Help with the derivation of the Gross-Pitaevskii equation

I'm having some difficulty deriving the gross pitaevskii equation using Heisenberg's equation of motion. I know that the many body hamiltonian is given by $$\hat{H}=\int \left ( \frac{\hbar ^2}{2m} \nabla \hat{\psi^{ \dagger}} \nabla \hat{\psi}\right )d r+\frac{1}{2} \int \hat{\psi^{\dagger}} \hat{\psi^{\dagger '}} V(r'-r) \hat{\psi} \hat{\psi '} dr'dr$$

from which we can use heisenbergs equation $$i\hbar\frac{\partial}{\partial t} \hat{\psi}(r,t)= \left [ \hat{\psi}(r,t),\hat{H} \right ]$$

to somehow obtain $$i\hbar\frac{\partial}{\partial t} \hat{\psi}(r,t)= \left [ -\frac{\hbar^2 \nabla ^2}{2m} +V_{\text{ext}}(r,t)+\int \hat{\psi^{\dagger}(r',t)V(r'-r)\hat{\psi^{\dagger}})r',t)} \right ] \hat{\psi}(r,t)$$

I dont understand how you obtain this equation from heisenbergs formula. Every textbook i've seen on the subject has left it as an exercise or jumped straight to that result, and was unsure on how to derive it explicitly.

$$\hat{H} \quad \text{by parts}=- \frac{\hbar ^2}{2m}\int (\hat{\psi^{\dagger} \nabla ^2 \hat{\psi}})dr+\frac{1}{2} \int \hat{\psi^{\dagger}} \hat{\psi^{\dagger '}} V(r'-r) \hat{\psi} \hat{\psi '} dr'dr$$ This gives $$i \hbar \frac{d}{dt} \hat{\psi}(r,t)= \left [ \hat{\psi},\hat{ \mathcal{H}} \right ]$$ $$= \left [\hat{\psi}, \frac{\hbar ^2}{2m}\int (\hat{\psi^{\dagger} \nabla ^2 \hat{\psi}})d\vec{r}+\frac{1}{2} \int \hat{\psi^{\dagger}} \hat{\psi^{\dagger '}} V(r'-r) \hat{\psi} \hat{\psi '} d\vec{r'}d\vec{r}\right ]$$ $$=\frac{\hbar ^2}{2m}\left [\hat{\psi}, \int \left ( \hat{\psi^{\dagger} \nabla ^2 \hat{\psi}})d\vec{r} \right ) \right ]+\frac{1}{2} \left [\hat{\psi}, \int \vec{\psi^{\dagger}}\vec{\psi^{\dagger '}}V(\vec{r'}-\vec{r}) \hat{\psi}\hat{\psi '} d\vec{r'} \right ]$$

$$=\frac{\hbar^2}{2m} \left ( \int \psi^{\dagger}[\psi^{\dagger},\nabla {\psi}]+\delta(r-r') \nabla{\psi} \right )$$

$$+ \frac{1}{2} \left ( \psi^{\dagger}\psi{\dagger '}V(r-r') \left [\psi,\psi \psi' \right ]+ \left [\psi,\psi^{\dagger} \psi^{\dagger '}V(r-r') \right ] \right )$$ Firstly, i am confused as to what the result of $$\psi^{\dagger} [\psi^{\dagger},\nabla \psi ]$$ is. Secondly, i have reduced the last line using the identity $$[A,BC]=B[A,C]+[A,B]C$$ to: $$\frac{1}{2} \left ( \psi^{\dagger}\psi{\dagger '}V(r-r') \left [\psi,\psi \psi' \right ]+ \left [\psi,\psi^{\dagger} \psi^{\dagger '}V(r-r') \right ] \right )= \frac{1}{2} \left ( \psi^{\dagger} \psi^{\dagger '} V(r-r') \psi \delta(r'-r \right )+(\psi^{\dagger} [\psi,\psi^{\dagger '}]+[ \psi,\psi^{\dagger}]\psi^{\dagger '})\psi \psi '$$

again not sure how to do these commutators (assuming what ive done is correct so far)

Use the equal-time non-relativistic Bose-field commutator $$[\psi({\bf x}),\psi^\dagger({\bf x}')]= \delta^3({\bf x}-{\bf x}').$$
• Integrate by parts to get the $\nabla$ off the $\psi^\dagger$, then use $[A,BC]= B[A,C]+ [A,B]C$, – mike stone Apr 14 at 19:19