THE INTERNATIONAL SERIES OF MONOGRAPHS ON PHYSICS ...

28 MICROSCOPIC PHYSICS
Thus there are two principal factors distinguishing liquid 4 He (or liquid 3 He)
from the weakly interacting gases. Liquid 4 He and 3 He can be in equilibrium at
zero pressure, and their chemical potentials at T = 0 are negative.
3.3.2 Model liquid state
Using our intuitive understanding of the liquid state, let us try to construct
a simple model of the quantum liquid. The model energy density describing
a stable isolated liquid at T = 0 must satisfy the following conditions: (i) it
must be attractive (negative) at small n and repulsive (positive) at large n to
provide equilibrium density of liquid; (ii) the chemical potential must be negative
to prevent evaporation; (iii) the liquid must be locally stable, i.e. either the
eigenfrequencies of collective modes must be real, or their imaginary part must
be negative. All these conditions can be satisfied if we modify eqn (3.23) for
a weakly interacting gas in the following way. Let us change the sign of the
first term describing interaction and leave the second term coming from vacuum
fluctuations intact, assuming that it is valid even at high density of particles. Due
to the attractive interaction at low density the Bose gas collapses forming the
liquid state. Of course, this is a rather artificial construction, but it qualitatively
describes the liquid state. So we have the following model:
ɛ(n) =− 1
2 αn2 + 2
5 βn5/2 , (3.32)
where α>0 and β>0 are fitting parameters. In principle, one can use the
exponents of n as other fitting parameters.
Now we can use the equilibrium condition to obtain the equilibrium particle
density n0(µ) as a function of the chemical potential:
dɛ
dn = µ → −αn0 + βn 3/2
0 = µ. (3.33)
From the equation of state one finds the pressure as a function of the equilibrium
density:
P (n0) =−˜ɛ(n0) =µn0(µ) − ɛ(n0(µ)) = − 1
2 αn20 + 3
5 βn5/2 0 . (3.34)
The two contributions to the vacuum pressure cancel each other for the following
value of the particle density:
2 5α
n0(P =0)= . (3.35)
6β
A droplet of the liquid with such a density will be in complete equilibrium in
empty space if µ 0. These conditions are satisfied since the chemical
potential and speed of sound are
µ(P =0)=− 1
6 n0α , (3.36)