# Gross-Pitaevskii equation invariants

Having the dimensionless GPE to model a BEC with harmonic plus optical potential, given by:

$$i\frac{\partial{\psi}}{\partial{t}} = -\frac{1}{2}\nabla^2\psi(x,t) + V(x)\psi(x,t) + \frac{4\pi a_s N}{\sqrt{\frac{\hbar}{mw_x}}}|\psi(x,t)|^2\psi(x,t)$$

where $$V(x) = \frac{1}{2}\left(x^2+\left(\frac{w_y}{w_x}\right)^2y^2+\left(\frac{w_z}{w_x}\right)^2z^2\right)+\frac{2\pi^2\hbar}{mw_x\lambda_0^2}S_0\left(\sin^2\left(\frac{2\pi}{\lambda}\sqrt{\frac{\hbar}{mw_x}}x\right) + \sin^2\left(\frac{2\pi}{\lambda}\sqrt{\frac{\hbar}{mw_x}}y\right) + \sin^2\left(\frac{2\pi}{\lambda}\sqrt{\frac{\hbar}{mw_x}}z\right)\right)$$

I would like to prove how both the normalization and the energy are conserved. They are defined as:

$$N\left(\psi(\cdot,t)\right) = \int{|\psi(x,t)|^2 dx}$$

$$E\left(\psi(\cdot,t)\right) = \int{\left[\frac{1}{2}|\nabla\psi|^2 + V(x)|\psi|^2 + \frac{\beta}{2}|\psi|^4\right]dx}$$

In order to demonstrate that the energy is conserved I was taught about deriving $$~E~$$ with respect to $$~t~$$ and see that it equals $$~0~$$, using for that the GPE and cancel terms. However, I am not getting to the desire result.

Any help with that derivation?

It seems your plan is good. I think a problem is with the kinetic term. You can use integration by parts to get the cancellation desired. I mean the following equalities $$\int f\,\nabla^2 g\, dx= - \int \nabla f\, \nabla g\, dx = \int \nabla^2f\, g\, dx \quad (1)$$ Consider the conservation of normalization: $$i\frac{\partial N}{\partial t} = \int \left(i\frac{\partial\psi^*}{\partial t}\psi + \psi^* i\frac{\partial\psi}{\partial t}\right) dx =$$ $$= \int\left[-\left(-\frac12\nabla^2\psi^*+V\psi^*+\beta\psi^*|\psi|^2\right)\psi +\psi^*\left(-\frac12\nabla^2\psi+V\psi+\beta|\psi|^2\psi\right)\right]dx =$$ $$= \frac12\int \left(-\nabla^2\psi^*\,\psi + \psi^* \nabla^2\psi\right) dx$$ Now apply first of equalities (1) to both terms in the last integral, we obtain $$i\frac{\partial N}{\partial t} = \int\left(\nabla\psi^*\nabla\psi -\nabla \psi^*\nabla \psi\right)dx = 0$$
• @Sergio, in this case, integration by parts means using equalities like (1). There are terms with $\nabla$ in the $i\partial E/\partial t$ expression. You can try to make these terms more symmetric about $\psi$ and $\psi^*$. Something like the transformation of $\int \psi^*\nabla^2\psi\,dx$ to $-\int \nabla\psi^*\nabla\psi\,dx$. – Gec Feb 19 at 15:22
• thank you so much! Now able to cancel every term in the energy except the kinetic. How is the $|\nabla\psi|^2$ defined? I thought this: $|\nabla\psi|^2 = |\nabla\psi| \cdot |\nabla\psi| = (\frac{\partial \psi}{\partial x})^2 + (\frac{\partial \psi}{\partial t})^2$ but obviosly this must be wrong. Cal I take the $\nabla$ operator out of the abs. value without any problem? Or how should I take it out? Thank you so much once again. – Sergio Feb 20 at 2:56
• @Sergio, $|\nabla\psi|^2\equiv \nabla\psi^*\nabla\psi$. – Gec Feb 20 at 5:24