Bose-Einstein Condensates in Rotating Traps and Optical ... - BEC

6.2 Tight binding regime 81
The leading contribution yields the single particle harmonic oscillator groundstate where σ =
s−1/4d/π = ¯h/m˜ω for the harmonic well with frequency ˜ω =2 √ sER/¯h, while the second
term is a small correction arising from the anharmonicity which slightly increases the width σ
of the groundstate.
Inserting the gaussian ansatz into the expressions (6.30) and (6.33) for the on-site contribution
to the energy and the chemical potential, we obtain
ε0
ER
µ0
ER
= d2
2π 2
1 1π2 + s
σ2 2d2 σ2 − 1π4
σ4
4d4
+ 1 gn d 1
√ , (6.51)
8π ER π σ
= ε0
+
ER
1 gn d 1
√ . (6.52)
8π ER π σ
Within the gaussian approximation, the inverse compressibility (6.44) can be rewritten as
κ −1 = gnd
√ 2πσ
and the tight binding effective coupling constant (6.45) takes the form
˜g = gd
√ 2πσ
(6.53)
(6.54)
with σ given by (6.50). For s =10,gn=0.5ER the approximation κ −1 =˜gn with ˜g given by
Eq.(6.54) differs from the exact value of κ −1 by less than 1%.
In section 5.4, we have found that the average density at the central site of a condensate
loaded in the combined potential of optical lattice and harmonic trap decreases like nl=0 ∼
(˜g/g) −3/5 ∼ s −3/20 . As already mentioned there, the decrease of the average density at
the trap center (5.31) is to be contrasted with the increase of the non-averaged peak density
n(r⊥ =0,z =0)∼ (˜g/g) 2/5 ∼ s 1/10 . We can now prove the latter statement using the
gaussian ansatz (6.46) with σ given by (6.49): The peak density then reads
n(r⊥ =0,z =0)=f(z =0) 2 nl=0(r⊥ =0;s) = 1
π1/2 (µ − µgn=0)
, (6.55)
σ ˜g
where we have used the approximative solution (µ − µgn=0)/˜g for the average density profile
(see Eq.(5.31)) with µ − µgn=0 and ˜g given by Eqs.(5.29) and (6.54) respectively. At s =0,
the density at the center is given by n(r⊥ =0,z =0)=µ/g where µ is the usual TF-value
of the chemical potential (see Eq.(5.29) with ˜g = g and µgn=0 =0). It follows that the ratio
between the peak densities at large s and at s =0equals √ 2(˜g/g) 2/5 .Fors =20and s =50,
this amounts to a 32% and 46% increase respectively.