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A superfluid is a fluid that flows without dissipation-a phenomenon that never ceases to cause excitement, because it defies common sense and raises the hope that one might capitalize on it. The classic example is liquid He 4 in its low-temperature phase, which can flow through the tiniest cracks without viscosity. Equally remarkable is a superconducting metal in which" supercurrents" can be maintained without an applied emf for months, and presumably indefinitely in principle. In this case one thinks of the charge carriers (the Cooper pairs) as a superfluid. Other examples include spin-aligned atomic hydrogen, which experimenters can now create in the laboratory. In this chapter, we shall concentrate on liquid He 4 as the prototype of a superfluid. The essence of superfluidity, as we shall see, is the existence of a condensate characterizable by a nonvanishing superfluid order parameter.
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Why Helium Does Not Solidify
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Helium is the only substance known to resist solidification under atmospheric pressure down to the lowest temperatures observed, and presumably to absolute zero. To solidify it requires an external pressure of at least 25 atm. This happy circumstance furnishes us with two naturally occurring quantum liquids, He 4 and its less abundant isotope He 3 The qualitative reasons for the fluidity of helium are two: (a) The molecular interactions between He atoms are weak, as evidenced by the fact that He is a noble gas; (b) the mass of He is the smallest among the noble gases. These circumstances lead to a large zero-point motion of the He atoms, so that it becomes impossible to localize the atoms at well-defined lattice sites. To understand these reasons, we must first have a few facts. The potential energy v( r) between two He atoms separated by a distance r has been calcui"ted by Slater and Kirkwood* on the basis of the electronic
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*J. C. Slater and J. G. Kirkwood, Ph s. Rev. 37, 682 (1931).
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Average interparticle distance
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Fig. 13.1
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Potential energy between two He atoms separated by distance r.
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structure of the He atom. They find that
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(5.67 X 10 6 ) e- 21.5(r/o)
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where a = 4.64 A and u(r) is in kelvins. A graph of u(r) is shown in Fig. 13.l. From the slope dP/dT of the experimental vapor pressure curve we can deduce the latent heat of vaporization of liquid helium. Extrapolation of experimental data shows that dP/dT > 0 at absolute zero. Hence liquid helium has a nonvanishing binding energy per atom at absolute zero. That is, the ground state of liquid helium is an N-body bound state that has a self-determined equilibrium density in the absence of external pressure. Consider a collection of H y atoms at absolute zero under no external pressure. The most probable configuration of the atoms is determined by the ground state wave function, which, according to the variational principle, must be such as to minimize the total energy of the system, with no external constraint imposed. Hence energy consideration alone determines the most probable configuration. We can then make the following qualitative argument. If a He atom is to have a well-defined location, it must be confined to within a distance ~x that is small compared to the range of the potential, say ~x :::: 0.5 A. By the uncertainty principle, we would then expect an uncertainty in energy (in units of
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Solid Line of A Ira nsition 25 atm 2.26 aIm
'--=-_ _L--, T x =2.18K
Fig. 13.2
Equation of state of He 4 (not to scale).
Boltzmann's constant) of the order of
- 1 ( - 11
)2 :::: 10K
This is comparable to the depth of the potential well. Hence the localization is impossible. The fact that no other noble element can remain in liquid form down to very low temperatures is explained by their much greater masses. The fact that H 2 , although lighter than He, solidifies at a finite temperature is explained by the strong molecular interactions between H 2 molecules. The argument we have given is independent of statistics and also explains why both He 4 and He 3 remain liquid down to absolute zero.