Feshbach resonance and the Gross-Pitaevskii equation

Question

I have a question about Feshbach resonance and the (generalized) Gross-Pitaevskii equation. If we consider a BEC at a finite temperature $T_{BEC}$ and a static mean-field thermal cloud at temperature $T\geq 0.4T_{BEC}$, then the equation of motion for the expectation value of the macroscopic wave function of the Bose gas, $\Phi(\vec{r},t)$, is

\begin{align} i\hbar\frac{\partial\Phi}{\partial t}=\left[-\frac{\hbar^2}{2m}\nabla^2+V_{trap}(\vec{r})+gn_c(\vec{r},t)+2g\tilde{n}(\vec{r},t)-iR(\vec{r},t)\right]\Phi\ , \end{align}

where $V_{trap}$ is the trapping potential, $n_c$ is the condensate density, $\tilde{n}$ is the noncondensate density, $R\sim g^2$ describes damping, $g=\dfrac{4\pi\hbar^2 a}{m}$, $a$ is the $s$-wave scattering length, and $m$ is the mass of the atoms in the BEC.

I know that Feshbach resonance changes the scattering length, such that \begin{align} a=a_{bg}\left(1-\frac{\Delta}{B-B_{peak}}\right)\ , \end{align} where $a_{bg}$ is the asymptotic background scattering length, $B$ is the magnetic field, $B_{peak}$ is the strength of the magnetic field on resonance, $\Delta=B_{zero}-B_{peak}$, and $B_{zero}$ is the strength of the magnetic field that gives $a=0$.

If I want to account for the effects of Feshbach resonance in the Gross-Pitaevksii equation, am I allowed to directly input the second formula into the first equation? Or do I need to include additional terms within my equation of motion? My concern is that when the magnetic field is chosen so that $a=0$, I will no longer have contributions arising from the condensate and non-condensate densities or damping. What do I do in the case that $a\to\pm\infty$?

Answer 1

• The GPE is a mean field theory. This means that the interaction term, the one where $g$ appears, is taken as an average over all states. Basically $\text{term} = \int \mathrm{d}k \,\psi_k^{\dagger} \,\text{operator}\, \psi_k$, which makes the maths easier and more tractable.
  Mean field theories, hence, always "leave terms out" - these are usually referred to as quantum fluctuations. It is not known a priori whether you have to include or not to explain your phenomena. From experience, the BEC transition is also explained within mean field theory. But the Mott transition is not. Quantum liquid droplets are stabilised by the Lee-Huang-Yang term, which is basically the quantum fluctuations beyond mean field theory.

• So yes, you can just plug the formula for $a$ in. But that will only give you "correct" answers for low $a$. At very high interactions, $a > 300 a_0$ or so let's say, you are in the strongly interacting regime, where more stuff may happen and you might need beyond mean field theory corrections. After all, the Mott transition that I mentioned before is driven by strong interactions.

• $a=0$ is the special case where your problem reduces to single-particle physics. So it makes sense there is no damping, and no repulsive potential from $n_c$ and $n$. It is a special case, and it makes sense. A BEC happens even at $a=0$, i.e. no interactions, because it is a single particle effect. $T_c$ is shifted with $a$, and more many-body phenomena happen, like repulsive potentials and damping etc.

• Experimentally, $a=0$ is used specifically to measure theoretically tractable clouds. Also, high $a$ also means high loss rates ($\propto n^2 a^4$, $n$ being density), so you would not have too many atoms left near the Feshbach resonance. Looking at the loss feature is actually one of the ways to find where the resonance is, along with looking (for instance) at the damping of the Rabi oscillation (caused by your $R$ term). Hence why strongly interacting regimes (high $a$) are hard to do experimentally with cold atoms in free space or in large traps, because the loss rate increases. Hence why the realm of optical lattices offers a greater chance of investigating strong interactions, because you can have low densities on-site, but high neighbour-neighbour couplings.