 Solution

For $|w|\gg |\Delta |$, the mixed states approach opposite-spin product states andThe omitted mixed-branch correction is of order ${\Delta }^2/|w|$. The four levels are generically nondegenerate, apart from special moment choices or crossings.

For equal-and-opposite signed moments, $g_2=-g_1$, $u=0$ and the parallel-spin level remains exactly twofold degenerate at $\Delta /4$; the other two approach $-\Delta /4\pm |g_1{\mu }_{0}B|$. Equal moment magnitudes are an extra assumption, not implied by opposite charges alone. If $g_1=g_2$, the differential Zeeman term vanishes and the singlet/triplet-zero block stays unmixed by the field. This is the strong-field limit of a two-spin exchange spectrum.