10.6 A Bose–Einstein condensate 235derivation can be found in the book by Pitaevskii and Stringari (2003).
Trapped atoms experience a harmonic potential
V(r)=1
2
M
(
ωx^2 x^2 +ω^2 yy^2 +ω^2 zz^2)
. (10.31)
For simplicity, we shall consider all three oscillation frequencies equal,
i.e. the isotropic potentialMω^2 r^2 /2, and use a variational method to
estimate the energy. We choose a trial wavefunction that is a Gaussian
function:
ψ=Ae−r
(^2) / 2 b 2
. (10.32)
Using this to calculate the expectation values of terms in eqn 10.29 gives
E=
3
4
ω{
a^2 ho
b^2+
b^2
a^2 ho}
+
g
(2π)^3 /^21
b^3. (10.33)
Differentiation of this expression shows that wheng= 0 the minimum
energy occurs whenb=aho,where
aho=√
Mω(10.34)
is the characteristic radius of the Gaussian ground-state wavefunction in
the quantum harmonic oscillator. A sodium atom in a trap with oscilla-
tion frequencyω/ 2 π= 100 Hz hasaho=2× 10 −^6 m.For this equilibrium
value ofb, the two terms representing the kinetic and potential energies
give an equal contribution to the total energy which isE=(3/2)ω,as
expected for the ground state of the quantum harmonic oscillator.^38 The^38 Each of the three degrees of freedom
variational method gives exact results in this particular case because the has a zero-point energy of^12 ω.
trial wavefunction has the same Gaussian form as the actual solution for
a harmonic oscillator.
Now we shall consider what happens wheng>0.^39 The ratio of the^39 Small Bose–Einstein condensates
with effectively attractive interactions
have been created, but they collapse
inward as the number of atoms grows
(see Exercise 10.11).
terms representing the atomic interactions and the kinetic energy is^40
(^40) The factor of (4/3)(4π/(2π) 3 / (^2) )that
arises has a numerical value of about
unity.
4
3(2π)^3 /^2g
a^3 hoωNa
aho. (10.35)
The nonlinear term swamps the kinetic energy whenN>aho/a.This
ratio equalsaho/a= 700 foraho=2μmanda= 3 nm, and so in this case
when the number of atoms in the condensateN>700 we can neglect the
kinetic energy.^41 We could estimate the condensate’s size from eqn 10.33^41 Typically, experiments haveN> 105
and the effectively repulsive interac-
tions make the condensate much larger
thanaho.The interactions between the
atoms in a dilute gas only have a small
effect (<10%) on the value ofTC.
by determining the value ofrat which the confining potential balances
the repulsive interactions, and this variational method is described in
Exercise 10.6; but it is remarkably easy to solve the Gross–Pitaevskii
equation when the kinetic energy term is neglected. In this so-called
Thomas–Fermiregime eqn 10.29 becomes simply
{
V(r)+g|ψ|^2
}
ψ=μψ. (10.36)So for the region whereψ= 0 we find
|ψ|^2 =μ−V(r)
g