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When we cool anything down we know it must order and the entropy go to zero. What about liquids and dilute gases which are inherently chaotic? With normal materials - no problem - they solidify and the chaos disappears (ideally anyway).
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When we cool anything down we know it must order and the entropy go to zero. What about liquids and dilute gases which are inherently chaotic? With normal materials - no problem - they solidify and the chaos disappears (ideally anyway). But the heliums remain liquid to absolute zero - what happens there?
To confine a helium at into a defined lattice site means that the wavelength l must be of order the interatomic spacing d. And thus the momentum is ~ h/l and the zero point energy p2/2m ~ h2/2l2m. If this is higher than the attractive potential well then the solid phase never forms. Both heliums have NO chemistry because the filled-shell interatomic forces are too small to stabilise the solid.
He4 is made up of bosons, so there is no problem in their all dropping into the same ground state to create a condensate but this particular condensate is hard to understand. He3 on the other hand is made up of fermions and can only go into a single ground state by forming Cooper pairs where two fermions couple to form a boson as in superconductivity. (The He3 condensate is thus composed of “soft” bosons, easily broken whereas He4 is made of “hard bosons” and we would need to ionize the atom to break it up.)
We’ll start by looking at the general properties of the condensate at absolute zero. That is, when there are no excitations to confuse the issue. Let us start with the simplest wavefunction:
rs, condensate density f, phase
We take the spatial gradient to find the momentum: The momentum then becomes just the gradient of the phase, f, which in a condensate is a global property of the liquid.
This is a crunch result for simple condensates. Since the superfluid velocity is given by the gradient of a phase, then the liquid is inherently irrotational as the curl of a gradient must be zero.
This leads us to the basic equations for simple superfluid dynamics. If we set the liquid as incompressible, then the divergence of the velocity must be zero (constant density thus no sources or sinks of superfluid). From above, since the velocity is given by the gradient of a phase the curl (or circulation) must also be zero. Thus we get: and
BUT, these are the equations for the electromagnetic fields in free space: or:
That means we get “pure potential flow” and the flow pattern can be calculated from standard forms in say electrostatics.
Pure potential flow ( , ) around a cylinder.
Which is the same flow field as putting an opposed linear dipole in a uniform field.
The superfluid absolutelycannot “absorb” angular momentum (or much ordinary momentum as we shall see later).
v .. v v
v .. v v
v .. v v
Ditto for simple s-wave superconductors (apart from anisotropy from lattice). D
However, He3 being p-wave paired (along with unconventional superconductors) is a bit more complicated.
We can cool the liquid to ~80mK This gives a purity of = ~1 in 104000
The liquid us therefore absolutely pure even before we think anything about the superfluidity aspect.
The superfluid state emerges as 3He atoms couple across the Fermi sphere to create the Cooper pairs. Pz Px Py
The superfluid state emerges as 3He atoms couple across the Fermi sphere to create the Cooper pairs. Pz Py Px
Since 3He atoms are massive, p-wave pairing is preferred, i.e. L = 1 which means S must also be 1. The ground state thus has S = 1 and L = 1 making the Cooper pairs like small dimers (and easier to visualise than the s-wave pairs in superconductors).
With S = L = 1 we have a lot of free parameters and the superfluid can exist in several phases (principally the A- and B-phases) .
With S = L = 1 we have a lot of free parameters and the superfluid can exist in several phases (principally the A- and B-phases) . Let us start with the A phase which has only equal spin pairs. That is Sz= ± 1. The momenta couple to give J = 1, thus Lz= ± 1. The directions of the S and L vectors are global properties of the liquid as all pairs are in the same state (this is the “texture” of the liquid). However, that causes problems for the pairs.
Assume the global L vector lies in the z-direction - We can easily have pairs like this:- L-vector That is fine as the constituent 3He fermion states can simply orbit the “equator” of the Fermi sphere:
However, if we try to couple pairs across the “poles” of the Fermi sphere there is no orbit that these pairs can make which gives a vertical L. Thus the liquid is a good superfluid in the equatorial plane and lousy at the poles – this is reflected in the A-phase energy gap:-
D The A-phase gap:- large round the equator, zero at the poles.
Thus the equal-spin pairs form a torus around the equator in momentum space, and there are no pairs at the poles. L-vector pairs The A phase is thus highly anisotropic. Also very odd excitation gas.
In the B phase we can also have opposite spin pairs (the L- and S-vectors couple to give J = 0) This now allows us to have Lz = Sz = 0 pairs which can fill in the hole left at the poles by the A phase, giving an “isotropic” gap:
D The B-phase gap:- equal in all directions. (because all spin-pair species allowed).
pairs pairs pairs pairs The equatorial equal-spin pairs torus is still there but along with the Lz = Sz = 0 pairs which now fill the gap at the poles. L-vector
pairs pairs pairs pairs The equatorial equal-spin pairs torus is still there but along with the Sz = 0 pairs which now fill the gap at the poles. L-vector pairs (which add up to a spherically symmetric total)
The A phase has a higher susceptibility than the B phase (because all pairs are ßß or ÝÝ no non-magnetic Ýß components). Thus by applying a magnetic field we can stabilise the A phase. The A phase is the preferred phase at T = 0 when the magnetic field reaches 340 mT.
Having made the five minute trip around the superfluid the context for what follows is: We can cool superfluid 3He to temperatures where there is essentially no normal fluid (1 in ~108 unpaired 3He atoms). We can cool and manipulate both phases to these temperatures by profiled magnetic fields. That means we can create a phase boundary between two coherent condensates, itself a coherent structure, at essentially T = 0. (This is the nearest analogue we have of a cosmological brane.) This brings us on to defects in general in the quantum fluids.
ROTATION Since for simple superfluids Ñ x v= 0, the condensate is completely irrotational. The superfluids thus provide the perfect “gyroscopic” materials since they sit unrotating in the “frame of the fixed stars”. Then what happens if we try to rotate a superfluid? - say by rotating the container. At some point we can exceed some critical velocity at the periphery and locally destroy the condensate - at which point we can create a vortex.
Plane of equal phase Vortex core (where rs goes to zero). 2p phase change around vortex
Plane of equal phase Vortex core (where rs goes to zero). 2p phase change around vortex