I need to find the approximate solution of nonlinear Schrodinger equation $$ i\hbar \partial_{t} \Psi + \frac{\hbar^{2}}{2m}\Delta \Psi - g |\Psi|^{2}\Psi - \frac{m\omega^2 (x^2 + y^2 + z^2)}{2}\Psi = 0 $$ by using variational method. $\Psi$ has norm $\int |\Psi|^{2}d^{3}\mathbf r = N$, where N is the number of particles.

It was saying to me that I must start from the trial function $$ \Psi (x, y, z) = h e^{-\frac{1}{2\omega^{2}}\left( x^2 + y^2 + z^2\right)}e^{\frac{i}{\hbar}\mu t}, $$

where $\mu$ plays role of chemical potential.

So by using the property of $\Psi$ I got $h = \frac{\sqrt{N}}{(\omega \pi )^{\frac{3}{4}}}$. Then, due to the variational method, I must compute functional $J(N) = \langle \Psi | \hat {H}| \Psi\rangle$ and determine N from relation $\frac{\partial J (N)}{\partial N} = 0$. But I don't understand this step, because $N$ already is given in the task and equal to the integer number. So I don't have parameters in the trial function which can minimize functional $J(N)$.

Can you help me? Maybe, there is the mistake in the trial function and it must be some parameter $l^2$ instead of $\omega^{2}$?


1 Answer 1


Minimizing the energy with respect to $N$ will give a relationship between $\mu$, $N$, and the other parameters of the system. This will fix the chemical potential to something like $\mu=gN$ (this is the correct value in the limit $\omega\to0$).

For sure the trial wave-function has a problem since $x^2/\omega^2$ is not dimensionless, but it seems quite obvious that you should just use the typical length scale given by the harmonic oscillator (that you called $l$, if I guess correctly).

  • $\begingroup$ Thank you. So you meaned that I get from $\langle \Psi |\hat {H}| \Psi \rangle $ expression $J(N, l)$ from the one side and $\mu $ from the another, which leads to fixing $\mu$? Is this step all that I must to do? $\endgroup$ Dec 4, 2013 at 20:00
  • $\begingroup$ @AndrewMcAdams: because you're looking for an eigenstate of $\hat H$, you have $\langle \Psi|\hat H|\Psi\rangle/\langle \Psi|\Psi\rangle=\mu$, and you want to minimize this to get the ground-state (or to as close to it as possible with this ansatz). $\endgroup$
    – Adam
    Dec 4, 2013 at 20:33

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