Hadronic production of a Higgs boson in association with two jets at ...

1.2. Electroweak Symmetry Breaking 15the potential takes the following form,V (φ) = − 12λ µ4 + 1 2 (2µ2 )φ 2 1 + O(φ3 i ). (1.39)We note that φ 1 gains the mass √ 2µ and φ 2 is the massless Goldstone boson. Untilnow the discussions of this section and that preceding it have been identical. However,this Lagrangian contains a covariant derivative linking φ to the electromagneticfield A µ , and we must also inspect what happens to this term in the Lagrangian asa result of the symmetry breaking.2∑|D µ φ| 2 1=2 (∂ µφ i ) 2 + √ 2eφ 0 A µ ∂ µ φ 2 + e 2 φ 2 0A µ A µ + . . . (1.40)i=1where . . . represent cubic and quartic interactions of the fields. The piece we aremost interested in isL mA = 1 2 m2 A Aµ A µ = e 2 φ 2 0 Aµ A µ (1.41)i.e. the photon has acquired a mass which is proportional to the vacuum expectationvalue φ 0 . This illustrates how the mechanism of spontaneous symmetry breakingcan be responsible for the W and Z vector boson masses. The question remains asto the specific gauge group to break to correctly generate the observed spectrum ofvector boson masses.1.2.3 The Higgs mechanismMerely breaking the group SU(2) does not generate the correct spectrum of massesobserved in nature, one can generate either three identical mass vector bosons ortwo identical and one massless vector boson depending on the representation of thescalar field. However when we couple the scalar to both SU(2) and U(1) fields wecan correctly generate massive bosons with different masses. A beautiful feature ofbreaking SU(2) × U(1) is that there is also one residual massless boson with a U(1)gauge symmetry. This naturally becomes electrodynamics, and as result the weakand electrodynamic forces can be unified into the larger gauge group.In terms of SU(2) × U(1) gauge theory the covariant derivative for φ isD µ φ = (∂ µ − igA a µ − i1 2 g′ B µ )φ (1.42)