What are phonons, rotons, and maxons, and what does their dispersion curve have to do with superfluidity? I understand that they are quasiparticles, but I'm not entirely sure what that implies. Are these concepts purely notational and/or mathematical? Or are these three quasiparticles fundamentally different from each other?
[Physics] Phonons, rotons, and maxons
quasiparticlesspectroscopysuperfluidity
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Your confusion confused me at first too, but I believe I can help with the main point. Your mistake comes in the phase "picture of a Cooper pair being broken into two quasiparticles." A Cooper pair does not break into two quasiparticles.
Just so we're on the same page, the BCS ground state is:
$$\prod_k \left(u_k+v_ka^\dagger_{k \uparrow}a^\dagger_{-k \downarrow}\right)|vac\rangle $$
In other words, at each value of $k$ we have a superposition of a Cooper pair (with amplitude $v_k$), and no Cooper pair (with amplitude $u_k$).
Making an excitation of two Bogoliubov quasiparticles looks like:
$\left(u_k+v_ka^\dagger_{k \uparrow}a^\dagger_{-k \downarrow}\right) \rightarrow \left(v_k^*-u_ka^\dagger_{k \uparrow}a^\dagger_{-k \downarrow}\right)$
(note that $u_k$ is real).
This is certainly not breaking a Cooper pair. Right at the Fermi surface* it is not changing any particle/hole occupations at all, since the magnitudes of $u_k$ and $v_k$ are the same. So that language seems quite sloppy to me. For an excitation that is deeper in the Fermi sea it does approach destroying a pair. This is okay because we are working in the grand canonical ensemble, and number is not conserved.
I'm not familiar enough with the experimental setup in the paper you link to for detailed comment, but in general the operator for tunnelling between a normal metal and a superconductor, which is what these conductance measurements really probe, has terms like: $$Tu_k(\alpha_k^\dagger)_1(a_{q})_2=Tu_k(u_ka^\dagger_k-v^*_ka_{-k})_1(a_{q})_2$$, which describes creation of a quasiparticle in superconductor 1 and destruction of an electron in normal metal 2. The point is that even when $|u_k|^2=|v_k|^2$ this term need not be zero. It doesn't conserve electron number, but again that is fine.
General conclusions are that: 1. Bogoliubov quasiparticles, while helpful in analysis, are tricky to get an intuitive sense for, and 2. It is important to remember that most discussions of superconductivity are implicitly assuming no number conservation.
*notice that what you call the "band edge" I am calling the "(gapped) Fermi surface," because I think is more conventional to reserve that term for the non-interacting band.
All these modes are oscillations in the conserved densities (particle number, energy, momentum, etc.) of an interacting many-body system in approximate thermal equilibrium.
Consider first ordinary (first) sound. An ordinary fluid has five conserved densities, mass (particle number), energy, and momentum. The corresponding hydrodynamic theory is then describes five modes of excitation. Three of these are diffusive (non-propagating): heat, and the two transverse components of the momentum density (shear modes). The longitudinal component of the momentum density couples to mass and energy density forms a pair ($\omega=\pm k$) of sound modes that propagate with velocity $c_s^2=(\partial P)/(\partial\rho)|_{s/n}$. This sound modes is weakly damped, but the damping grows at the mean free path increases (typically, as the temperature is lowered).
At low temperature most fluids solidify, but some substances (most notably $^4He$, a boson, and $^3He$, a fermion) remain liquid and become quantum fluids. A Bose fluid (like $^4He$) eventually becomes superfluid, and the sound mode of the normal fluid continues into the superfluid, where we can eventually understand it as a quantized excitation of the superfluid (a phonon). In a Fermi fluid (like $^3He$) ordinary sound becomes strongly damped, but there is a particle-hole mode (which we can understand as an oscillation of the Fermi surface) that has the quantum numbers of ordinary sound, and is known as zero sound. The transition is smooth, but it is apparanant from the behavior of sound velocity and attenuation.
The image below is for liquid helium three, and shows a transition from zero to first sound at approximately 10 mK.
In a superfluid there is another hydrodynamic mode, associated with the Goldstone mode $\varphi$ related to spontaneous $U(1)$ breaking. The superfluid velocity is the gradient of this field, $\vec{v}_s=\hbar\vec\nabla\varphi/m$. Excitations of this mode mix with ordinary sound, and diagonalizing the system of equations leads to two propagating modes, known as first and second sound. The first sound mode is the one that connect to ordinary sound at the critical temperature $T_c$, whereas the speed of second sound goes to zero.
In liquid helium ordinary sound is to good approximation a density wave, whereas second sound is an oscillation of the normal fluid against the superfluid in which the density is approximately constant, but the entropy density oscillates. As a result, I can excite first sound with a vibrating plate, and second sound with a pulsed heater (and then check that the velocities behave as expected).
Below is an image of the two modes in an ultracold atomic gas (top panel: first sound; lower panel: second sound). We observe that second sound is slower, and that it cannot enter the normal fluid regime (dashed line).
Finally, third and fourth sound only exist in special geometries (thin films and channels).
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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.