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

4.1 Solution of the Schrödinger equation 39
2m¯v/qB
0.5
0.25
0
−0.25
−0.5
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
0.5
0.25
0
−0.25
a)
b)
−0.5
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
¯hk/qB
Figure 4.5: Group velocity (4.15) as function of quasi-momentum ¯hk of a particle in the optical
lattice potential V = sERsin 2 (πx/d) with a) s =1and b) s =5. Solid lines: Exact numerical
results. Dashed lines: Tight binding result (4.31) with δ as obtained from (4.33) using the
numerical data for m ∗ (4.18).
The current density does not depend on the spatial coordinate x since we are dealing with a
stationary state solution of the Schrödinger equation. Hence, each Bloch state is characterized
by a certain value of the current density Ij(k). One can show easily (see discussion in chapter
6.1) that
Ij(k) = 1 ∂εj(k)
, (4.17)
L ¯h∂k
where L is the length of the system. It is interesting to note that Eq.(4.17) implies the result
(4.15) since the group velocity must fulfill the equation ¯vj(k) =IL.
For small quasi-momenta k → 0, the lowest energy band depends quadratically on k, its
curvature defining the effective mass
1
m ∗ := ∂2 εj(k)
¯h 2 ∂k 2
With this defintion, current and group velocity for k → 0 are given by
Ij=1(k) → 1 ¯hk
L m∗ ¯vj=1(k) → ¯hk
m∗
. (4.18)
k=0
(4.19)
(4.20)