field, off resonant with the $F=1 \rightarrow F = 2$ transition, may be used to tune the energy levels of the $F=1$ hyperfine sublevels. In this sense, it plays the same role as the quadratic Zeeman shift. By changing the detuning, one can change the effective $q$ from positive ($E_{m_{F} = 0} < E_{m_{F} = \pm1}$) to negative ($E_{m_{F} = 0} > E_{m_{F} = \pm1}$), where $E$ denotes the energy of the hyperfine sublevels.

It is therefore of experimental and theoretical interest to consider what happens to the $m_{F} = 1, 0$ states at negative $q$. For the $m_{F} =1$ state, the spin mode becomes gapped by the quadratic Zeeman energy, and the thermodynamic instability disappears. This is to be expected as a negative $q$ favors a ground state with $\langle \textbf{S}\cdotp\textbf{S} \rangle = 1$.

The polar state however is more interesting. From (\[eq:7\]), it is clear that for ferromagnetic interactions ($c_{2} < 0$), the first term on the right hand side is always positive, and spectrum becomes gapped. The second term however is negative, signaling a finite wave-vector instability. Analyzing (\[eq:8\]) (see Fig.\[fig:-2\]), we find that increasing $|q|$ pushes the instability to larger wave-vectors, but frequency of the most 