 Solution

Use the explicit convention $H=J_1{\sum }_{⟨ij⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j+J_2{\sum }_{⟨⟨ij⟩⟩}{\mathbf{S}}_i\cdot {\mathbf{S}}_j$, counting every bond once and taking dimensionless spins of length $S$. On the simple cubic lattice there are six nearest and twelve face-diagonal next-nearest neighbours.

For type I all six nearest bonds are antiparallel and all twelve next-nearest bonds parallel, so $E_I/N=S^2(-3J_1+6J_2)$. Choose the parallel direction of type II as $z$. It has two parallel and four antiparallel nearest bonds, and four parallel $xy$ face diagonals but eight antiparallel $xz,yz$ diagonals. Thus $E_{II}/N=S^2(-J_1-2J_2)$ andConsequently, for the stated antiferromagnetic exchanges,This comparison is also consistent with the full classical spiral energy. With $u_i=\cos (q_{i}a)$,It is affine in each $u_i\in [-1,1]$, so some minimum occurs at a cube vertex, giving the competing collinear states above. At the boundary, the coefficient of $u_z$ vanishes for $u_x=u_y=-1$, producing the larger degeneracy $(\pi /a,\pi /a,q_z)$ and permutations. It is not an isolated two-state crossing. The first panel of the model figure below shows the competing energies.