rotating at its Larmor frequency is

$$\begin{aligned}
\label{eq:1}
{\cal H} = \int d\vec{\textbf{r}}~ \psi_{a}^{\dagger}\left(-\frac{1}{2m} + U( \vec{r}, t) + q \textbf{S}_{z} \cdotp \textbf{S}_{z}\right)\psi_{a} + \frac{c_{0}}{2}\psi^{\dagger}_{a}\psi_{a^{'}}^{\dagger}\psi_{a^{'}}\psi_{a} 
+ \frac{c_{2}}{2}\psi^{\dagger}_{a}\psi^{\dagger}_{a^{'}}\vec{\textbf{S}}_{ab}\cdotp \vec{\textbf{S}}_{a^{'}b^{'}}\psi_{b^{'}}\psi_{b}\end{aligned}$$

where $a = (-1, 0, 1)$, is the quantum number for the z-component of the spin, $\psi_{\sigma}(\textbf{r})$ is the field annihilation operator obeying bosonic commutation relations, and $U (\vec{r}, t)$ is the trapping potential. Here $\vec{\textbf{S}}$ denotes the dimensionless vector spin operator. Throughout, we set $\hbar = 1$. Boldface is used to denote matrices, and arrows denote vectors.

The interaction strengths, expressed in terms of the scattering lengths in the spin-$2$ and $0$ channel ($a_{2}, a_{0}$) are $c_{0} = 4\pi(a_{0} + 2a_{2})/3m$ and $c_{2} = 4\pi(a_{2} - a_{0})/3m$. A negative $c_{2}$ favors a ferromagnetic state with $\langle \vec{\textbf{S}} \rangle = 1$, while for positive $c_{2}$, the equilibrium state is one with $\langle \vec{\textbf{S}} \rangle = 0$, where all the atoms are in the $m_{F} = 0$ state, or simply an incoherent mixture.

Additionally, one considers the quadratic Zeeman effect $q \propto B^{2} $, which favors a state with $m_{F} = 0$ ($\langle \vec{\textbf{S}}\cdotp \vec{\textbf{S}} \rangle = 0$). Thus for a gas with negative $c_{2}$ such as $^{87}$Rb, the spin dependent contact interaction 