= Solution
Write $|\uparrow\rangle,|\downarrow\rangle$ for the two $S_3$ <eigenstates>, of <eigenvalues> $\pm\hbar/2$. The <spin-spin interaction> can be evaluated using
$$
\mathbf S^{(1)}\cdot\mathbf S^{(2)}=S_3^{(1)}S_3^{(2)}+\frac12\left(S_+^{(1)}S_-^{(2)}+S_-^{(1)}S_+^{(2)}\right).
$$
For $|\beta|\gg|\alpha|$, take $H_0=\beta(S_3^{(1)}-S_3^{(2)})/\hbar$ and $V=\alpha\mathbf S^{(1)}\cdot\mathbf S^{(2)}/\hbar^2$. The unperturbed <energy eigenvalues> are $\beta,-\beta,0,0$ on $|\uparrow\downarrow\rangle,|\downarrow\uparrow\rangle,|\uparrow\uparrow\rangle,|\downarrow\downarrow\rangle$, respectively. For an isolated normalized <energy eigenstate> $|n\rangle$, <first-order nondegenerate perturbation theory> gives $\Delta E_n^{(1)}=\langle n|V|n\rangle$. The two antiparallel <spins> have expectation $-\alpha/4$, so
$$
\boxed{E_{\uparrow\downarrow}=\beta-\frac\alpha4+O(\alpha^2/|\beta|),\qquad E_{\downarrow\uparrow}=-\beta-\frac\alpha4+O(\alpha^2/|\beta|).}
$$
For the zero-<energy> <degenerate eigenspace>, <degenerate perturbation theory> requires diagonalizing the restriction of $V$. On the ordered <basis> $(|\uparrow\uparrow\rangle,|\downarrow\downarrow\rangle)$ it is $(\alpha/4)I$: the <ladder operators> cannot take one parallel state to the other. Indeed neither parallel state couples to anything else under $H$, so \b[the remaining two levels are exactly $\alpha/4$], retaining their twofold <quantum degeneracy>.
For $|\alpha|\gg|\beta|$, take $H_0=\alpha\mathbf S^{(1)}\cdot\mathbf S^{(2)}/\hbar^2$ and $V=\beta D$, where $D=(S_3^{(1)}-S_3^{(2)})/\hbar$. The <spin-one-half triplet state> <basis> and <spin-one-half singlet state> are
$$
|t_+\rangle=|\uparrow\uparrow\rangle,\quad |t_0\rangle=\frac{|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle}{\sqrt2},\quad |t_-\rangle=|\downarrow\downarrow\rangle,\quad |s\rangle=\frac{|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle}{\sqrt2}.
$$
The identity $\mathbf S^{(1)}\cdot\mathbf S^{(2)}=(\mathbf J^2-3\hbar^2/2)/2$ gives unperturbed <energy eigenvalues> $E_t^{(0)}=\alpha/4$ and $E_s^{(0)}=-3\alpha/4$. Direct action gives
$$
D|t_\pm\rangle=0,\qquad D|t_0\rangle=|s\rangle,\qquad D|s\rangle=|t_0\rangle.
$$
Consequently the restriction of $V$ to the entire triplet <degenerate eigenspace> is zero, so no first-order splitting is missed. The singlet expectation is also zero. In <second-order nondegenerate perturbation theory>, $\Delta E_n^{(2)}=\sum_{r\ne n}|\langle r|V|n\rangle|^2/(E_n^{(0)}-E_r^{(0)})$. Within the triplet, the corresponding second-order effective <matrix> is $\beta^2|t_0\rangle\langle t_0|/\alpha$, already diagonal in this <basis>. Thus, through second order,
$$
\boxed{E_{t_\pm}=\frac\alpha4,\qquad E_{t_0}=\frac\alpha4+\frac{\beta^2}{\alpha},\qquad E_s=-\frac{3\alpha}{4}-\frac{\beta^2}{\alpha}.}
$$
The last two expressions have higher corrections beginning at $O(\beta^4/|\alpha|^3)$, as the exact calculation in part (d) also verifies. For positive $\alpha$, coupling to the lower singlet raises the triplet's $M=0$ level and lowers the singlet; the signs are precisely the <energy> denominators above.
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