Superfluidity. Krzysztof Myśliwy. October 30, Theoretical Physics Proseminar

Transcription

1 Superfluidity Krzysztof Myśliwy Theoretical Physics Proseminar October 30, 2017

2 Outline The λ transition Phenomenology of He-II Landau theory- a semi-phenomenological approach Feynman s explanation- from first principles What next?

3 A bait for the press about liquid helium There exists a substance which, under some specific thermodynamic conditions, has the following properties: It remains liquid even at very low temperatures: its solidification requires extremely high pressures, even near absolute zero. On the other hand, it does not boil, even if the pressure is practically zero. It behaves as if it violated the second law of thermodynamics (details to follow). This description refers to a specific phase of an isotope of helium, 4 He, the so-called Helium-II. To be more precise, let us take a look at the phase diagram of He in the p T space.

4 Phase diagrams

5 The λ point Figure 1 : Left: liquid 4 He in T > T λ, right: the same probe below the λ point. The cooling was performed along the (He-I) gas coexistence line. You can see clearly that the boiling stopped.

6 Why λ? Figure 2 : Note the T 3 dependence for small temperatures, which will be of importance later on.

7 Let s face some basic facts The most remarkable property of He-II is its ability to flow through extremely thin capillaries with no resistance. Any measurement of its viscosity by observing such linear flow yields values well below the experimental error- one can conclude that the viscosity of He-II is zero. Indeed, if stirred, He-II will move on for days. If placed into an open beaker, He-II will creep up the walls of the vessel and simply escape!

8 Left: above T λ, right: below T λ. The bottom of the vessel is made up of a membrane, whose pores are about a micron in diameter. He-II can leak through even finer pores easily.

9 The fountain and mechanocaloric effects If we connect two vessels of He-II with an extremely thin pipe and slightly increase the pressure in one of the vessels, we will observe that as He-II flows to the other vessel, the first will heat up while the one into which the fluid inflows cools. Conversely, if we will increase the temperature in one of the vessels, an inflow of He-II into that vessel begins until the level of the fluid, and so the pressure, reaches a certain value. Note that here we have a spontaneous flow of matter from a colder body to a hotter body while work is performed on the environment, what apparently violates the second law of thermodynamics in Kelvin s statement.

10 Viscosity and dissipation in kinetic theory If we dare to explain superfluidity, we must ensure ourselves that we are comfortable with how viscosity relates to statistical mechanics:

11 Thermally inactive fluid Just as viscosity follows from dissipative transport of momentum, head conductivity follows from dissipative transport of energy. The fact that He-II does not boil can be explained by assuming its heat conductivity to be practically infinite. We have thus enough hints to claim that superfluid helium behaves just as if no thermal degree of freedom were available in the system.

12 Not viscous? Surprisingly, if we introduce a rotating cylinder into He-II, we will notice that its angular velocity will diminish, just as if friction were present, which suggests that He-II may be viscous.

13 Two-fluid theory The most plausible phenomenological explanation for all the remarkable and apparently contradictory behaviours of He-II is to assume that it consists of two components: the normal and the superfluid. The normal component has the same properties as a normal, viscous fluid. The superfluid component has zero viscosity and has zero specific entropy. There is no momentum exchange between these fluids- they flow frictionless into one another. The total flux of He-II is hence given by j = ρ s v s + ρ n v n The ratio ρ s /ρ n is a function of temperature and is identically zero above the λ point.

14 BEC? Perhaps the superfluid component consists of some sort of a Bose-Einstein condensate. It qualitatively fits the scheme of a thermally nonactive substance. For an ideal Bose gas with the same density as the density of liquid 4 He near the λ point, the condensation temperature is equal to T c = 3.14K This would also explain the appearance of a phase transition. However, there are dissimilarities-ρ s /ρ n (T ) vs. n 0 N (T ) and the specific heat curve.

15 Quantum image of dissipation-excitations The normal component seems to be able to dissipate energy thermally. We can easily visualize how dissipation occurs classicaly as we readily imagine random collisions and interparticle interactions. The quantum picture is somewhat less intuitive. All we know for sure is that liquid helium is essentially an interacting systems of Bose particles. Its hamiltonian may be at least written down H = 2 2m i + v(r i r j ) i i<j

16 Quasi-particles Solving the Schrödinger equation for the Hamiltonian from the previous slide explicitly is a hopeless task. However, in certain cases one can prove that the spectrum in consideration is equal to that of a gas of non-interacting free quasi-particles. These one-particle states are thus described by the particle momenta p, but their energy is a general function of p: ɛ p = e 0 + a 1 p + a 2 p Examples: lattice vibrations in solids, Bogoliubov transformations...

17 Landau s theory Landau in his late twenties, about five years before the experimental discovery of superfluidity by his Soviet fellow Piotr Kapica, and ten years before his theoretical explanation.

18 Phonons and rotons Basing on the fact that the specific heat of He-II behaves as T 3 for T 0, Landau (1941) postulated that at low temperatures, the dominating form of excitations are phonons, which have a linear dispersion relation: ɛ = cp. Indeed, for the ideal phonon gas we have U = lim ɛ p < n p >= p d 3 V cp p (2π ) 3 1 e βcp = V π2 T c 3 and C v = ( U T ) V = 2π2 V 15 3 c 3 T 3

19 Rotons The spectrum of excitations as proposed by Landau, later established experimentally by neutron scattering experiments.

20 Rotons, cont. The parabolic-like fragment of the spectrum curve corresponds to the so-called rotons -excitations with a nonrelativistic-particle-like dispersion relation ɛ k = + 2 (k k 0 ) 2 2σ which were introduced by Landau in order to explain the experimentally established second sound velocity as a function of temperature.

21 Critical velocity Here comes the Landau argument for superfluidity. Suppose we observe a fluid (in general, a viscous one) moving through a capillary with velocity v. Let us change the reference frame to the one in which the capillary moves with velocity v, and the fluid is initially at rest. The fluid will eventually start to move because of viscosity. It cannot be moved all at once, but first single elementary excitations (e.e.) will gradually emerge. Let ɛ p and p be the energy and momentum of an e.e.- they are equal to the energy E 0 and momentum P 0, respectively, of the fluid in its frame. We return to the capillary frame by performing the Galilean transformation and obtaining E = E 0 + P 0 v + Mv 2 2 P = P 0 + M v By assumption E = ɛ p + p v + Mv 2 2.

22 We see that the excitation causes the change of the fluid energy by ɛ + p v. It is clear that this change must be negative, for the energy dissipates; it is maximally negative if p and v are antiparallel so v > ɛ p This means that if the velocity flow is sufficiently small and the spectrum curve is linear for small p, the excitations will not arise. This observation is in agreement with experiment. One sees easily geometrically that the Landau spectrum allows for superfluidity to occur. On the contrary, the spectrum of an ideal Bose gas does not. This is why interparticle interactions are indispendable in superfluidity.

23 Feynmans s approach Richard Feynman devised the existence of phonons and rotons from first principles- that is, from second-year quantum mechanics.

24 Feynman curve

25 Phonons and rotons in Feynman s theory An utterly ingenious explanation of phonons and rotons in Feynman s theory, from Huang, Statistical Mechanics.

26 Towards a new physics There is a fundamental discrepancy: the observed critical velocities are well below those predicted. This has led to the introduction of the so-called quantum vortices, which have since been observed in ultracold Bose-Einstein condensates. But before that, this has led to the outbreak of new physical ideas in condensed matter physics: quantum vortices, two-dimensional superfluidity, Biereziński-Thouless-Kosterlitz transition (NOBEL 2016)...

27 For further self-study Superfluid hydrodynamics- a fascinating field of its own Bogoliubov theory of a weakly interacting Bose gas BEC-superfluid correspondence Theory of λ transition-still a mystery Quantum vortices

28 References Landau Lifshitz, Statistical Physics, vol. 2, and Hydrodynamics Feynman R.P., Statistical Mechanics: A Set of Lectures Huang K., Statistical Mechanics

29 Appendices

30 Thermocaloric effect in two-fluid theory If the superfluid component carries zero entropy and hence no heat, then if it leaks of the beaker through the capillary, the heat present in the beaker is distributed among a smaller number of particles, which results in a rise in the temperature [c v > 0].

31 The fountain effect If the superfluid flow carries no entropy, the total entropy of both vessels remains constant. Mechanical equilibirium is established when the total energy is a minimum (or the total entropy is a maximum). The process happens at constant entropies and the particle distribution among the vessels is the sole quantity that changes, hence ( ɛ 1 N ) s 1 = ( ɛ 2 N ) s 2 µ(p 1, T 1 ) = µ(p 2, T 2 ) By virtue of the Gibbs-Duhem relation we have ( µ p ) T = 1 ρ and ( µ T ) p = s, so by expanding the equilibrium condition in small pressure and temperature differences we arrive at the equation p = ρs T which explains both the mechanocaloric and fountain effects.

32 A curiosity: Why does He remain a liquid all the way down to absolute zero? This may be explained quite easily. Using quantum-chemical methods, one is able to calculate the interparticle potential of two interacting helium atoms. The depth of the potential well is about 9 K. However, the zero-point oscillation energy 1/m. Helium is thus light enough to have the zero-point oscillation energy greater than the depth of the interparticle potential well- hence, no localised bound state can be formed. Other molecules and atoms are heavier and/or interact more strongly, giving rise to a deeper potential well. This reasoning applies only to a sufficiently dilute system, where only two-body interactions may be taken into account. Thus at higher pressures the solidifiaction will eventually occur.

33 Phonon-roton gas as the normal component If the gas of quasi-particles moves within the fluid with velocity v, we can again change the frame to the frame of the gas. In this frame the energy equals E = E 0 P 0 v + Mv 2 2. If another excitation of energy ɛ(p)(in the original frame) arises, the momentum changes into P 0 + p, so the energy is now equal ɛ p v. The distribution function in the gas frame is thus equal n(ɛ p p v), where n(x) is the Bose distribution function. Thus the total momentum of the gas P = pn(ɛ p p v) d 3 p (2π ) 3 v 3 ( dn d 3 p dɛ )p2 (2π ) 3 This allows us to define the normal fluid density, since P = j = ρ n v. This is how Landau theory supports the two-fluid model.

34 It thus suffices to find ɛ(p), compute n(ɛ pv) and find the normal fluid density ρ n = 1 3 ( dn dɛ )p2 d3 p.it gives a good (2π ) 3 estimate for ρ n (T ) for small temperatures. Close to the λ point, the theory breaks down due to the fact that the non-interacting quasi-particle model becomes meaningless.

35 Ground-state wave function First of all, the ground state wave function φ has no nodes and is unique, so that it can be chosen to be real and positive. For a system of bosons, it must be symmetric with respect to permutations of the atomic coordinates. The energy is a minimum when the gradients of φ are small, and when the value of φ is small if the value of the interaction potential energy is large. Hence, we conclude that φ reaches its maximum for these confugurations where the atoms are evenly distributed and do not overlap.

36 Excited states The first-excited state ψ must be orthogonal to the ground state. Since φ is always positive, this means that ψ is positive for half of the configurations and is negative for the other half for dγψ φ, configurations the scalar product, to vanish. The more nodes ψ possesses, the more kinetic energy it has. So ψ corresponds to a density wave- or indeed a sound waveof a very long wavelength (small k). But this is indeed a phonon-like excitation. Note that this argument fails when the system is fermionic- a permutation of atoms involves a change in sign, which causes that ψ would change its sign more frequently than in a simple sound wave.

37 Equations So Feynman postulates a trial excited state wave function ψ = i f (r i )φ being some sort of a modulation of the ground state. By applying the variational principle ψ = min ψ (ψ Hψ) (ψ ψ) one arrives at ψ = ( i e i k r i )φ and ɛ k = 2 k 2 2mS(k).

38 Feynman curve Here S(k) = e i k r g( r)d 3 r, where g( r) is the two-point correlation function- a quantity that is measurable experimentally. By the use of experimental data, Feynman obtained a curve very similar to that of Landau, with the roton part slightly elevated.