Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 305 2 a Solution Created 2026-10-03 Updated 2026-10-05
Take , , metric , and a Higgs potential with , . The gauge-scalar electroweak interaction isHere . With the printed plus-sign convention for , define ; thus . The minus sign in the nonlinear gauge field strength follows from this convention.
The minima have , with . A gauge transformation rotates the vacuum to . In unitary gauge, the three angular Goldstone bosons are removed, leaving and one real Higgs boson. The generator annihilates the vacuum, because its lower component has and hypercharge . This identifies the unbroken electromagnetic gauge group.
The quadratic terms from the Higgs field kinetic term areDefine the Weinberg angle and physical fields byThe neutral combinations areThe neutral gauge-boson mass matrix is . It has eigenvalues zero and , with the zero eigenvector giving . Expanding the Higgs potential about its minimum gives . Therefore the tree-level electroweak gauge-boson masses and scalar mass areAlso and . The three removed Goldstone bosons supply the longitudinal polarization vector degrees of freedom of the massive electroweak gauge bosons.
Replacing by in the neutral mass term gives all the Higgs boson couplings to Z bosons in unitary gauge:There are exactly the cubic and quartic elementary tree-level Feynman diagrams. Differentiating with respect to the identical fields gives Feynman rules and , respectively. There is no elementary vertex for the real neutral radial field.
Tree-level electroweak gauge-boson masses 2026-10-05
For the Higgs doublet vacuum , the neutral gauge-boson mass matrix in is . Its zero eigenvector is proportional to , the photon; the perpendicular vector is the Z boson. The charged fields have the stated W boson mass. Thus , where is the Weinberg angle. The unbroken charge generator annihilates the vacuum and explains the zero eigenvalue.
