Bogoliubov Excitations of Inhomogeneous Bose-Einstein ...
Bogoliubov Excitations of Inhomogeneous Bose-Einstein ...
Bogoliubov Excitations of Inhomogeneous Bose-Einstein ...
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4.1. Hydrodynamic limit I: ξ = 0<br />
The free Green function � G0 = (2π) d δ(k − k ′ ) � G0(k, ω) is diagonal in k,<br />
�G0(k, ω) = � ω 2 − c 2 k 2 + i0 � −1 . (4.7)<br />
The full Green function � G is expanded in the same way as in equation (3.30),<br />
�G = � G0 + � G0 � V � G0 + . . .. Again, we take the disorder average and find<br />
�G(k, ω) =<br />
�<br />
�G0(k, ω) −1 − � � −1<br />
Σ(k, ω) . (4.8)<br />
The poles <strong>of</strong> this average Green function at ω 2 = c 2 k 2 + � Σ(k, ω) now determine<br />
the effective dispersion relation. In the Born approximation, the<br />
self-energy � Σ(k, ω) evaluates as<br />
�Σ(k, ω) = V 2<br />
0 σ d<br />
� d ′ d k<br />
(2π) dCd(|k − k ′ |σ) (k′ ·k) 2<br />
m2 �G0(k ′ , ω). (4.9)<br />
The self-energy corrects the quadratic dispersion relation (�ω) 2 = c2k2 +<br />
�Σ(k, ω). To leading order, we can use the on-shell approximation and replace<br />
�Σ(k, ω) with � Σ(k, ck), such that ω = ck � 1 + � Σ(k, ck)/(2ck) � . This converts<br />
to the inverse mean free path (kls) −1 = −Im� Σ(k, ck)/(c2k2 ) and the relative<br />
correction <strong>of</strong> the speed <strong>of</strong> sound Λ = Re� Σ(k, ck)µ 2 /(2c2k2V 2<br />
0 ).<br />
Before discussing these results in detail, we verify that these results are<br />
in agreement with the result (3.39) from section 3.4 in the limit ξ → 0. The<br />
envelope functions reduce to w (2) = O(kξ), w (1)<br />
k ′ k = −k′ ·k ξ/ √ 2kk ′ = −y (1)<br />
k ′ k .<br />
With this, we indeed find<br />
lim<br />
ξ→0 ΣB (k, ɛk) = � � Σ(k, ck)<br />
. (4.10)<br />
2ck<br />
Next, let us consider the central results inverse mean free path and correction<br />
to the speed <strong>of</strong> sound.<br />
4.1.2. Transport length scales<br />
Mean free path<br />
The imaginary part <strong>of</strong> the self-energy � Σ leads to the inverse mean free path<br />
1<br />
kls<br />
= V 2<br />
0<br />
4µ 2(kσ)d<br />
�<br />
dΩd<br />
(2π) d−1Cd<br />
� � 2<br />
2kσ sin θ/2 cos (θ). (4.11)<br />
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