= Solution
Use $|j\rangle$ for the six printed basis vectors. The three singly occupied triplets are
$$
|T_+\rangle=|1\rangle,\qquad
|T_0\rangle=\frac{|2\rangle+|3\rangle}{\sqrt2},\qquad
|T_-\rangle=|4\rangle.
$$
Each has \b[<energy> zero]: either Pauli exclusion prevents hopping to <double occupancy> or the two contributions cancel. Define
$$
|S\rangle=\frac{|2\rangle-|3\rangle}{\sqrt2},\quad
|D_+\rangle=\frac{|5\rangle+|6\rangle}{\sqrt2},\quad
|D_-\rangle=\frac{|5\rangle-|6\rangle}{\sqrt2}.
$$
The state $D_-$ has <energy> $U$. In the $S,D_+$ subspace the <Hamiltonian> is $\bigl(\begin{smallmatrix}0&-2t\\-2t&U\end{smallmatrix}\bigr)$, with
$$
E_\pm=\frac{U\pm\sqrt{U^2+16t^2}}2.
$$
For $U\gg|t|>0$, the fourth low-energy eigenstate is the normalized singlet
$$
\boxed{|\psi_-\rangle=\mathcal N\left(|S\rangle-\frac{E_-}{2t}|D_+\rangle\right)
\simeq\mathcal N\left(|S\rangle+\frac{2t}{U}|D_+\rangle\right),}
$$
where $\mathcal N=[1+(E_-/(2t))^2]^{-1/2}$. Its <energy> is
$$
\boxed{E_-=-\frac{4t^2}{U}+O(t^4/U^3).}
$$
The \b[four lowest energies are $-4t^2/U,0,0,0$ to leading order], with the singlet and triplets just given. The upper pair lie near $U$ and $U+4t^2/U$.
The low-energy <Hamiltonian> can be written
$$
\boxed{H_{\rm eff}=J\left(\mathbf S_1\cdot\mathbf S_2-\frac14\right),\qquad
J=\frac{4t^2}{U}>0,}
$$
using dimensionless spin-half operators. It places the singlet below the triplet by $J$, hence favors antiferromagnetic alignment. A <lattice> of such bonds can support <antiferromagnetism>, although its ordering pattern also depends on geometry and competing couplings. The physical <energy> gain comes from virtual charge fluctuations, not from a permanent <double occupancy> in the large-$U$ limit.
\Image[/past-exam-of-the-advanced-quantum-condensed-matter-physics-course-of-the-university-of-cambridge/2015/paper-1-hubbard-spectrum.png]
{title=Exact singlet-triplet spectrum of the two-site repulsive Hubbard model and the large-U exchange scale}
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