Elementary excitations

In the low-temperature regime, the excited states of the system can be calculated from the ``classical'' frequencies of the linearized GP equation. Let us look for solutions in the form of small oscillations of the order parameter around the stationary value.

$$\Phi({\bf r},t) = e^{-i\mu t/\hbar} [\phi({\bf r})+u({\bf r})e^{-i\omega t}+v^{\star}({\bf r})e^{i\omega t}]$$ (1.15)

By keeping terms linear in the complex functions $u$ and $v$, equation (1.11) becomes

$$\left\{ \begin{array}{rll} \hbar \omega u({\bf r}) & = & [H_0-\mu... ...g\phi^2({\bf r})] v({\bf r}) + g\phi^2({\bf r})u({\bf r})\\ \end{array}\right.$$ (1.16)

where $H_0 = -\hbar^2\triangle/2m + V$. In a uniform gas, the amplitudes $u$ and $v$ are plane waves and the resulting dispersion law takes the Bogoliubov form

$$(\hbar \omega)^2 = \left(\frac{\hbar^2k^2}{2m}\right) \left(\frac{\hbar^2k^2}{2m} + 2gn\right),$$ (1.17)

where ${\bf k}$ is the wave vector of the excitations and $n = \vert\phi\vert^2$ is the density of the gas. For large momenta the spectrum coincides with the free-particle energy $\hbar^2 k^2/2m$. At low momenta equation (1.17) yields the phonon dispersion $\omega = ck$, where the sound velocity $c$ is given by the formula

$$c = \sqrt{\frac{gn}{m}}$$ (1.18)