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Assume that we have an electrically neutral interacting gas (or liquid) of Bose or Fermi particles in a superfluid state. For simplicity, assume that the particles interact via an assigned central potential (the fact that the particles may have integer or half-integer spin is irrelevant to the mutual interaction).

If the system is bosonic, then it should have a repulsive interaction between particles, so that it is "stable" and its Landau critical velocity is bigger than zero (the non-interacting Bose gas is not superfluid because its Landau critical velocity is zero). On the contrary, for a Fermi system, the interaction should be attractive, so we can form Cooper pairs that form the condensate.

Question: provided that the Cooper pairing of the Fermi system has total spin $S=0$ (like in standard BCS), which are the qualitative differences between a Fermi and a Bose superfluid close to absolute zero? Can we expect differences in the nature of the quasiparticles and collective modes? Any good reference to some specialized article or book that makes a clear comparison between the Bose and Fermi cases?

"Close to absolute zero" means that $T \ll T_c$, where $T_c$ is the critical temperature for the superfluid transition. I am asking about that hydrodynamic behaviour because I know that the Tisza-Landau model works for $^4$He, but I am not sure if (and how) it can be derived for a Fermi system.

Comments: let me quickly summarize the info in the comments. At very low energy, the elementary excitations for both cases (Fermi and Bose) are Goldstone modes that should be bosonic collective modes ("phonons"). The high-energy part of the spectrum is not universal (e.g., if the interaction in the Bose system is strong, like in liquid $^4$He, then there may be roton minimum). At the moment the Fermi case is less clear to me (both the low-energy and high-energy parts). In fact: if the phonon-like dispersion relation is valid at low-energy also for the Fermi case, where is the famous "gap" in the spectrum?

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    $\begingroup$ For me, the question is not completely clear. As I understand, you would like to consider fermions with attractive/repulsive interaction and find low-energy theory for such fermions. Next, you would like to compare it with case of bosons $\endgroup$
    – Artem Alexandrov
    Commented Jan 11, 2022 at 14:55
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    $\begingroup$ If yes, a rough answer is following: 1) for bosons, you have $\omega\propto p$, 2) for fermions, you have $\omega_p\propto p^2$. In my view, the rigorous approach is to decouple fermions interaction in appropriate channel (Cooper, Coulomb, exchange) and find excitations spectrum of apperaed low-energy theory. It is discussed in Altland & Simons book. "Old" approach is discussed in Landau Course, 9 vol. $\endgroup$
    – Artem Alexandrov
    Commented Jan 11, 2022 at 14:59
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    $\begingroup$ @ArtemAlexandrov Thank you for asking. Yes, I'd like to know which are the qualitative differences in the spectrum and in the elementary excitations present in a Bose superfluid or in a Fermi superfluid (both neutral and close to T=0). For simplicity, I am assuming that in both cases the order parameter is a scalar (if this is relevant). You understood it well: I just want to change the Bose-Fermi character of the particles, to pin the pure effect of statistics on the spectrum and excitations. $\endgroup$
    – Quillo
    Commented Jan 11, 2022 at 15:02
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    $\begingroup$ At very low energy, the elementary excitations for both cases are Goldstone modes. Fermionic excitations become gapped. $\endgroup$
    – Meng Cheng
    Commented Jan 11, 2022 at 16:14
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    $\begingroup$ @Quillo there is a roton quasiparticle, but as I know there is no rigorous theory for rotons in spectrum of Bose system $\endgroup$
    – Artem Alexandrov
    Commented Jan 11, 2022 at 18:50

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