An ordered phase breaks a discrete Z₂ symmetry — like an Ising ferromagnet where spins can only point up or down. According to Goldstone's theorem, what low-energy excitations does this phase produce?
ATwo Goldstone modes, one associated with each of the two broken ground states
BOne gapless Goldstone mode, because one symmetry is broken
CNo Goldstone modes, because Goldstone's theorem applies only to broken continuous symmetries
DA gapless mode at k=0 and a gapped mode at large k
Goldstone's theorem specifically requires the broken symmetry to be continuous. For a discrete symmetry like Z₂ (up or down), there is no way to continuously interpolate between the two ground states — you cannot make a long-wavelength 'twist' from up to down. The theorem's logic requires continuously varying the order parameter spatially, generating low-energy deformations; discrete symmetries don't permit this. An Ising ferromagnet has an energy gap to excitations — you must flip spins by a discrete step, which costs a finite energy. This is one of the clearest distinctions between Ising (discrete, gapped) and Heisenberg (continuous, gapless) magnets.
Question 2 Multiple Choice
In a Heisenberg ferromagnet below its Curie temperature, the spins are aligned along the z-axis, breaking the continuous rotational symmetry. Which statement correctly describes magnon excitations?
AMagnons have a minimum energy gap, similar to the energy gap in a semiconductor, required to create any spin excitation
BMagnons are long-wavelength spin deformations (spin waves) whose energy vanishes as the wavevector k → 0, making them gapless Goldstone modes
CMagnons are localized spin flips whose energy is set by the exchange coupling constant between nearest neighbors
DNo low-energy excitations exist in the ordered phase because the ground state is stable
A magnon (spin wave) is a smooth, spatially varying rotation of the magnetization direction — a long-wavelength 'twist' of the order parameter. The key property is that the energy of a magnon vanishes as its wavelength goes to infinity (k → 0): it costs nothing to uniformly rotate all spins together (which just moves to another equally valid ground state). These gapless excitations are the Goldstone modes of the broken rotational symmetry. Their dispersion ω ∝ k² (for ferromagnets) or ω ∝ k (for antiferromagnets) goes to zero at k = 0, confirming the gapless character.
Question 3 True / False
Phonons in a crystal lattice are Goldstone modes arising from the breaking of continuous translational symmetry by the periodic arrangement of atoms.
TTrue
FFalse
Answer: True
A perfect crystal breaks continuous translational symmetry — atoms settle into specific lattice positions that are not invariant under arbitrary spatial translations. Long-wavelength sound waves (acoustic phonons) are the corresponding Goldstone modes: coherent slow displacements of the lattice in which atoms slide smoothly from position to position over a long length scale. Their dispersion ω = vk vanishes as k → 0, confirming they are gapless. This is a general result: any time continuous translational symmetry is broken by an ordered structure, acoustic phonons are the Goldstone modes.
Question 4 True / False
The energy of a Goldstone mode is independent of wavelength — it has a flat dispersion relation, meaning most wavelengths require the same energy to excite.
TTrue
FFalse
Answer: False
Goldstone modes have precisely the opposite dispersion: their energy vanishes as wavelength increases (k → 0). This is what 'gapless' means — the energy cost goes to zero in the long-wavelength limit. Formally, ω → 0 as k → 0, not a constant. Short-wavelength Goldstone modes do require more energy (twisting the order parameter over a short distance costs more than a slow twist over a long distance), but the long-wavelength modes are essentially free to excite. A flat dispersion would describe a gapped mode at finite energy, which is the opposite of a Goldstone mode.
Question 5 Short Answer
Explain why an Ising ferromagnet (discrete Z₂ symmetry) has no Goldstone modes while a Heisenberg ferromagnet (continuous SO(3) symmetry) does. What is the physical reason for the difference?
Think about your answer, then reveal below.
Model answer: In the Heisenberg ferromagnet, continuous rotational symmetry is broken: you can smoothly interpolate between ground states by slowly rotating the magnetization direction. A long-wavelength twist of the order parameter costs energy proportional to k², which goes to zero as k → 0 — producing gapless magnon modes. In the Ising ferromagnet, the broken symmetry is discrete: spins are either up or down, with no continuous interpolation between them. You cannot create a smooth long-wavelength deformation; to change a spin you must flip it discretely, which always costs a finite energy. No gapless modes result.
The physical essence of Goldstone's theorem is the ability to 'slide' between degenerate ground states continuously, generating long-wavelength, low-cost deformations of the order parameter. Continuous symmetry makes this possible; discrete symmetry does not. This is why the theorem's statement specifies 'continuous symmetry' — it is not a technical restriction but the physical heart of the argument. The counting rule — one Goldstone mode per broken continuous symmetry generator — follows directly from this logic: each independent direction in which you can continuously deform the order parameter produces one gapless mode.