How do I find the exchange energy from the effective Hamiltonian?

I've just recently started studying noise in singlet-triplet qubits and the exchange interaction (as a function of detuning $${\epsilon}$$ and tunnel coupling $${t_c}$$) seems fundamental in many studies. With reference to some recent literature (1, 2), the form of exchange energy should be proportionate to $$J({\epsilon},t_c) \approx {\frac{2t_c^2}{\left|U-{\epsilon_0}\right|}}$$ in the far-detuned regime.

To better understand this, I constructed the Hubbard Hamiltonian in the {$$S_{11},T_0,T_+,T_-,S_{20}$$} basis with the form:

$${H=H_{\epsilon}+H_{t_c}+H_U+H_Z}$$

where $$H_{\epsilon}={-\epsilon_i}\sum_{i,{\sigma}} c^\dagger_{i,{\sigma}}c_{i,{\sigma}}$$, $$H_{t_c}=t_c\sum_{i,j,{\sigma}} c^\dagger_{i,{\sigma}}c_{j,{\sigma}}$$ for $$i \neq j$$, $$H_U = \sum_{i} U_i n_{i\uparrow} n_{i\downarrow}$$, and $$H_Z = \sum_{i} {\frac{E_{Z_i}}{2} (n_{i\uparrow}-n_{i\downarrow})}$$ is the Zeeman term.

Doing the Schrieffer-Wolff transformation leaves me with the below effective Hamiltonian:

$$H_{\text{eff}}= \begin{bmatrix} {-\frac{4t_c^2}{U-{\epsilon}}} & {\frac{{\delta}E_Z}{2}} \\ {\frac{{\delta}E_Z}{2}} & 0 \\ \end{bmatrix}$$

where $${{\epsilon}={\epsilon_1}-{\epsilon_2}}$$ and $${{\delta}E_Z=E_{Z_1}-E_{Z_2}}$$.

From what I (think I) understand, the exchange energy $$J$$ should then be $$E_{\text{ex}}=E_{\text{singlet}}-E_{\text{triplet}}={-\frac{4t_c^2}{U-{\epsilon}}}-0=\left({-\frac{4t_c^2}{U-{\epsilon}}}\right)$$

Is there an explanation for the extra factor of 2 I've ended up with, or have I calculated the exchange energy wrongly?

Thanks everyone!

E.g., $$\hat{S}_i^z=\frac{\hbar}{2} \begin{bmatrix}c_{i,\uparrow}^\dagger & c_{i,\downarrow}^\dagger\end{bmatrix}\hat{\sigma}_z\begin{bmatrix}c_{i,\uparrow} \\ c_{i,\downarrow}\end{bmatrix}= \frac{\hbar}{2}\begin{bmatrix}c_{i,\uparrow}^\dagger & c_{i,\downarrow}^\dagger\end{bmatrix} \begin{bmatrix}1&0\\ 0&-1\end{bmatrix} \begin{bmatrix}c_{i,\uparrow} \\ c_{i,\downarrow}\end{bmatrix}= \frac{\hbar}{2}\left(c_{i,\uparrow}^\dagger c_{i,\uparrow} - c_{i,\downarrow}^\dagger c_{i,\downarrow}\right)$$