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

Question

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!

Answer 1

Firstly, you need to finish the canonical transformation by writing the second quantized Hamiltonian in the new basis. Then you can expressed the spin operators in terms of the creation/annihilation operators (i.e., write their second quantized operators using the usual prescription) and move on to Heisenberg-like spin Hamiltonian, where the exchange interaction will be obvious.

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