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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

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