 Solution

For spin one-half, $s_{\pm }=s_x\pm i{s}_y=(\hslash /2)({\sigma }_x\pm i{\sigma }_y)$, henceThey raise or lower the spin projection by one unit of $\hslash$ and annihilate the corresponding endpoint. The exchange identity ${\mathbf{\sigma }}_1\cdot {\mathbf{\sigma }}_2={\sigma }_{1z}{\sigma }_{2z}+2({\sigma }_{1+}{\sigma }_{2-}+{\sigma }_{1-}{\sigma }_{2+})$ preserves total spin projection.

Write $\Delta =\Delta E$, $u={\mu }_{0}B(g_1+g_2)$ and $w={\mu }_{0}B(g_1-g_2)$. In the listed order $(\uparrow \uparrow ,\downarrow \downarrow ,\uparrow \downarrow ,\downarrow \uparrow )$, direct action givesThe printed matrix instead uses $(\uparrow \uparrow ,\uparrow \downarrow ,\downarrow \uparrow ,\downarrow \downarrow )$. Permuting the listed indices to $(1,3,4,2)$ recovers its central block and diagonal entries. This is a basis-order mismatch in the source, not a different Hamiltonian.

The two-spin exchange spectrum with unequal signed magnetic moments isUse the products $g_j{\mu }_0$ as the signed magnetic coefficients in the stated Hamiltonian. Opposite physical charges must be represented in those signs; they do not by themselves imply equal moment magnitudes. Here ${\mu }_0$ denotes the source's magneton, not vacuum permeability.