I feel like the answer should be "no" since all superfluids are not strictly BEC since they can undergo a Kosterlitz–Thouless transition in 2D, for example. I believe the ideal gas isn't superfluid, but is there any experimental evidence of a BEC without superfluid properties? I've been searching with no luck.
Bose-Einstein Condensate – Are All Bose-Einstein Condensates Superfluid?
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1) Some of the assumptions of the Gross-Pitaevskii equation (GPE) are:
- all atoms are in the same condensate wave function,
- the condensate is at $T=0$,
- collisions between atoms are sufficiently low energy that the interactions can be well described by the $s$-wave scattering length, so that the interaction can be written $g\delta(\mathbf{x}_i-\mathbf{x}_j)$.
Generalized GPEs can also be solved, allowing for thermal and quantum depletion (some atoms not in the condensate) and allowing for other forms of interaction, such as dipolar.
2) The interaction term, $g|\Psi(\mathbf{x})|^2$, is in addition to the external potential $V_\mathrm{ext}(\mathbf{x})$, the effective potential is the sum of both: $V_\mathrm{ext}(\mathbf{x})+g|\Psi(\mathbf{x})|^2$. The condensate density is $n_0(\mathbf{x})=|\Psi(\mathbf{x})|^2$, so the interaction term is $gn_0(\mathbf{x})$ which is the potential due to interaction with the condensate itself.
More detail in response to the OP's comment:
The interaction potential between two atoms can usually be written as $V(\mathbf{r}_{ij})$ where $\mathbf{r}_{ij} =\mathbf{x}_i-\mathbf{x}_j$. For neutral atoms without a significant magnetic dipole moment, the dominant interaction is van der Waals so $V(\mathbf{r}_{ij})\propto r_{ij}^{-6}$.
When considering the scattering between two atoms, we can do a partial wave expansion (matching incoming and outgoing wave functions and expanding in terms of Legendre polynomials, e.g. "Quantum Mechanics", Ch. 17, Landau and Lifshitz). For slow particles with van der Waals interaction, the $s$-wave term is dominant and the interaction can be simplified to $V(\mathbf{r}_{ij}) = g \delta(\mathbf{r}_{ij})$ where $g=4\pi\hbar^2 a_s/m$ and $a_s$ is the $s$-wave scattering length. To get a feel for the scattering length, in the the $s$-wave approximation, the cross section is $\sigma=4\pi a_s^2$, so $a_s$ is a length scale for the interaction.
The interaction potential in the GPE can be written $$\int d\mathbf{x'} V(\mathbf{x}'-\mathbf{x})|\Psi(\mathbf{x'})|^2$$ When $V(\mathbf{x}'-\mathbf{x})=g\delta(\mathbf{x}'-\mathbf{x})$, this simplifies to $$\int d\mathbf{x'} g\delta(\mathbf{x}'-\mathbf{x})|\Psi(\mathbf{x'})|^2 = g|\Psi(\mathbf{x})|^2$$
3) The external potential $V_\mathrm{ext}(\mathbf{x})$ is generally due to applied optical or magnetic fields, and is often approximately a harmonic oscillator. The oscillator strength may be very strong in some directions creating quasi one or two dimensional confinement. A particle in a box is not possible yet (the atoms would interact with the "walls"), but the external potential may be locally approximately uniform near the center of the trap. Lattice potentials are also common, where (in addition to harmonic confinement) the atoms are trapped in a standing wave created by counterpropogating lasers resulting in a periodic potential. Many other shapes are possible, such as toroids.
A good reference is the book "Bose-Einstein Condensation in Dilute Gases" by Pethick and Smith. This slightly dated review is also good (free arXiv version here): section III is relevant to your question 2.
Yes, Bose-Einstein condensates are affected by gravity. Most condensates are formed in laser traps and often (especially in the early experiments) the lasers must be turned off to get a good image of the condensate, with the consequence that many images of condensates (again, especially from the early experiments) show them falling.
An example (source):
Best Answer
You can have superfluids that are not BECs and BECs that are not superfluid. Let me quote a text, "Bose-Einstein Condensation in Dilute Gases", Pethick & Smith, 2nd edition (2008), chapter 10: