Before electroweak symmetry breaking, the theory has four massless gauge bosons: W1, W2, W3 (from SU(2)_L) and B (from U(1)_Y). After breaking, the physical states are W+, W-, Z, and the photon. How are the Z and the photon related to W3 and B?
AZ = W3 and photon = B
BThe Z and photon are orthogonal linear combinations of W3 and B, mixed by the weak mixing angle theta_W: Z = cos(theta_W) W3 - sin(theta_W) B, and A (photon) = sin(theta_W) W3 + cos(theta_W) B
CThe Z is a bound state of W3 and B
DW3 becomes the Z and B becomes the photon, with no mixing
Symmetry breaking mixes W3 and B because the Higgs doublet carries both SU(2) and U(1) charges. The physical photon and Z are the mass eigenstates — orthogonal rotations of W3 and B by the weak mixing angle theta_W (approximately 28.7 degrees). The photon is the combination that leaves the vacuum invariant (coupling to the unbroken U(1)_EM), and the Z is the orthogonal combination (which acquires mass). The mixing angle is determined by the ratio of the SU(2) and U(1) gauge couplings: tan(theta_W) = g'/g.
Question 2 True / False
The weak interaction violates parity (it distinguishes left from right) because only left-handed fermions couple to the W boson. This parity violation is built into the gauge structure of SU(2)_L.
TTrue
FFalse
Answer: True
The subscript L in SU(2)_L indicates that only left-handed fermion fields form doublets under SU(2) — right-handed fermions are singlets that do not couple to W bosons. This means the W boson couples only to left-handed particles (and right-handed antiparticles), violating parity maximally. This is not an incidental feature but is fundamental to the gauge structure: you cannot make the weak interaction parity-conserving without changing the gauge group. Parity violation was discovered experimentally by Wu in 1957 and explained theoretically by the V-A (vector minus axial vector) structure of the weak current.
Question 3 True / False
At energies far above the electroweak scale (E >> 246 GeV), the distinction between the electromagnetic and weak interactions disappears.
TTrue
FFalse
Answer: True
Above the electroweak scale, all four gauge bosons are effectively massless (their masses are negligible compared to the collision energy), and the full SU(2)_L x U(1)_Y symmetry is restored. Weak and electromagnetic interactions become comparable in strength (the apparent weakness of the weak force at low energies is due to the large W and Z masses suppressing the propagator, not an intrinsically smaller coupling). This is directly observed at the LHC, where W and Z production cross sections are large and the electroweak gauge structure is directly probed.
Question 4 Short Answer
Explain why the photon remains massless after electroweak symmetry breaking, while the W and Z acquire masses. What symmetry protects the photon?
Think about your answer, then reveal below.
Model answer: The Higgs doublet breaks SU(2)_L x U(1)_Y to U(1)_EM. The three broken generators (corresponding to W+, W-, and Z) give rise to three Goldstone bosons that are eaten by these gauge bosons, making them massive. The unbroken generator — the electric charge Q = T_3 + Y/2, a specific linear combination of the SU(2) generator T_3 and the hypercharge Y — corresponds to the photon. Because U(1)_EM remains an exact symmetry, the photon has no mass: gauge invariance of the residual U(1) forbids a photon mass term. The Higgs vacuum is electrically neutral (Q|vac> = 0), which is why it does not break electromagnetic gauge invariance.
This is a beautiful example of the general principle: the number of massive gauge bosons equals the number of broken generators. SU(2)_L x U(1)_Y has 4 generators, U(1)_EM has 1, so 3 generators are broken and 3 gauge bosons become massive. The one surviving gauge boson — the photon — remains massless because its symmetry is unbroken.