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Perturbed solution

We start from the three-dimensional energy functional (1.110) of a homogeneous weakly-interacting Bose gas in the presence of a $\delta $-function perturbation (an impurity) moving with a constant velocity ${\bf V}$

\begin{displaymath}
E=\int\left(\frac{\hbar^2}{2m}\vert\nabla\psi\vert^2+(\mu-{g...
... V}t))\vert\psi \vert^2+\frac{g}2\vert\psi\vert^4\right) d^3x,
\end{displaymath} (7.1)

where $\psi$ is the condensate wave function, $\mu $ is the chemical potential, $m$ mass of a particle in the condensate, $g=4\pi\hbar^2a/m$ and ${g_{i}}=2\pi\hbar^2b/m$ are particle-particle and particle-impurity coupling constants, with $a$ and $b$ being the respective scattering lengths7.1. We will assume that the interaction with impurity is small and we will use perturbation theory. By splitting the wave function into a sum of the unperturbed solution and a small correction $\psi ({\bf r},t)=\phi_0+\delta\psi ({\bf r},t)$ and linearizing the time-dependent GP equation with respect to $\delta\psi$, we obtain an equation describing the time evolution of $\delta\psi$
$\displaystyle i\hbar \frac{\partial}{\partial t}\delta\psi=
\left(-\frac{\hbar^...
...) \delta\psi +
g\phi_0^2\,\delta\psi^*+{g_{i}}\,\delta ({\bf r}-{\bf V}t)\phi_0$     (7.2)

In a homogeneous system $\phi_0$ is a constant fixed by the particle density $\phi_0=\sqrt{n}$ and $\mu =gn=mc^2$.

The perturbation follows the moving impurity, i.e. $\delta\psi$ is a function of $({\bf r}-{\bf V}t)$, so the coordinate derivative is related to the time derivative

$\displaystyle \partial\delta\psi({\bf r}-{\bf V}t)/\partial t=-{\bf V}\vec{\nabla}\delta\psi ({\bf r}-{\bf V}t)$     (7.3)

We shall work in the frame moving with the impurity ${\bf r}^{\prime}={\bf r}-{\bf V}t$ and the subscript over ${\bf r}$ will be dropped.

Eq. (7.2) for a perturbation in a homogeneous system can be conveniently solved in momentum space. In order to do this we introduce the Fourier transform of the wave function $\delta\psi_{{\bf k}}=\int e^{-i
{\bf k\cdot}{\bf r}}\delta\psi ({\bf r})\,d^3x$. Eq. (7.2 ) becomes

\begin{displaymath}
\left\{
\begin{array}{ccc}
\left(-\hbar{\bf k\cdot V}+\frac{...
...ta\psi_{{\bf -k}})^*+{g_{i}}\,\phi_0&=&0\\
\end{array}\right.
\end{displaymath} (7.4)

Here the second equation is obtained by doing the substitution ${\bf k}\rightarrow
-{\bf k}$ and and complex conjugation. We also use property of the Fourier transformation $\delta (\psi^*)_{{\bf k}} = \delta (\psi_{-{\bf k}})^*$. The system of linear equations (7.4) can be easily solved
\begin{displaymath}
\delta\psi_{{\bf k}}={g_{i}}\phi_0\frac{\hbar {\bf k\cdot V}...
...\frac{\hbar^2k^2}{2m}
\left(\frac{\hbar^2k^2}{2m}+2\mu\right)}
\end{displaymath} (7.5)


next up previous contents
Next: Total energy Up: Three-dimensional system Previous: Three-dimensional system   Contents
G.E. Astrakharchik 15-th of December 2004