Superfluid Helium-4 has a very well studied excitation structure -- at very low momenta, there is a low energy excitation, the phonon, that corresponds to a periodic density fluctuation in the superfluid with well defined wave-number and an energy $E = c \hbar k$ (c being the speed of sound in the superfluid). Though others might quibble with me over vocabulary, I prefer to call phonons "collective excitations" and reserve "quasiparticle" for excitations that correspond to renormalized single-particle excitations (like an electron in a Fermi Liquid).
In either case though, what is meant is simply that the excitation is long-lived, or, by the uncertainty principle, that it has a sharply defined energy, and here the phonon does.
This linear relation breaks down at higher momenta, where the $E,k$ curve turns down, then turns back up. Near the local maximum, there are sharp excitations with an inverse-parabolic dispersion. These are the maxons. Near the local minimum we have a sharp excitation with a parabolic dispersion, and these are the rotons.
Historically rotons were introduced by Landau, with a guess for their dispersion ($E=\Delta+(p-p_0)^2/2m$), not merely as a mathematical device, but because a superfluid described only by the phonon dispersion fails to capture the actual thermodynamics observed in Helium-4. On the other hand, Landau wasn't concerned about a qualitative, microscopic picture of what rotons are, in the sense that we know what phonons mean for the local density of the system.
So, I am actually not too sure what rotons are in that same microscopic sense, and the same for maxons, although it is apparent that a gas of phonons and a gas of rotons are very different things. On the other hand, this is somewhat paradoxical since all three of these excitations are just different parts of the same dispersion curve, so to say they are fundamentally different is a phrase I wouldn't use without a great deal of caution...
Lastly, knowing all of the excitations present in superfluid helium is very important for calculating its thermodynamic properties, etc., however I think (and hopefully someone will correct me if I'm wrong!) the only excitation that is intrinsic to superfluidity is the phonon mode - this is because the superfluid state breaks a continuous gauge symmetry of the normal fluid, and thus phonons are the massless Goldstone modes of the superfluid state.
Both of them are super-interesting to study:-)
Superfluid is a low-temperature state of a quantum many-body system of electrically neutral particles (e.g. atoms). Superfluids have some amazing properties. For example, there is no dissipation (i.e. friction) and the flow is irrotational (up to quantum vortices). Theoretically it is described by a macroscopic wave-function with its absolute value related to the superfluid density and the gradient of its phase defines the superfluid velocity. In addition, in every superfluid there is also a physical Goldstone mode which costs nothing to excite.
Superconductor is like a superfluid, but the elementary particles are electrically charged (e.g. electrons). This leads to the Meissner effect, i.e. it costs energy for the magnetic field to penetrate into a superconductor. There is also no physical soft mode in the energy spectrum. Theoretically all this is explained by the Higgs mechanism.
In both cases the lowest energy quantum state is macroscopically populated which is know as Bose-Einstein condensation.
Best Answer
Superfluidity phase transition occurs when all the constituent atoms of a sample begin to occupy the same quantum state. This happens when the atoms are placed very close together and cooled down so much that their quantum wave functions begin to overlap and the atoms lose their individual identities, behaving more like a single super-atom than an agglomeration of atoms.
When (helium-4) cooled to very low temperatures, a superfluid-ready set of bosons, atoms with an even number of nucleons, forms into a Bose-Einstein condensate, a superfluid phase of matter. When fermions, atoms with an odd number of nucleons such as the helium-3 isotope, are cooled down to a few Kelvin, this is not sufficient to cause this transition.
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