# Why does the chemical potential show up in the Gross-Pitaevskii equation?

In the case of a weakly interacting Bose-Einstein condensate the Gross-Pitaevskii equation is the motion equation that governs the evolution of the N-particle wavefunction. Starting from the Time-Dependent Gross-Pitaevskii equation: $$i\hbar\dot{\Psi}(x,t) = \left(-\frac{\hbar^2}{2m}\nabla^2_x + V(x,t) + g|\Psi(x,t)|^2 \right) \Psi(x,t)$$

Assuming the system to be time-independent the solution can be expressed as: $$\Psi(x,t) = \psi(x) \phi(t)$$, where, both fields satisfy: $$i\hbar\dot{\phi(t)} = -\mu \phi(t)$$ $$\mu\psi(x) = \left(-\frac{\hbar^2}{2m}\nabla^2_x + V(x,t) + g|\psi(x)|^2 \right)\psi(x)$$

My question is very simple but I don't see the point: how do we know that the constant of the separation of variables method corresponds to the chemical potential? Is related with the fact that $$N=\int dx |\psi(x)|^2$$ ?

• Commented Sep 10, 2022 at 15:23

## 1 Answer

Energy eigenstates are stationary states, so their time dependence is only contained in a phase factor.

The GPE described a BEC at $$T=0$$, so all atoms in one state with one energy. That energy is the ground state energy $$E_0$$ which is equal to the chemical potential $$E_0 = \mu$$. This ensures that the ground state has infinite occupancy because: $$f(E) = \frac{1}{\mathrm{e}^{\frac{E-\mu}{k T}}-1}, \quad\text{so} \quad f(E_0) \rightarrow \infty \quad \text{as}\quad E_0 \rightarrow \mu.$$

Hence, the phase factor of the energy eigenstate of the GPE will just be: $$\phi(t) = \exp \left ( \frac{-\mathrm{i}E_0t}{\hbar} \right ) = \exp \left ( \frac{-\mathrm{i}\mu t}{\hbar} \right ).$$

Substituting the latter equation into the time-dependent GPE gives you the time-independent equation.