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

162 Condensate fraction
This expression can be evaluated analytically. One obtains the well known result
∆Ntot
=
Ntot
8 1
3 π1/2 (a3n) 1/2 . (12.20)
In view of the discussion of the case s = 0below, it is important to point out that the integrand
in (12.19) contributes most at p ∼ mc where the dispersion (12.17) passes from the phononic
regime ¯hω ∼ p to the single particle regime ¯hω ∼ p2 . Yet the convergence is very slow and
the integral is saturated by momenta much larger than mc [156], where elementary excitations
have single particle character.
Quantum depletion of the 2D uniform gas
Let us consider a uniform system of axial size d and transverse size L and let us assume that
themotionisfrozeninthez-direction. We set Ψ(r) = 1
√ d Ψ2(x, y). Under these conditions
the stationary GPE takes the form
− ¯h2
The 2D density is given by
The 2D coupling constant emerges as
2m ∇2⊥ + g
2
|Ψ2|
d
Ψ2 = µΨ2 . (12.21)
n2 = |Ψ2| 2 = Ntot/L 2 . (12.22)
g2 = g ¯h2 a
=4π . (12.23)
d m d
The groundstate solution is given by
µ = g2n2 . (12.24)
The calculation of the Bogoliubov spectrum is analogous to the 3D case. The result for
the depletion changes only because of the change in the dimensionality of the integral over
momenta. We rewrite the derivation here for clarity. The Bogoliubov equations for the system
in the groundstate read
p2 2m +2g2|Ψ2| 2
− µ up + g2|Ψ2| 2 vp , =¯hω(p)up
(12.25)
p2 2m +2g2|Ψ2| 2
− µ vp + g2|Ψ2| 2 up = −¯hω(p)vp , (12.26)
where p 2 = p 2 x + p 2 y and we have assumed that the Bogoliubov amplitudes take the form
1 1
up(x, y) =Up √
d L ei(pxx+pyy)/¯h , (12.27)
1 1
vp(x, y) =Vp √
d L ei(pxx+pyy)/¯h . (12.28)