T. Tsuneto
Superfluidity in liquid 3 He, discovered in 1972, is due to pairing, as is the case for superconductivity in ordinary metals. However, the pair is in the 3 P state so that a new aspect of superfluidity appears with internal degrees of freedom. It offers quite an interesting real example of ‘spontaneous symmetry breaking’. This aspect will be emphasised in the following discussion. Fermi liquid 3 He A 3 He atom is a fermion with nuclear spin ½. Although 3 He gas liquefies at low temperatures due to the weak van der Waals force, it does not solidify even at T = 0 unless the pressure exceeds 33.5 atm (3.4 × 10 6 Pa) as shown in the phase diagram (Fig. 6.1 (a)). While Bose liquid 4 He becomes superfluid at about 2K, 3 He is Fermi-degenerate at less than T F ∼ 0.1 K and shows the properties of a normal Fermi gas. For example, the specific heat and the susceptibility are proportional to T at T < T F . It is only at an ultra-low temperature in the mK-region that the system undergoes a transition to the superfluid state. Figure 6.1(b) is the phase diagram showing the superfluid phases with variables T,P and external magnetic field H in the mK-region. It should be noted that there are two thermodynamically distinct phases called the A phase and the B phase even at H = 0, and that for H ≠ 0 the A 1 phase appears.
Cambridge University Press eBooks•Dec 10, 1998•0 citations