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.