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.

Added work(with the suggestion from the comments):

$$\hat{H} \quad \text{by parts}=- \frac{\hbar ^2}{2m}\int (\hat{\psi^{\dagger} \nabla ^2 \hat{\psi}})dr+\frac{1}{2} \int \hat{\psi^{\dagger}} \hat{\psi^{\dagger '}} V(r'-r) \hat{\psi} \hat{\psi '} dr'dr $$ 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)


2 Answers 2


Let us calculate each term at a time. The first thing we will take care of is the kinetic term. For this we partial integrate it and neglect the boundary term, which will appear.

$ \int_{\Omega} dx (\nabla_x \Psi^\dagger(x,t) ) (\nabla_x \Psi(x,t)) = [\Psi^\dagger(x,t) \nabla_x \Psi(x,t)]_{\delta \Omega} - \int_{\Omega} dx \Psi^\dagger(x,t) (\Delta_x \Psi(x,t)) $

The next observation is that $[\Psi(r,t),(\Delta_x \Psi(x,t))] =0$

With this we can calculate the kinetic term using $[A,BC]= B[A,C] + [A,B]C $.

$ [\Psi(r,t), \int dx (\nabla_x \Psi^\dagger(x,t) ) (\nabla_x \Psi(x,t))] =-[\Psi(r,t), \int dx \Psi^\dagger(x,t) (\Delta_x \Psi(x,t)) ] \\ =- \int dx [\Psi(r,t),\Psi^\dagger(x,t) ] (\Delta_x \Psi(x,t)) =- \int dx \delta(r-x) (\Delta_x \Psi(x,t)) = - (\Delta_r \Psi(r,t)) $

The term $V_{ext}$ needs to be added to the quadratic part of the Hamiltonian otherwise it will not appear in the GP-eq. Calculating its contribution is straight forward.

Let us now turn to the interacting part and look at the operator content of it.

$ [\Psi(r,t), \Psi^\dagger(x,t) \Psi^\dagger(y,t) \Psi(y,t) \Psi(x,t)] \\ = \Psi^\dagger(x,t) \Psi^\dagger(y,t) [\Psi(r,t),\Psi(y,t) \Psi(x,t)] + [\Psi(r,t), \Psi^\dagger(x,t) \Psi^\dagger(y,t) ] \Psi(y,t) \Psi(x,t) \\ =\Psi^\dagger(x,t) [\Psi(r,t), \Psi^\dagger(y,t) ] \Psi(y,t) \Psi(x,t) + [\Psi(r,t), \Psi^\dagger(x,t)] \Psi^\dagger(y,t) \Psi(y,t) \Psi(x,t) \\ = \Psi^\dagger(x,t) \Psi(y,t) \Psi(x,t) \delta(r-y) + \Psi^\dagger(y,t) \Psi(y,t) \Psi(x,t)\delta(r-x) $

Therefore we can write

$ [\Psi(r,t), \frac{1}{2}\int dx \int dy \Psi^\dagger(x,t) \Psi^\dagger(y,t) \Psi(y,t) \Psi(x,t) ]\\ =\frac{1}{2} \int dx \int dy V(x-y) [ \Psi^\dagger(x,t) \Psi(y,t) \Psi(x,t) \delta(r-y) + \Psi^\dagger(y,t) \Psi(y,t) \Psi(x,t)\delta(r-x) ] \\ = \frac{1}{2}\int dx V(x-r) \Psi^\dagger(x,t) \Psi(x,t) \Psi(r,t) + \frac{1}{2}\int dy V(r-y) \Psi^\dagger(y,t) \Psi(y,t) \Psi(r,t)\\ = \frac{1}{2}\int dx [V(x-r)+V(r-x)] \Psi^\dagger(x,t) \Psi(x,t) \Psi(r,t) $

If we assume that $V(x)=V(-x)$ we get

$ [\Psi(r,t), \frac{1}{2}\int dx \int dy \Psi^\dagger(x,t) \Psi^\dagger(y,t) \Psi(y,t) \Psi(x,t) ] = \int dx V(x-r) \Psi^\dagger(x,t) \Psi(x,t) \Psi(r,t) $

Collecting all terms for the eqm. we find

$ i\hbar \partial_t \Psi(r,t) = [ -\frac{\hbar^2 \Delta}{2m} + \int dx V(x-r) \Psi^\dagger(x,t) \Psi(x,t) ] \Psi(r,t) $

  • $\begingroup$ thankyou very much, this is an excellent answer! $\endgroup$
    – user352879
    Nov 19, 2020 at 7:37
  • $\begingroup$ What are $x$ and $y$ for the interacting part? $x=r$ and $y=r'$? $\endgroup$ May 12, 2023 at 18:55

Use the equal-time non-relativistic Bose-field commutator $$ [\psi({\bf x}),\psi^\dagger({\bf x}')]= \delta^3({\bf x}-{\bf x}'). $$

  • $\begingroup$ Integrate by parts to get the $\nabla$ off the $\psi^\dagger$, then use $[A,BC]= B[A,C]+ [A,B]C$, $\endgroup$
    – mike stone
    Apr 14, 2020 at 19:19
  • $\begingroup$ Okay great I will try that and post back with any questions thankyou $\endgroup$
    – user352879
    Apr 14, 2020 at 19:25
  • $\begingroup$ updated with the hints, but still confused on how to apply the recommended identity $\endgroup$
    – user352879
    Apr 14, 2020 at 20:19
  • $\begingroup$ So group your terms in commutators and use the formula I gave you. $\endgroup$
    – mike stone
    Apr 14, 2020 at 21:11
  • $\begingroup$ I am confused on what my "B" and "C" are, and how to group them into commutators exactly $\endgroup$
    – user352879
    Apr 14, 2020 at 21:20

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