A real scalar field has the potential V(phi) = -mu^2 phi^2/2 + lambda phi^4/4. The Lagrangian is invariant under phi -> -phi (Z_2 symmetry). However, the minima of V are at phi = +/- mu/sqrt(lambda), not at phi = 0. Why is phi = 0 not the vacuum?
AQuantum fluctuations destabilize the phi = 0 state
BThe phi = 0 configuration is a local maximum of V, not a minimum — the system rolls to one of the degenerate minima at +/- mu/sqrt(lambda), and whichever minimum is chosen breaks the Z_2 symmetry
CThe Z_2 symmetry is explicitly broken by higher-order terms
DThe phi = 0 vacuum is forbidden by the uncertainty principle
The potential V(phi) = -mu^2 phi^2/2 + lambda phi^4/4 has V''(0) = -mu^2 < 0, so phi = 0 is a local maximum (unstable). The true minima are at phi_0 = +/- mu/sqrt(lambda), where V'' > 0. The system settles into one of these minima, say phi_0 = +mu/sqrt(lambda). This ground state is NOT invariant under phi -> -phi (which would map phi_0 to -phi_0), so the Z_2 symmetry is spontaneously broken. The Lagrangian retains the full symmetry, but the vacuum does not.
Question 2 Multiple Choice
For a complex scalar field with the Mexican hat potential V = -mu^2 |phi|^2 + lambda |phi|^4, the vacuum manifold is a circle |phi| = v = mu/sqrt(2 lambda). Fluctuations along the circle cost no energy. What do these zero-energy excitations correspond to?
APhotons
BGoldstone bosons — massless scalar particles that are the quantum excitations along the flat direction of the potential, one for each spontaneously broken continuous symmetry generator
CGhost particles required for gauge fixing
DTachyons
The potential is flat along the bottom of the Mexican hat (the circular valley at |phi| = v). Excitations along this flat direction require no energy to excite, so they correspond to massless particles — the Goldstone bosons. For the U(1) symmetry of a complex scalar field, there is one broken generator and one Goldstone boson. In contrast, excitations in the radial direction (perpendicular to the valley) have a restoring force and correspond to a massive particle. This is the content of Goldstone's theorem: each spontaneously broken continuous symmetry generator produces one massless scalar boson.
Question 3 True / False
Spontaneous symmetry breaking means the symmetry is gone from the theory entirely — neither the Lagrangian nor the vacuum state respects it.
TTrue
FFalse
Answer: False
In spontaneous symmetry breaking, the Lagrangian retains the full symmetry — it is the vacuum (ground state) that does not. This is a crucial distinction from explicit symmetry breaking (where a term in the Lagrangian violates the symmetry). Because the Lagrangian is symmetric, Ward identities and other consequences of the symmetry still hold, even though the vacuum picks a preferred direction. The symmetry is 'hidden' rather than 'gone.' The Goldstone theorem, the Higgs mechanism, and the structure of the particle spectrum all follow from the interplay between the symmetric Lagrangian and the asymmetric vacuum.
Question 4 Short Answer
Explain why spontaneous symmetry breaking is necessary for the Standard Model to describe massive particles, and what goes wrong without it.
Think about your answer, then reveal below.
Model answer: The Standard Model is based on the gauge group SU(3) x SU(2) x U(1). Gauge invariance forbids explicit mass terms for the gauge bosons (a term m^2 A_mu A^mu is not gauge invariant) and for fermions with chiral couplings (a Dirac mass term m psi-bar psi mixes left- and right-handed components that transform differently under SU(2)). Without symmetry breaking, all gauge bosons and fermions would be massless. Spontaneous symmetry breaking (via the Higgs mechanism) generates masses while preserving the gauge structure of the Lagrangian, which is necessary for renormalizability. The W and Z bosons acquire mass by 'eating' three Goldstone bosons, and fermions acquire mass through Yukawa couplings to the Higgs field.
This is why the Higgs mechanism is essential, not optional, in the Standard Model. A world without spontaneous symmetry breaking would have massless W and Z bosons (making the weak force long-range), massless electrons (making atoms impossible), and a radically different universe.