Questions: Goldstone Theorem

Question 1

A theory has a global SU(3) symmetry that is spontaneously broken to SU(2). How many Goldstone bosons appear?

• A. 3
• B. 5
• C. 8
• D. 6

Answer: (B)

SU(3) has 8 generators and SU(2) has 3 generators. The number of broken generators is 8 - 3 = 5, so Goldstone's theorem predicts 5 massless scalar bosons. Each broken generator corresponds to a flat direction in the field space, and excitations along each flat direction produce one Goldstone boson. The unbroken SU(2) generators correspond to massive modes (or to the symmetry that still constrains the spectrum).

Question 2

Pions (pi+, pi-, pi0) are often called 'pseudo-Goldstone bosons' of QCD. They are very light (approximately 140 MeV) but not exactly massless. Why aren't they exactly massless as the Goldstone theorem would predict?

• A. Because the pion is a composite particle, not an elementary scalar
• B. Because the chiral symmetry SU(2)_L x SU(2)_R of QCD is not an exact symmetry — it is explicitly (though softly) broken by the small but nonzero up and down quark masses, making the Goldstone bosons 'pseudo' with small but nonzero masses proportional to sqrt(m_q)
• C. Because confinement modifies the Goldstone theorem
• D. Because pions interact with each other, which generates a mass

Answer: (B)

In the limit of zero quark masses, QCD has an exact SU(2)_L x SU(2)_R chiral symmetry that is spontaneously broken to SU(2)_V (isospin) by the quark condensate <q-bar q> != 0. This produces three exactly massless Goldstone bosons — the pions. The physical quark masses (m_u ~ 2 MeV, m_d ~ 5 MeV) explicitly break chiral symmetry, giving the pions small masses: m_pi^2 proportional to m_q. The pion mass (140 MeV) is much smaller than other hadron masses (proton: 938 MeV) because it is suppressed by the small quark masses. This is the most dramatic physical manifestation of the Goldstone theorem.

Question 3

Goldstone's theorem applies only to global symmetries. When a local (gauge) symmetry is spontaneously broken, the would-be Goldstone bosons are 'eaten' by the gauge bosons.

• T. True
• F. False

Answer: True

This is the Higgs mechanism. When a gauge symmetry is spontaneously broken, the Goldstone bosons do not appear as physical massless particles. Instead, they become the longitudinal polarization components of the gauge bosons, which thereby acquire mass. The gauge boson goes from two polarization states (transverse, massless) to three (two transverse plus one longitudinal, massive). The number of degrees of freedom is conserved: the Goldstone scalar becomes the third polarization of the gauge boson. This is sometimes described as the gauge boson 'eating' the Goldstone boson and getting 'fat' (massive).

Question 4

Prove that the Goldstone boson is massless by considering small fluctuations around the vacuum in a theory with spontaneously broken U(1) symmetry.

Model answer: Write phi = (v + rho(x)) e^{i theta(x)/v}, where v is the vacuum expectation value, rho is the radial fluctuation, and theta is the angular fluctuation. The potential V depends only on |phi|^2 = (v + rho)^2, so it is independent of theta entirely. Since V has no theta-dependence, the mass term for theta (which would be proportional to d^2V/d theta^2 at the vacuum) is exactly zero. The kinetic term for theta survives: |partial_mu phi|^2 contains (partial_mu theta)^2/2, which is the kinetic term of a massless scalar field. Therefore theta is a massless scalar — the Goldstone boson. The field rho has mass m_rho^2 = d^2V/d rho^2 |_{rho=0} = 2 mu^2 > 0.

This argument generalizes to any symmetry group: for each broken generator, there is a direction in field space along which the potential is flat (no restoring force), producing a massless excitation. The number of flat directions equals the number of broken generators. This is the intuitive content of Goldstone's theorem.