1549380323-Statistical Mechanics Theory and Molecular Simulation

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440 Quantum ideal gases


The occupation of the ground state undergoes a transition from afinite value at
T = 0 to zero atT =T 0 , and for all higher temperatures, it remains zero. Now,
〈f ̄ 0 〉/〈N〉represents the probability that a particle will be found in the groundstate
and is, therefore, the fraction of the total number of particles occupying the ground
state on average. ForT << T 0 , this number is very close to 1, and atT = 0, it
becomes exactly 1, implying that atT= 0, all particles are in the ground state. This
phenomenon, in which all particles “condense” into the ground state, is known as
Bose–Einsteincondensation. The temperature,T 0 , at which “condensation” begins is
known as theBose–Einstein condensation temperature.
Bose–Einstein condensates were first realized experimentally usinglow-temperature
(170 nano-Kelvin) magnetically confined rubidium atoms (Andersonet al., 1995).
These and other experiments have sparked considerable interestin the problem of cre-
ating Bose–Einstein condensates for technological applications, including superfluidity.
Indeed, Bose–Einstein condensation is a striking example of a large-scale cooperative
quantum phenomenon.
Note that there is also a critical density corresponding to the Bose–Einstein con-
densation temperature, which is given by the solution of


ρλ^3
g

=R(3/2). (11.6.39)


Eqn. (11.6.39) can be solved to yield


ρ=

gR(3/2)
λ^3

=gR(3/2)

(


mkT 0
2 π ̄h^2

) 3 / 2


≡ρ 0 , (11.6.40)

and the occupation number, expressed in terms of the density is


〈f ̄ 0 〉
〈N〉

=





1 −(ρ 0 /ρ) ρ > ρ 0

0 ρ < ρ 0

. (11.6.41)


The divergent term in eqn. (11.6.18),−(λ^3 /V) ln(1−ζ), becomes, forζvery close
to 1,
λ^3
V


ln(V/a)∼

lnV
V

, (11.6.42)


which clearly vanishes in the thermodynamic limit, sinceV∼〈N〉. Thus, the pressure
simplifies even forζvery close to 1, and the equation of state can be written as


P


gkT

=





g 5 / 2 (1)/λ^3 ρ > ρ 0

g 5 / 2 (ζ)/λ^3 ρ < ρ 0

, (11.6.43)


whereζ is obtained by solvingρλ^3 /g=g 3 / 2 (ζ). It is interesting to note that the
pressure is approximately independent of the density forρ > ρ 0. Isotherms of the
ideal boson gas are shown in Fig. 11.5. Here,v 0 is the volume corresponding to the
critical densityρ 0. The figure shows thatP ∼T^5 /^2 , which is quite different from

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