I'm familiar with the two standard proofs of Goldstone theorem, namely the 1PI effective action approach and the current-algebra-ish operator proof. Then I was trying to view Goldstone modes from a wavefunctional perspective but got myself quite confused.
Consider the Mexican hat potential with a complex scalar ϕ. If one writes in polar coordinates ϕ=ρeiθ, then the quadratic term of the Lagrangian has the form ∼(∂μθ)2, and people usually at this point conclude there's a massless mode corresponding to angular excitation. However, if we go one step further and make explicitly what these excitations are, there seems to be trouble.
In the wavefunctional picture, a state is a functional Ψ[θ(p)] if we work in momentum space, and for free fields an energy eigenstate a solution to the collection of infinite number (labeled by p) of harmonic oscillator Schroedinger equations
(−d2dθ2p+ω2(p)θ(p)2)Ψ[θ(p)]=EpΨ[θ(p)]⋯(1)
A occupation number state |np1,np2,…⟩ (npi particles with momentum pi) is represented by the functional
Ψ{npi}[θ(p)]=∏p≠piψ0(θ(p))∏piψnpi(θ(pi))⋯(2),
where ψn is the nth excited state of a harmonic oscillator.
The above is true for all quadratic theories, and if there were no restrictions on θ, (∂μθ)2 type of action indeed gives gapless excitations for p≈0. However there is a restriction on θ, that is θ∼θ+2π, so we are really dealing with oscillators on a circle, which means there is a gap due to periodicity no matter how small p is. Why do we still have Goldstones then?