How to explain BEC of noninteracting boson in 2nd quantization? How to spontaneously break $U(1)$ symmetry of free boson? Every textbook of statistical mechanics will derive the Bose-Einstein condensation(BEC) of free boson. I know how to derive in this way. It gives the heat capacity of free boson is $C\propto T^{3/2}$. But for $He^4$ that is interacting boson we know $C\propto T^3$. So to explain this, every condensed matter textbook will use 2nd quantization to rewrite the orginal many-body hamitonian in field formalism. For interacting boson, the Hamitonian of field is 
$$H=\int d^3 x  \frac{1}{2m} |\nabla\phi|^2 + U |\phi(x)|^2 + g |\phi(x)|^4$$
For Helium 4, $U<0$ and $g>0$, so field configuration of minimum energy is $\phi = const.\neq0$, thus there is a spontaneous symmetry breaking of $U(1)$ , and  due to Goldstone theorem it gives a massless excitation which explains the $C\propto T^3$.
It's all reasonable above. However how to use 2nd quantization back to explain BEC of noninteracting boson?
My questions:


*

*For free boson, the field Hamitonian is now 
$$H=\int d^3 x  \frac{1}{2m} |\nabla\phi|^2  $$
The saddle point of field configuration is now $\phi(x)=const.$ for $\forall const. \in \mathbb{C}$, not necessary to be nonzero. How to explain the macroscopic occupation number of ground state?

*How to use  Hamitonian of field to explain the BEC of noninteracting boson in potential well, like harmonic potential well? In this case,
$$H=\int d^3 x  \frac{1}{2m} |\nabla\phi|^2 + \frac{1}{2}  x^2 |\phi(x)|^2$$
Certainly the minimum energy configuration is $\phi(x)=0$.

*Certainly the above two cases must be able to have BEC. Then for free boson and noninteracting boson in potential well, does BEC spontaneously breaking the $U(1)$ symmetry? If yes, there must exist massless excitation due to Goldstone theorem, so why the heat capacity of free boson is not propotional to $T^3$? If no, it will contradict with Landau's paradigm of phase transition that SSB results in the 2nd order phase transition. How to explain? Related to my another question.
 A: As was stated in the comments, the fact that $\phi = 0$ minimizes the classical action is not sufficient to establish absence of SSB. In fact, even calculating the expectation value $\langle \phi(x) \rangle$ is not sufficient; in fact, this expectation value is always zero when calculated with respect to the grand-canonical ensemble (it's a fun exercise to prove this).
Rather, the correct way to diagnose spontaneous symmetry-breaking is through the long-rangedness of the two-point correlation
$$ \lim_{|x - y| \to \infty} \langle \phi^{\dagger}(x) \phi(y) \rangle$$
When there is SSB, this goes to a nonzero value, otherwise it goes to zero. In fact, by expressing $\phi(x)$ as the Fourier transform of the fermion annihilation operator $a_k$ in momentum space, one finds that this limit is actually equal to the macroscopic occupation of the $k=0$ state, which is non-zero even for free bosons.
As for the question of the Goldstone bosons, a non-interacting BEC does have gapless excitations; you just add one more boson in a state other than the $k=0$ state. But since they are non-relativistic particles, the dispersion relation is $E \sim k^2$. This correctly gives a $\sim T^{3/2}$ specific heat. To get a $\sim T^3$ specific heat you would need a linear dispersion relation $E \sim |k|$, which is what happens in the interacting case. In non-relativistic systems, Goldstone's theorem doesn't necessarily require that the Goldstone bosons have linear dispersion; another example of a case where you get a quadratic dispersion is the spin-wave excitations of ferromagnets.
