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From the Gross-Pitaevskii equation \begin{equation}i\hbar\frac{\partial\psi}{\partial t}=\left(-\frac{\hbar^2}{2m}\nabla^2+V+g|\psi|^2\right)\psi\end{equation} using the variational relation \begin{equation}i\hbar\frac{\partial\psi}{\partial t}=\frac{\partial\varepsilon}{\partial \psi^*}\end{equation} we find the energy density \begin{equation}\varepsilon=\frac{\hbar^2}{2m}|\nabla\psi|^2+V|\psi|^2+\frac{g}{2}|\psi|^4\end{equation} The energy would be $E=\int d^3r \varepsilon$ and this is a prime integral of the motion, meaning it is a conserved quantity.

My questions are:

1) How do we get the variational relation?

2)How can we prove that $E$ is a conserved quantity?

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    $\begingroup$ 1) You may read the chapter 7 of Pethick and Smith's book on BEC. 2) $\epsilon$ is energy density which is not conserve, only the total energy $\endgroup$
    – unsym
    Commented Jul 9, 2014 at 23:24
  • $\begingroup$ 1) Even if I read that book already I didn't remember that. Thank you. 2) Of course you are right, my mistake; I will amend the question $\endgroup$
    – Semola
    Commented Jul 9, 2014 at 23:37
  • $\begingroup$ Do you know a nice way to show the conservation of $ E$ different from brute force computation? $\endgroup$
    – Semola
    Commented Jul 9, 2014 at 23:41
  • $\begingroup$ After an integration by parts, $\epsilon$ is almost the operator that gives the time-evolution, and one could perhaps try to copy the proof of the Ehrenfest theorem. $\endgroup$ Commented Jul 10, 2014 at 0:35
  • $\begingroup$ I think it won't work because the exponential would depend on $|\psi|^2$ as well. $\endgroup$
    – Semola
    Commented Jul 11, 2014 at 13:40

2 Answers 2

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The answer to the second question is actually quite straightforward:

by computing $\partial_t E$ and using what given in the Gross-Pitaevskii for $\dot{\psi}$ and $\dot{\psi^*}$ one can check that all the terms cancel out so that $\partial_tE=0$.

To derive the expression for the energy one could also start considering a lagrangian giving the Gross-Pitaevskii when thrown inside the Euler-Lagrange machinery and derive the Hamiltonian in the usual way, so that one can go around the variational relation.

However it would be still interesting to know where that variational relation is coming from.

Does anyone have an idea? Also I am sure there must be a more elegant way to show that the energy is conserved, something less dumb than forced term by term computation.

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    $\begingroup$ Do you think Dirac-Frenkel variational principle (which can be derived as a saddle point approximation for the coherent state Feynman path integral) with the wave functional ansatz $$|\{\psi(\bar{r},t)\}>=e^{\int_{V}d^3_{}\bar{r}\psi(\bar{r},t)\Psi_{}^{\dagger}(\bar{r})}|\ \text{vac}>$$ will lead to the Gross-Pitaevskii equation? $\endgroup$
    – Sunyam
    Commented Oct 6, 2017 at 15:16
  • $\begingroup$ maybe this can help physics.stackexchange.com/q/623285/226902 $\endgroup$
    – Quillo
    Commented Apr 7, 2021 at 19:35
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here is a way to generally prove the energy functional is a conserved quantity for a dynamical system. To make the notation more comfortable, I would like to use discrete modes to illustrate. Suppose we have a system whose state is characterized by a set of complex numbers c1,c2,... , its dynamics governed by $i\partial_t c_i=h_i(c_1,c_1^*,...)$, i=1,2,3...
If we can find a function $E(c_1,c_1^*,...)$ satisfying
$\frac{\partial E}{\partial c_i^*}=h_i(c_1,c_1^*,...)=i\partial_t c_i$ for all i, and E being real
Then clearly E is a conserved quantity through evolution of the system, because
$\frac{\partial E}{\partial t}=\sum_i \frac{\partial E}{\partial c_i^*}\frac{\partial c_i^*}{\partial t}+\frac{\partial E}{\partial c_i}\frac{\partial c_i}{\partial t}$
$=\sum_i i\frac{\partial c_i}{\partial t} \frac{\partial c_i^*}{\partial t}-i\frac{\partial c_i^*}{\partial t}\frac{\partial c_i}{\partial t}=0$
In terms of GPE, coefficient of discrete modes $c_i(t)$ becomes wave function at a spatial point $\psi(\vec{r},t)$.

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