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Bogoliubov Excitations of Inhomogeneous Bose-Einstein ...

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2. The <strong>Inhomogeneous</strong> <strong>Bogoliubov</strong> Hamiltonian<br />

Im[γ (s) (θ)]<br />

f<br />

f<br />

2<br />

0<br />

− f<br />

2<br />

−f<br />

0<br />

π<br />

4<br />

π<br />

2<br />

kξ = 0.2<br />

0.5<br />

1.0<br />

2.0<br />

5.0<br />

Figure 2.8.: Elastic scattering amplitude (2.61) in units <strong>of</strong> f = γ0 V0/(8ξµ) for different<br />

values <strong>of</strong> kξ, at fixed potential radius kr0 = 0.5. Symbols: Results from numerical<br />

integration <strong>of</strong> the full GP equation. Solid curves: analytical prediction (2.61). With<br />

increasing k, the node moves to the left, according to (2.60). The overall amplitude has<br />

a maximum at kξ ≈ 1.<br />

• The inverse group velocity (∂k/∂ɛk) behaves like the constant c −1 for<br />

sound waves (kξ ≪ 1) and decreases as k −1 for particles (kξ ≫ 1).<br />

The product <strong>of</strong> both contributions therefore has limiting behavior k and k −1 ,<br />

respectively, with a maximum around the crossover kξ ≈ 1 from phonons<br />

to particles.<br />

2.4.5. One-dimensional setting<br />

In a one-dimensional setting, only forward scattering and backscattering<br />

are present. This allows easily computing the transmission <strong>of</strong> a <strong>Bogoliubov</strong><br />

excitation across an impurity. Now we consider the reflection <strong>of</strong> an incident<br />

<strong>Bogoliubov</strong> excitation γ (0)<br />

k ′ = δk ′ kγ0 and compute the elastically reflected<br />

wave γ (r)<br />

k ′ = δk ′ (−k)γr in the Born approximation. The elastically scattered<br />

wave is again singled out by the imaginary part <strong>of</strong> the Green function, such<br />

that (2.61) applies<br />

Im γr<br />

γ0<br />

= L 1<br />

2V−2k<br />

4µξ<br />

3π<br />

4<br />

θ<br />

π<br />

kξ<br />

k2ξ2 . (2.62)<br />

+ 1<br />

For a point like scatterer, the right hand side is proportional to the weight<br />

<strong>of</strong> the impurity L 1<br />

2Vk ≈ � dxV (x) =: 8µξB. From the elastically reflected<br />

wave, we obtain the reflection amplitude r = 2Imγr/γ0. The factor 2 is<br />

because the reflected wave exists only on one side <strong>of</strong> the impurity. For<br />

a narrow potential, the transmission intensity T = 1 − |r| 2 is shown in<br />

figure 2.9. The impurity is perfectly transparent at long wave lengths. At the<br />

46

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