# Is gapless spectrum necessary for superfluity

In 3D interacting Boson system, the condensate leads to a gapless spectrum and superfluidity. The same gapless spectrum for 2D with quasi-condensate (correct me if it is wrong ).

I am wondering if it is a necessary condition? I remember the correct way to define superfluidity is to show superfluid density non-zero. Is gapless spectrum necessary to get non-zero superfluid density?

Landau's formula of superfluid density gives the normal density is $\rho_n=-\int \frac{d^dp}{(2\pi)^d}\frac{p^2}{d}\frac{\partial}{\partial\epsilon_p}\frac{1}{e^{\beta\epsilon_p}-1}$ ,then the superfluid density is $\rho_s=\rho-\rho_n$.

At BEC, using BDG quasi-particle spectrum, one could calculate superfluid density by this formula. Is it possible to show $\rho_s=0$ using this formula with a general gapped spectrum (not necessary parabolic) in normal state above BEC?

• What non BEC superfluid are you interested in,I’m having trouble understanding what you’re asking. – Shane P Kelly Nov 12 '17 at 20:45
• @Shane How can I get the superfluid change from zero to non-zero while phase changing from non-BEC to BEC using this formula. Without going to the detail of the spectrum, the difference is spectrum changing from gapped to gapless. – p.s Nov 12 '17 at 20:56
• I’m Having trouble understanding your grammar. It sounds like it should depend on what’s causing the gap. I assume your bosons are conserved? If they are and are non interacting, then that formula should work. – Shane P Kelly Nov 12 '17 at 23:08
• @Shane I am talking about interacting bosons, I saw a proof on Stoof's "ultracold quantum fields" showing parabolic spectrum gives $ρ_n$=$ρ_{non-condensate}$, so superfluid density vanish without condensate. – p.s Nov 13 '17 at 6:45