Bogoliubov Excitations of Inhomogeneous Bose-Einstein ...
Bogoliubov Excitations of Inhomogeneous Bose-Einstein ...
Bogoliubov Excitations of Inhomogeneous Bose-Einstein ...
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2. The <strong>Inhomogeneous</strong> <strong>Bogoliubov</strong> Hamiltonian<br />
r ′ ). With this, we expand the grand canonical Hamiltonian Ê = ˆ H − µ ˆ N<br />
(2.11) in orders <strong>of</strong> δ ˆ Ψ and δ ˆ Ψ † . The linear part vanishes, because Φ(r) is<br />
the ground-state <strong>of</strong> the Gross-Pitaevskii equation. The relevant part is then<br />
the quadratic part ˆ F <strong>of</strong> the Hamiltonian Ê<br />
Ê = E0 + ˆ F [ ˆ δn, δ ˆϕ]. (2.28)<br />
Third-order and forth-order terms in the fluctuations are neglected. Here,<br />
E0 = E[n0(r), ϕ0(r)] is the <strong>Bogoliubov</strong> ground-state energy. The quadratic<br />
Hamiltonian is found as<br />
�<br />
ˆF = d d � �� 2 �<br />
r ∇<br />
2m<br />
δˆn<br />
2Φ(r)<br />
� 2<br />
+<br />
� ∇ 2 Φ(r) �<br />
4Φ 3 (r) δˆn2 + Φ 2 (r)(∇δ ˆϕ) 2<br />
�<br />
+ g<br />
2 δˆn2<br />
�<br />
.<br />
(2.29)<br />
The potential V (r) does not appear directly in the equations <strong>of</strong> motion<br />
for the excitations. Instead, it enters via the condensate function Φ(r),<br />
according to the nonlinear equation (2.19).<br />
With (2.29), the problem is reduced to a Hamiltonian that is quadratic<br />
in the excitations. To this order, there are no mixed terms <strong>of</strong> δˆn and δ ˆϕ,<br />
but the Heisenberg equations <strong>of</strong> motion (2.10) for δ ˆϕ and δˆn<br />
∂δ ˆϕ<br />
∂t<br />
1 � �<br />
= δ ˆϕ, Fˆ ,<br />
i�<br />
∂δˆn<br />
∂t<br />
are coupled, because � δˆn(r), δ ˆϕ(r ′ ) � = i δ(r − r ′ ).<br />
2.3.1. The free <strong>Bogoliubov</strong> problem<br />
1 � �<br />
= δˆn, Fˆ i�<br />
(2.30)<br />
Before including the external potential, we consider the excitations <strong>of</strong> the<br />
homogeneous system V (r) = 0, n0(r) = µ/g. In this case, the <strong>Bogoliubov</strong><br />
Hamiltonian (2.29) reduces to<br />
ˆF (0) �<br />
=<br />
= �<br />
k<br />
d d � 2 �<br />
r<br />
2m<br />
��<br />
∇ δˆn<br />
2 √ �2 n∞<br />
+ (∇ √ n∞δ ˆϕ) 2<br />
�<br />
�<br />
ɛ 0 kn∞δ ˆϕkδ ˆϕ−k + (2µ + ɛ 0 k) δˆnkδˆn−k<br />
4n∞<br />
+ 2g n∞<br />
� δˆn<br />
2 √ n∞<br />
� 2�<br />
�<br />
, (2.31)<br />
with ɛ 0 k = �2 k 2 /(2m). The Fourier representation is already diagonal in k,<br />
but the equations <strong>of</strong> motion (2.30) are still coupled. This can be resolved by<br />
the <strong>Bogoliubov</strong> transformation [71], a transformation that couples density<br />
30