Superfluidity

He-4 below the lambda transition temperature 2.7 K can be analyzed as a two-fluid liquid composed of He atoms in the normal state and He atoms in the macroscopic superfluid state. Superfluidity is a property of He-4 in the liquid state because He-4 atoms obey Bose- Einstein statistics. Many He-4 atoms can be in the same macroscopic quantum state that is, the same momentum states for these atoms moving in the superfluid. If so, then why can He-3 at low temperatures also become a superfluid?

Answer

The odd number of constituents in He-3 (two protons, one neutron, and two electrons) classifies it as a fermion that obeys Fermi-Dirac statistics. So no two He-3 atoms can share the same quantum state defined by the four momentum (energy and three-momentum) and spin. The surprise in the early 1970s was that He-3 can magnetically couple with another He-3 to form a boson and become a superfluid liquid at the extremely low temperature of 2.7 milli kelvins. The He-3 pairs form one momentum macro state. Because the component He-3 atoms are not bosons, there should be some small width to the macro state momentum in addition to the small width because the He atoms are composed of fermions.

The pairs of atoms are magnetic, so the He-3 superfluid is more complex than its He-4 counterpart. In fact, superfluid He-3 exists in three different phases related to different magnetic or temperature conditions. In the A phase, for example, the superfluid is highly anisotropic that is, directional like a liquid crystal.