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My question is in the title. I do not really understand why water is not a superfluid. Maybe I make a mistake but the fact that water is not suprfluid comes from the fact that the elementary excitations have a parabolic dispersion curve but for me the question remain. An equivalent way to ask it is: why superfluid helium is described by Gross-Pitaevsky equation and it is not the case for water?

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  • $\begingroup$ Recent work actually suggests that water may have a superfluid liquid phase $\endgroup$
    – user20145
    Jan 23, 2013 at 15:24
  • $\begingroup$ @x you have to substantiate this claim by a reference or link and a quote, at least from the abstract. $\endgroup$
    – anna v
    Jan 23, 2013 at 15:31

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You refer to the Landau criterion for superfluidity (there is a separate question whether this is really the best way to think about superfluids, and whether the Landau criterion is necessary and/or sufficient). In a superfluid the low energy excitations are phonons, the dispersion relation is linear $E_p\sim c p$, and the critical velocity is non-zero. In water the degrees of freedom are water molecules, the dispersion relation is quadratic, $E_p\sim p^2/(2m)$, and the critical velocity is zero.

The Gross-Pitaevskii equation applies (approximately) to Helium, because in the superfluid phase there is a single particle state which is macroscopically occupied. The GP equation describes the time evolution of the corresponding wave function. In water there are no macroscopically occupied states. You can try to solve the full many-body Schroedinger equation, but at least approximately this problem reduces to cassical kinetic theory.

I think the best criterion for superfluidity is irrotational flow: The non-classical moment of inertia, quantization of circulation, and persistent flow in a ring. Again, these don't appear in water because there is no spontaneous symmetry breaking, and no macroscopically occupied state.

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  • $\begingroup$ So now my question is why there is no macroscopically occupied state for water ans there is one for helium? In general we don't try to solve Schrödinger equation for helium in order to obtain GP equation, isnt'it? And how can I obtain a classical kinetic equation for water starting from Schrödinger ? $\endgroup$
    – PanAkry
    Sep 24, 2012 at 6:56
  • $\begingroup$ A rough criterion is the condition for Bose condensation in an ideal gas, $n\lambda^3\sim 1$, where $n$ is the density and $\lambda$ is the thermal wave length. Note that your question is in some sense backwards: Helium is the exception, water is the rule. Most ordinary fluids solidify instead of becoming superfluid at low $T$. $\endgroup$
    – Thomas
    Sep 24, 2012 at 12:38
  • $\begingroup$ I'm not sure that one should consider ordinary molecules as quasiparticles leading to dissipation because the Landau criterion originates from investigation of a phonon branch. Instead, we have to use the same formalism of phonons to be able to compare liquids' behavior. In non-superconducting liquid a phonon dispersion law is $E=c|p|$. So applying Landau criterion we obtain a peculiar critical velocity equal to $c$ - the speed of sound. It doesn't mean necessarily that the liquid is superconducting below $c$. It rather means that it goes into a different state above $c$ (supersonic flow). $\endgroup$
    – Olexot
    Jul 6, 2020 at 19:27
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Because water is liquid at much too high a temperature. Helium is only superfluid near absolute zero. To have a superfluid, you need the quantum wavelength of the atoms given the environmental decoherence to be longer than the separation between the atoms, so they can coherently come together.

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