 Solution

Fix canonical single-particle order $(1\uparrow ,1\downarrow ,2\uparrow ,2\downarrow )$ and write the occupation ket as $|n_1{n}_2{n}_3{n}_4⟩$. The six states in the printed order are $|1010⟩$, $|1001⟩$, $|0110⟩$, $|0101⟩$, $|1100⟩$, $|0011⟩$. The last state is double occupancy on site 2.

For a fermionic raising and lowering operator acting on mode $j$, the sign is $(-1{)}^{{\sum }_{k<j}{n}_k}$, and annihilation requires that mode occupied while creation requires it empty. Starting from the second basis state, the hop that reaches the last state isIt therefore contributes $H_{6,2}=-t$. Starting from the third state instead,because the occupied $2\uparrow$ mode precedes $2\downarrow$. The Hamiltonian's leading $-t$ makes $H_{6,3}=+t$. Finally $n_{2\uparrow }{n}_{2\downarrow }|0011⟩=|0011⟩$, so $H_{6,6}=U$. No other displayed basis state reaches $|0011⟩$ in one allowed hop. These signs verify the three nonzero entries without assigning distinguishable-particle hopping signs to fermions.