# Chemical potential of a BEC

I know that the chemical potential of a BEC can be calculated with $$\mu=\frac{\partial E}{\partial N}$$, where $$E$$ is the energy and $$N$$ is the number of particles. For the Gross-Pitaevskii equation with a time-dependent potential $$V(x,t)$$,

\begin{align} i\hbar\dot\psi=-\frac{\hbar^2}{2m}\nabla^2\psi+V(x,t)\psi^2+g\psi^4\ , \end{align}

we have the energy

\begin{align} E=\int d^3x\left[-\frac{\hbar^2}{2m}\nabla^2\psi+V(x,t)\psi^2+g\psi^4\right] \end{align}

and $$N=\psi^2$$. Therefore, $$\mu=g|\phi|^2+V(x,t)$$. If $$V(x,t)=V(x)$$, the chemical potential is apparently a constant (according to (1.39) in this thesis), but I am unsure how to see that. Is $$\mu$$ still constant even with time-dependence incorporated into the potential, $$V$$? I know that I'm missing something, but I'm not sure what I'm missing.

• I'm not quite sure I'm able to follow your train of thought. If $\mu=g|\phi|^2+V(x,t)$ (but I doubt there should be an $x$ in there), then $\mu$ is clearly time dependent because of the potential. – noah May 10 at 22:54
• What does 'BEC' stand for? – Gert May 10 at 23:13