How does the used wavefunction look like for the mean-field approximation in BEC?

Question

In order to derive out Gross-Pitaevskii equation in BEC, mean field approximation is used.

$$ \hat{\psi}=\langle\hat{\psi}\rangle+\delta\hat{\psi}' $$

and $\langle\hat{\psi}\rangle$ is called as wave function of condensate. My question is how the field operator acts on the state for calculating the average.

Note: I use second quantization to build up the equation. (Look at the below derivation.)

The Hamiltonian of the condensate system is:

$$ \hat{H}=\int d\vec{r} \psi^{\dagger}{(\vec{r})}\Big[-\frac{\hbar^2}{2m} \nabla^2+V_{ext}(\vec{r})\Big]\psi{(\vec{r})}+\frac{1}{2}\int d\vec{r}d\vec{r}'\psi^{\dagger}{(\vec{r})}\psi^{\dagger}{(\vec{r}')}V(\vec{r}-\vec{r}')\psi{(\vec{r})}\psi{(\vec{r}')} $$

$\psi{(\vec{r})}=\sum_{i}\phi_i(\vec{r})a_i$ is a field operator. $\phi_i(\vec{r})$ is the eigenstate of single-particle Hamiltonian. Now in Heisenburg representation:

$$ i\hbar\frac{\partial}{\partial t}\psi(\vec{r})=[\psi(\vec{r}),H] $$

Now, $V=g\delta(\vec{r}-\vec{r}').$ So after the lenghty calculation, I get:

$$ i\hbar\frac{\partial}{\partial t}\psi(\vec{r})=\Big[-\frac{\hbar^2}{2m} \nabla^2+V_{ext}(\vec{r})\Big]\psi{(\vec{r})}+g\psi^{\dagger}{(\vec{r})}\psi{(\vec{r})}\psi{(\vec{r})} $$

Answer 1

In the mean-field approximation, we can write the ground state many-body wavefunction as

$|\mathrm{gs}\rangle \sim e^{\int dx \varphi(x)\hat{\psi}(x)}|0\rangle $

where $\hat{\psi}$ is the field operator, and $\varphi(x)$ is the "wavefunction", $|0\rangle$ is the vacuum state with no particle. You can then directly compute $\hat{\psi}(x)|\mathrm{gs}\rangle=\varphi(x)|\mathrm{gs}\rangle$.