppose

that W3 may be a mixture:

µ µ µ W3 = cos θwZ + sin θwA

17 Here θw is the Weinberg angle or weak mixing angle

2 sin θw = 0.23117(16)

● The combination orthogonal to W3 is

Bµ = sin θ Zµ + cos θ Aµ − w w In the , Bµ is the gauge boson field for an additional Abelian gauge symmetry. Thus the overall electroweak symmetry is SU(2) U(1). × ● The for the U(1) gauge interaction is g! = g. The coupling % of any fermion to Bµ is proportional to its weak , Y , defined by

Y = Q I − 3

where Q is the charge (in units of e ) and I3 is the third component of the 1 | | weak isospin, i.e. 2 for the upper/lower component of a weak isospin doublet 1 ± (Iw = 2 ) and 0 for a singlet (Iw = 0).

18 Electroweak Quantum Numbers

Particles Q I3 Y ν , ν , ν 0 1 1 eL µL τL 2 − 2 e , µ , τ 1 1 1 L L L − − 2 − 2 νeR, νµR, ντR 0 0 0 e , µ , τ 1 0 1 R R R − − 2 1 1 uL, cL, tL + 3 2 6 d , s , b 1 1 1 L L L − 3 − 2 6 2 2 uR, cR, tR + 3 0 3 d , s , b 1 0 1 R R R − 3 − 3 N.B. Hypercharge is the average charge of the weak isospin multiplet. ● Weak isospin doublets are

νe νµ ντ uL cL tL , , , , ,  eL   µL   τL   dL   dL   bL              19 ● The electroweak Lagrangian becomes

= 1 Gµν G 1 Bµν B + iΨγµD Ψ LEW − 4 j jµν − 4 µν µ where Bµν = ∂µBν ∂ν Bµ and now − D = ∂ + igW I + ig!B Y µ µ µ · µ 1 with I = 2 τ for doublets (0 for singlets), and Y is as given in the table. ● General gauge transformation is

Ψ Ψ! = exp (igω I + ig!ω Y ) Ψ → · 0 µ where ω0,...,3 are 4 arbitrary real functions of the - coordinates x . ● The terms in involving the neutral gauge are LEW µ Ψγ g(cos θ Z + sin θ A )I + g!( sin θ Z + cos θ A )Y Ψ − w µ w µ 3 − w µ w µ

Hence the co-upling to the photon is g sin θw I3 + g! cos θw Y , which mu.st be equal to eQ = e I3 + e Y

g sin θ = g! cos θ = e ⇒ w w 20 ● Note that since Q is the same for left- and right-handed states, the photon couples to the currents

µ µ ψ¯Lγ ψL + ψ¯Rγ ψR = 1 ψ¯γµ(1 γ5)ψ + 1 ψ¯γµ(1 + γ5)ψ 2 − 2 = ψ¯γµψ Hence photon interactions are conserving (no γ5). ● The coupling of the Z0, on the other hand,

e 2 2 g cos θw I3 g! sin θw Y = cos θw I3 sin θw Y − sin θw cos θw − 2e + 2 , = I3 sin θw Q sin 2θw − involves a current that contains γ5 parity+ violation. , ⇒ ψ¯ γµ(I sin2 θ Q)ψ + ψ¯ γµ( sin2 θ Q)ψ L 3L − w L R − w R = 1 ψ¯γµ[(I sin2 θ Q)(1 γ5) sin2 θ Q(1 + γ5)]ψ 2 3L − w − − w = 1 ψ¯γµ(I 2Q sin2 θ I γ5)ψ 2 3L − w − 3L 21 The ● The electroweak theory we have discussed so far is perfectly self-consistent but it cannot be correct because all the fermions and gauge bosons have to be massless to preserve gauge-invariance of the Lagrangian. ● Recall that a fermion mass term would convert left-handed particles into right-handed ones and vice-versa:

mψ¯ψ = mψRψL + mψLψR

● Since the left-handed fermions are weak isospin doublets and the right-handed 1 ones are singlets, we need to replace m by a doublet (Iw = 2 ) type of quantity to restore gauge invariance, e.g. a Yukawa interaction

gf Φ†ψRΨL + gf ΨLψRΦ

φ1 where Φ = is a weak isospin doublet of scalar fields.  φ2    22 ● For first-generation leptons, for example, we find ¯ ¯ geφ1†ψeR ψνe + geφ2†ψeR ψeL + h.c. and thus for an electron mass we need

φ1 0 Φ0 = =  φ2   v/√2      with gev/√2 = me.

● Φ0 is the of the Higgs field Φ. Note that φ2 must be neutral, with Q = 0 , I = 1 Y = 1 3 − 2 ⇒ Higgs 2 Then φ1 must have Q = +1, so in general

+ φ1 φ = 0  φ2   φ     

23 0 ● Of course, the particular choice of Φ0 = breaks gauge invariance,  v/√2  so we do not seem to have achieved much. But weshall see that this type of symmetry breaking is ‘natural’, and does not spoil the good properties of gauge theories. ● The interactions between the Higgs field and the gauge fields are generated by the usual ‘kinetic’ term in the Higgs (Klein-Gordon) part of the Lagrangian

µ = D Φ† (D Φ) + . . . (see later) LH µ where Dµ is the electroweak+ covar,iant derivative,

µ µ µ µ D = ∂ + igW I + ig!B Y ·

● For the vacuum Higgs field we have explicitly (since v is constant)

√ +µ µ iv 2gW D Φ0 = µ µ 2√2  g!B gW  − 3   24 Now

µ µ g µ µ g!B gW3 = (sin θw B cos θw W3 ) − cos θw − g = Zµ −cos θw Hence 2 2 v g µ + 1 µ = 2W − W + Z Z + . . . LH 8 µ cos2 θ µ ' w ( 0 which corresponds to W± and Z mass terms

1 mW = 2 vg = mZ cos θw

2 2 mW N.B. sin θw = 1 m2 . − Z

25 Parameters of the Standard Model ● Standard (Glashow-Weinberg-Salam) Model describes electroweak interactions

at present energies in terms of three parameters g, g! (or θw), v. ● These are most accurately measured from ❖ the fine structure constant 2 2 2 e g sin θw 1 α = = = 137.03599976(50) − 4π 4π { }

❖ the Fermi weak coupling constant 2√ g 2 1 5 2 GF = = = 1.16639(1) 10− GeV− 2 2 8mW √2v ×

❖ the Z0 boson mass gv MZ = = 91.1882(22) GeV 2 cos θw

26 N.B. These relations are subject to higher-order (electroweak and strong) corrections. ● In addition there are further parameters of the Higgs sector, including Yukawa couplings. ● The Fermi constant is the effective 4-fermion coupling for the decay

µ− e− + ν¯ + ν → e µ - -i g e - -i 2 2 e 2 2 q-mW νe - - - µ w µ νe q2 m2 -i g νµ W -i G F νµ 2 2 2

5 2 G = 1.166 10− GeV− v = 246 GeV F × ⇒

27 ● The Yukawa couplings can be deduced from the fermion masses:

√2 g = m f f v The numerical values of the Yukawa couplings (and thus the fermion masses) are presently not understood (1 parameter per fermion). N.B. m = 175 GeV g = 1.0. t ⇒ t Is this accidental??

28 Spontaneous Symmetry Breaking ● The advantage of the Higgs mechanism for mass generation is that the

gauge-symmetry-breaking vacuum Higgs field Φ0 can arise ‘spontaneously’, even though the Lagrangian is exactly gauge-invariant. ● Let’s study first for simplicity the spontaneous breaking of a global (i.e. space-time independent) Abelian symmetry. Consider a classical complex scalar field φ with Lagrangian density

µ 2 2 = (∂ φ∗)(∂ φ) µ φ∗φ λ(φ∗φ) L µ − − Clearly is invariant w.r.t. L iα iα φ e φ , φ∗ e− φ∗ → → for any real constant α. ● The Hamiltonian density is

∂φ 2 = + φ∗ φ + V (φ) H ∂t ∇ · ∇ / / / / / / 29 / / where the ‘potential energy’ V (φ) is

V (φ) = µ2 φ 2 + λ φ 4 | | | | V

Imφ Reφ ❖ This potential has a minimum at the origin and hence the minimum-energy (vacuum) state has φ = 0 (classically). ❖ After second there will be zero-point fluctuations, but the vacuum expectation value φ = 0 φˆ 0 will still be zero. 0 - | | . ❖ The curvature of V (φ) at the minimum tells us the mass of the scalar bosons created and annihilated by the field (c.f. Klein-Gordon

30 equation): 1 d2V m2 = = µ2 2 dφ2 /φ=φ0 / / ❖ The term λ φ 4 represents a 4-boson inte/ raction that can be treated as a | | perturbation (coupling constant λ). ∝ ● Now suppose we change the sign of the first term in V (φ):

V = µ2 φ 2 + λ φ 4 − | | | |

V

Imφ Reφ 31 ❖ The potential has a ‘hump’: the minimum now lies anywhere on the circle

µ2 φ = eiθ (0 θ 2π) 02λ ≤ ≤

❖ The physical vacuum can be any one of these (infinitely many) degenerate vacua. But choosing a particular one (a value of θ) breaks the U(1) symmetry, since eiαφ will be a different vacuum (c.f. a falling rod or ferromagnet). Since a particular vacuum is realized, there is spontaneous symmetry breaking. ❖ The two principal curvatures of V (φ) at the vacuum solution are now different. One is zero (around the circle), indicating a massless boson (a ), in agreement with Goldstone’s theorem (1961): Spontaneous breaking of global symmetry massless boson. ⇒ N.B. We shall see this is not true in gauge theories. ❖ The other curvature (radial) is non-zero, indicating that a massive boson is also present.

32 ● Without loss of generality (because of gauge invariance) we can define the vacuum expectation value to be real:

v µ2 φ0 = , v = √2 0 λ Then setting 1 φ = [v + σ(x) + iη(x)] √2 we expect σ and η to represent massive and massless boson fields, respectively. ● Substituting in the Lagrangian, we find

= 1 [(∂µσ)(∂ σ) + (∂µη)(∂ η)] L 2 µ µ + 1 µ2[(v + σ)2 + η2] 1 λ[(v + σ)2 + η2]2 2 − 4 = 1 [(∂µσ)(∂ σ) m2 σ2] + 1 (∂µη)(∂ η) 2 µ − σ 2 µ + const. + interaction terms

where mσ = √2µ.

33 ● Now suppose our model scalar field theory is a U(1) gauge theory (Higgs model, 1964), i.e. the Lagrangian is invariant w.r.t.

iα(x) iα(x) φ e φ , φ∗ e− φ∗ → → for any real function α(x). We know this means there must be a gauge field Bµ(x) and the Lagrangian becomes

1 µν µ 2 2 = B B + D φ∗D φ + µ φ∗φ λ(φ∗φ) L − 4 µν µ − where Bµν = ∂µBν ∂ν Bµ − µ µ µ D = ∂ + ig!Y B Y being the corresponding hypercharge. ● The spontaneous symmetry breaking in the Higgs field φ works as before, giving a vacuum with Bµ = 0, φ = φ = 0. Choosing φ = v/√2 (real), this 0 % 0 generates a mass term for the gauge boson corresponding to mB = g!Y v:

1 µ 1 2 µ 2 (D v)(Dµv) = 2 (g!Y v) B Bµ 34 ● We seem to have created a degree of freedom: a massive vector field (S=1) has 3 polarization states in contrast to the 2 of a massless field. Longitudinal polarization (helicity 0) is possible, as well as transverse (helicity 1). ± ● This is because one degree of freedom of the Higgs field has disappeared. In a gauge theory we can always set the massless field η(x) to zero by a gauge transformation: choosing η(x) tan α(x) = −v + σ(x) we have

iα 1 φ! = e φ = (cos α + i sin α) (v + σ + iη) √2

1 2 2 1 = (v + σ) + η = (v + σ!) √2 √2 1 η2 where σ! = σ + + . . . = real 2v (This is called the ‘unitary gauge’.)

35 ● Thus we can say that the gauge field has ‘eaten’ the Goldstone boson in order to create its extra polarization state. ● The situation in the electroweak case is a little more complicated. The original SU(2) U(1) symmetry (with 4 generators) is spontaneously broken down to × U(1)em [not the original U(1)], with 1 . There should thus be 3 + 0 Goldstone bosons, but these are eaten by the W , W− and Z to produce their longitudinal polarization states. This leave one massless gauge boson – the photon – and one massive, neutral scalar boson – the .

36 Properties of the Higgs Boson ● Note that the mass of the Higgs boson H0, represented by the field σ, is

MH = mσ = √2µ . This is not determined directly by current data, which fixes only the combination of Higgs parameters µ v = = 246 GeV √λ

● The fact that Higgs bosons were not (?) produced at LEP implies

MH > 114.1 GeV (95% C.L.)

● Higher-order corrections depend (weakly) on µ and these suggest that

M < 200 GeV H ∼

37 ● Latest results from the LEP Electroweak Working Group:

2 High Q except mt 68% CL 200 ] GeV [

t 180 m mt (Tevatron)

Excluded 160 2 3 10 10 10 mH [GeV]

38 6 Theory uncertainty (5) Δαhad = 5 0.02761±0.00036 0.02749±0.00012 4 incl. low Q2 data 2 3 Δχ

2

1 Excluded 0 30 100 500

mH [GeV]

39 ● The couplings of the Higgs boson to fermions are all given by the corresponding Yukawa couplings, which are fixed by the fermion masses:

1 gf (v + σ)ψ¯ψ ψ √2 v mf = gf ⇒ √2 vertex factor ψ ⇒ g m i f = i f − √2 − v ● Thus the Higgs boson likes to decay into (and be produced by) the heaviest available fermion, H0 b¯b if M < 2m . → H t ● The coupling to gauge bosons is obtained by replacing v by (v + σ) in the term that produced the gauge boson masses:

2 (v + σ) 2 µ + 1 µ = g 2W − W + Z Z LH 8 µ cos2 θ µ ' w ( 2 1 2 1 2 2 µ + = MW + 2 vg σ + 4 g σ W − Wµ + , 40 vg2 g2 + 1 M 2 + σ + σ2 ZµZ 2 Z 2 cos2 θ 4 cos2 θ µ ' w w ( corresponding to the vertices: W Z

2 1ivg2gµν ivg gµν 2 2 W 2cos θW Z W Z

2 1 ig2gµν ig gµν 2 W 2 Z 2cos θW

+ 0 0 ● Thus decays to W W− and Z Z are expected if MH is large enough.

41 ● Branching ratios of the Higgs boson to various final states:

42