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Weak condensate case

Firstly, let us consider the case of a weak condensate so that the effect of the nonlinear term is small in comparison to the linear ones, $ \vert\varrho\psi_{0}\vert\lesssim\vert\triangle\psi_{0}\vert$. Since $ \Omega$ is a constant we observe that the Laplacian term acts to balance the external potential term (like in the linear Schrödinger equation) and the nonlinear term can be, at most, as big as the linear ones

$\displaystyle \Omega \sim \frac{1}{r_{0}^{2}} \sim U(r_{0}) \gtrsim \varrho,$

where $ r_{0}$ is the characteristic size of the condensate (it is defined as the condensate `` reflection'' point via the condition $ \Omega
=U(r_{0})$, see below).

Now for a WKB description to be valid we require $ kr_{0}\gg 1$, i.e. we require the characteristic length-scale of our wavepackets to be a lot smaller than that of the large-scales. Using this fact we find

$\displaystyle k^{2} \gg \frac{1}{r_{0}^{2}} \sim U(r_{0}) \gtrsim \varrho.$

Therefore, the condensate correction to the frequency, given by ([*]), is small. In other words the wavepacket does not `` feel'' the condensate. Indeed, from $ k_{max}^{2} =
U(r_{R})$ we have $ U(r_{R}) \gg U(r_{0})$ and this implies that $ r_{R}
\gg r_{0}$ (where $ r_{R}$ is the wavepacket reflection point, see figure 2). Thus, the condensate in this case occupies a tiny space at the bottom of the potential well and hence does not affect a wavepacket's motion. Therefore, a wavepacket moves as a `` classical'' particle described by the Ehrenfest equations ([*]) and ([*]). In fact, in this case it would be incorrect to try to describe the small condensate corrections via our WKB approach because these corrections are of order $ \varrho \sim
\varepsilon^{2}$ (the $ \varepsilon^{2}$ terms being ignored in a WKB description).


next up previous
Next: Strong condensate case Up: Applicability of WKB descriptions Previous: Applicability of WKB descriptions
Dr Yuri V Lvov 2007-01-23