I am trying to understand the difference in the Bogoliubov spectrum for superconductors and superfluids. In the context of superconductivity, the spectrum of the low-energy excitations is given by $$E_{\mathbf{k}}=\sqrt{\varepsilon^2_{\mathbf{k}}+\Delta^2_{\mathbf{k}}}, $$ where $\varepsilon_{\mathbf{k}}$ is the dispersion of the single-particle model and $\Delta_{\mathbf{k}}$ is the gap. If we are in the superconducting phase, then a gap will open up, and we see that $E_\mathbf{k}$ will also become gapped.
Now let us think about the Bose superfluid case. Here, the Bogoliubov spectrum is given by $$E_\mathbf{k} = \sqrt{\varepsilon^2_{\mathbf{k}}+2Un_0\varepsilon_{\mathbf{k}}}, $$ where $U$ is the on-site interaction and $n_0$ is the density. In the latter case, we see that the spectrum is gapless (assuming we set $\varepsilon_{\mathbf{k}=0}=0$).
Here is my question: how can one understand this on an intuitive level? I suspect one must invoke particle statistics. The explanation might also have something to do with the vacuum state: in the fermionic case, we are creating excitations on top of the Fermi surface of the metal with $\varepsilon_{\mathbf{k}}=0$.
