Ideal Bose gas and Goldstone modes In an ideal Bose gas there is a symmetry breaking phase transition, namely Bose-Einstein condensation. In a weakly interacting Bose gas or in helium-4 there is a longitudinal phonon because of the symmetry breaking, which leads to a linear dispersion relation for small energies and momenta.
I would expect to have something similar also in the ideal Bose gas case, instead there is no phonon-like dispersion relation. Where is the Goldstone mode corresponding to the symmetry breaking?
From another point of view, it would be strange to have a collective excitation if particles don't interact with each other. Does this have anything to do with the absence of a Goldstone mode? What other hypothesis (that is not satisfied in the ideal Bose gas) is needed in order to have a Goldstone mode derived from a symmetry breaking?
 A: Looking at Chapter 6 of Altland-Simons book, you can read about the non interacting Bose gas and the weakly interacting Bose gas, and I think that the anser might be in those pages.
In my opinion the key point is that the non interacting system is pathological in the sense that there is no solution that minimizes the action, and therefore "expanding the action around the minimum" makes no sense, so the Goldstone modes can't be properly defined. The introduction of a weak interaction regularizes everything, introduces a minimum and makes it possible to expand the action around it.
More details are given below, but I strongly suggest to check the book.
Non interacting pathological case
The action in Fourier space for the field $\psi_{k,n}$ reads
$$
S[\bar{\psi},\psi] = - \beta \bar{\psi}_{0,0} \mu \psi_{0,0} + \sum_{k,n} \bar{\psi}_{k,n} \left(-i\omega_n + \xi_k \right) \psi_{k,n}
$$
where I have separated the $k=0,n=0$ term from the rest of the sum. Here $\omega_n$ are Matsubara frequencies, $\xi_k = \varepsilon_k - \mu$, with $\varepsilon_k = k^2/2m$ the energy corresponding to momentum $k$ and $\mu$ the chemical potential.
The equation for $\psi_{0,0}$ is simply $-\beta \mu \psi_{0,0} = 0$, and it has the trivial solution $\psi_{0,0}=0$ as long as $\mu \neq 0$. As you may know, above the critical temperature we have $\mu<0$, so the solution is trivial. However below the critical temperature we have $\mu=0$ for consistency (see the book) and the equation has no solutions.
We conclude that below the critical temperature, the action is unbounded with respect to the variable $\psi_{0,0}$. In other words there is no minimum around which we can expand, no ground state and no Goldstone mode. How can we determine the value of $\psi_{0,0}$ then? Well it is determined by imposing that the total number of particle is fixed to the value $N$, so $\psi_{0,0}$ is not even a dynamical variable, but is just a parameter of the theory. In other words, since $\mu$ is constrained to $0$ for consistency, you need a new parameter to set the correct number of particles, and this is precisely $\psi_{0,0}$.
The weakly interacting Bose gas as a solution
The picture above is clearly not satisfactory, because we would like to treat $\psi_{0,0}$ as a dynamical variable, find the minimum of $S$ with respect to it and so on. If we introduce a weak interaction of strength $g/V$, with $V$ the volume of the system as follows:
$$
T S[\bar{\psi}_0,\psi_0] = - \bar{\psi}_{0} \mu \psi_0 + \frac{g}{V} |\psi_0|^2,
$$
then the pathology is solved for all $g > 0$, no matter how small it is. Notice the cheaper notation $\psi_0 = \psi_{0,0}$. If you compute the minimum of the action, you find two solutions: $\psi_0 = 0$ and $|\psi_0| = \sqrt{\mu V/g}$. The latter makes sense when $\mu>0$, which is now possible below the critical temperature due to the interaction (again see the book for the details) and it turns out to be the minimum.
