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 $\phi$. If one writes in polar coordinates $\phi = \rho e^{i\theta}$, then the quadratic term of the Lagrangian has the form $\sim (\partial_\mu \theta)^2$, 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 $\Psi [\theta(p)]$ 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

$$\Big(-\frac{d^2}{d\theta^2_p}+\omega^2(p) \theta(p)^2\Big)\Psi [\theta(p)] =E_p\Psi [\theta(p)] \cdots (1)$$

A occupation number state $|n_{p_1}, n_{p_2},\ldots\rangle$ ($n_{p_i}$ particles with momentum $p_i$) is represented by the functional

$$\Psi_{\{n_{p_i}\}} [\theta(p)] = \prod_{p\neq p_i}\psi_0(\theta(p))\prod_{p_i}\psi_{n_{p_i}}(\theta(p_i)) \cdots(2),$$

where $\psi_n$ is the nth excited state of a harmonic oscillator.

The above is true for all quadratic theories, and if there were no restrictions on $\theta$, $(\partial_\mu \theta)^2$ type of action indeed gives gapless excitations for $p\approx0$. However there is a restriction on $\theta$, that is $\theta\sim\theta+2\pi$, 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?