Absence of Spontaneous U(1) Symmetry Breaking in Finite Superfluid Bose Gases
Whether a phonon in a superfluid Bose gas is a Goldstone boson depends on applying the strict definition of spontaneous symmetry breaking to the ground state rather than to the order parameter. For a finite system of weakly interacting spinless bosons the U(1) rotation returns the ground state up to a phase and the field-operator expectation vanishes, so no symmetry is broken and the phonon is an ordinary quantized collective mode, not a Goldstone boson. Apparent degeneracy in the infinite-gas limit arises from uncertainty in the particle number, not from the U(1) invariance of the Hamiltonian.
Is a phonon excitation of a superfluid Bose gas a Goldstone boson? Maksim Tomchenko
Addresses whether phonons in a superfluid Bose gas are Goldstone bosons. Using the strict definition of spontaneous symmetry breaking -- the ground state, not merely the order parameter, must fail to…