# Bogolyubov excitations in BEC

The dynamics of a Bose-Einstein condensate is usually described by the time-dependent Gross-Pitaevskii equation (TDGPE) $$i\hbar\frac{\partial\psi(x,t)}{\partial t} = -\frac{\hbar^2\nabla^2}{2m}\psi(x,t)+V\psi + g |\psi(x,y)|^2\psi(x,t)$$ where $V$ is the external potential and $g$ the interaction strength.

To find the excitation spectrum for small perturbation, it is possible to consider the ansatz $$\psi(x,t) = e^{-i \mu t/\hbar} ( \psi_0(x) + \delta\psi(x,t) ) \;,$$ where we can use a Fourier expansion $$\delta\psi(x,t) = \sum_i [ u_i(x) e^{-i\omega_i t}+v_i^*(x) e^{i\omega_i t}] \;.$$ Inserting this ansatz in the TDGPE gives a system of equations: one time-independent GPE for $\psi_0$, and the two bogoliubov equations for $u_i$ and $v_i$.

Now my questions:

1) Is my above explanation of the procedure correct?

2) Sometimes it is mentioned that $\delta\psi(x,t)$ needs to be orthogonal to $\psi_0(x)$, to conserve the particle number. Therefore I would expect $\int dx (\psi(x)\delta\psi(x,t)) = 0$. If we ask for $\int dx \vert\psi(x,t)\vert^2 = 1$ and $\int dx \vert\psi_0(x,t)\vert^2 = 1$, to first order we find $\int dx (\psi(x)\delta\psi^\ast(x,t) + \psi^\ast(x)\delta\psi(x,t)) = 0$. How do I understand this as an orthogonality constraint? What will be the constraint on the $u_i$ and $v_i$?

3) Since excitations are expected to be bosonic quasiparticles, one wants bosonic commutation relations to be satisfied (this appears in second quantisation). It is known that this implies that $u_i$ and $v_i$ must obey $\int dx (u_i^2 - v_i^2) = 1$. Is there an argument to impose this in the approach I used above (without going to second quantisation)?