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Let $R=\sqrt{J^2+D^2}$ and $k_0=\arctan (D/J)$. Since $J\cos k+D\sin k=R\cos (k-k_0)$, the continuum minimum of the dispersion relation isFor $D>0$, the energy gap closes at a positive $u_c$, at wave vector $k_0$. Below it, the quadratic expansion about the polarized state has a negative mode and is no longer a stable ground state approximation; higher-order terms and a different ordered reference state are needed.

Two qualifications are required by the actual assumptions. First, $D=0$ is permitted, in which case $u_c=0$; it is a concrete counterexample to the printed strict inequality $u_c>0$. Second, a finite ring only has discrete momenta. Its finite-ring spin-wave threshold with antisymmetric exchange isThis approaches $u_c$ in the thermodynamic limit and equals it when $k_0$ is allowed. For example, $L=4$ and $0<D<J$ give $u_c(L)=0$, since the maximum of $1,D/J,-1,-D/J$ is one. Thus the continuum formula and its positive threshold require the large-ring interpretation. The two symmetry-related polarized ground states already have equal exact energy; the calculated energy gap refers to excitations about one chosen orientation.