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

2.1 Single vortex line configuration 11
explicit d-dependence of the healing length inside the logarithm, in order to account for the
density dependence of the size of the vortex core.
Eq.(2.6) shows that no excitation energy is carried by the system when d = R⊥ . Of course
this estimate, being derived using the TF-approximation, is not accurate if we go too close
to the border. If the surface of the condensate is described by a more realistic density profile
the energy is nevertheless expected to vanish when the vortex line is sufficiently far outside
the bulk region. Within the simplifying assumptions made above we can conclude that the
configuration with d = R⊥ corresponds to the absence of a vortex, while the transition from
d = R⊥ to d =0describes the nucleation path of the vortex.
Angular momentum of the vortex configuration
The inclusion of vorticity is not only accompanied by an energy cost but also by the appearance
of angular momentum. The simplest way to calculate the angular momentum associated with
a displaced vortex is to assume axi-symmetric trapping (ε =0) and to work with the TFapproximation.
In this limit the size of the vortex core is small compared to the radius of the
condensate so that one can use the vortex-free expression n(r) =µ[1−(r⊥/R⊥) 2 −(z/Z) 2 ]/g
for the density profile of the condensate, where r2 ⊥ = x2 +y2 and g =4πa¯h 2 /m is the coupling
constant, fixed by the positive scattering length a. Then, one can write the angular momentum
in the form
Lz = m dz dr⊥r⊥n(r⊥,z) vvortex · dl , (2.7)
where the line integral is taken along a circle of radius r⊥. Use of Stokes’ theorem gives the
result [49]
d
Lz(d/R⊥) =N¯h 1 −
2 5/2
, (2.8)
where d is the distance of the vortex line from the symmetry axis. Eq. (2.8) shows that the
angular momentum per particle is reduced from the value ¯h as soon as the vortex is displaced
from the center and becomes zero for d = R⊥.
Vortex energy in the rotating frame
Eq.(2.6) makes evident that a macroscopic energy is required to achieve a transition to a
vortex-state. In a rotating trap of the form (1.3) the system may nevertheless like to acquire
the vortex configuration. This happens if there is a total energy gain in the rotating frame
where the system has energy
E(Ω) = E − ΩLz . (2.9)
Here E is the energy in the laboratory frame, Lz is the angular momentum, and Ω is the
angular velocity of the trap around the z-axis.
In an axi-symmetric trap (ε =0) Eq.(2.9) yields the energy [49, 27]
2 3/2
2 5/2
d
d
Ev(d/R⊥, Ω,µ)=Ev(d =0,ε=0,µ) 1 −
− ΩN¯h 1 −
. (2.10)
R⊥
R⊥
R⊥