Ground state energy

Within the formalism of the mean-field theory it is easy to obtain the ground state energy from the stationary solution of the Gross-Pitaevskii equation (1.11). To this purpose the condensate wave function should be written as $\Phi({\bf r},t) = \phi({\bf r}) \exp(-i\mu t/\hbar)$, where $\mu$ is the chemical potential and the function $\phi({\bf r})$ is real and normalized to the total number of particles $\int\vert\phi({\bf r})\vert^2 {\bf dr} = N$. Then the Gross-Pitaevskii equation becomes

$$\left(-\frac{\hbar^2}{2m}\triangle+V({\bf r})+g\phi^2({\bf r})\right)\phi({\bf r}) = \mu \phi({\bf r})$$ (1.12)

It has the form of a nonlinear Schrödinger equation. In the absence of interactions ($g = 0$) it reduces to the usual single-particle Schrödinger equation with the Hamiltonian $-\hbar^2\triangle/2m + V({\bf r})$. In the uniform case, $V = 0$, $\phi({\bf r})$ is a constant $\phi({\bf r}) = \sqrt{n}$ and the kinetic term in () disappears. The chemical potential is given by

$$\mu = gn$$ (1.13)

At $T = 0$ the chemical potential is the derivative of the energy with respect to the number of particles $\mu = \partial E / \partial N$. Substitution of (1.2) into (1.13) and simple integration gives the ground state energy per particle

$$\frac{E}{N} = 4\pi(na^3)\frac{\hbar^2}{2ma^2}$$ (1.14)