Consider the dimensionless Bogoliubov-de Gennes equations for the excitations in a BEC: $$ \begin{cases} \big(-\frac{1}{2}\nabla^2+2gn({\bf r}) - \mu -\omega\big) u({\bf r})-n({\bf r})g v({\bf r}) = 0\\ \big(-\frac{1}{2}\nabla^2+2gn({\bf r}) - \mu +\omega\big) v({\bf r})-n({\bf r})g u({\bf r}) = 0, \end{cases} $$ where $g$ is the value of the contact interaction between the atoms, $n({\bf r})$ is the condensate density, $\mu$ is the chemical potential and $\omega$ is the eigenvalue of the system which should describe the frequency of the oscillating excitation wave functions $u({\bf r})$ and $v({\bf r})$.

First suppose we consider a homogeneous system. Then the condensate density is a constant: $n({\bf r}) = n_0$ (and from the Gross-Pitaevski equation one has $\mu = n_0 g$) . Going to Fourier space $u({\bf r}) \propto e^{i {\bf k\cdot r}}$ (and similarly for $v({\bf r})$, one easily solves for $\omega$ by demanding that the determinant of the system vanishes and one finds $4\omega^2 = k^2(k^2 + 4n_0g) $.

Now suppose we consider a `realistic' radially symmetric system where $\lim_{r\rightarrow\infty}n(r) = 0$, i.e., the condensate density is only finite in a finite region of the space. In this case the coupling between the two equations vanishes and one has $u''(r) = -2\omega u(r)$ and $v''(r) = 2\omega v(r)$. Making a similar ansatz for u and v, you find $k^2 = \pm 2\omega$. Hence, one solution has an imaginary part and is unstable.

I do not know how to interpret this result. I have the feeling that something is wrong in my reasoning, because I would expect similar behavior for large $r$ in the two cases.

Note: the reason I am asking this is because I want to numerically solve a somewhat more complicated form of the BdG-equations for a radially symmetric system. I have already obtained a numerical solution for $n(r)$ from the Gross-Pitaevski equation, but in order to solve for $u(r)$ and $v(r)$ I need proper boundary conditions on $r=0$ and $r\rightarrow \infty$. That is why I was thinking about this limit.

  • $\begingroup$ The linear stability analysis deals with fluctuations on top of the ground state. In other words, the assumption is that the fluctuations are small with respect to the ground state. Hence, I doubt whether this approach can be used when the condensate density vanishes. $\endgroup$ – Abhijit May 18 '16 at 7:14

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