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Rosa added Assuming_that_the_two_dots__.tex
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Assuming that the two dots feel a strong interdot Coulomb interaction then the total charge in the system can be only $0$ or $1$. The Hamiltonian for the two dots is
\begin{equation}
H_{i}=\sum_{i}\epsilon_i d^{\dagger}_{i} d_{i} +U n_{1}n_{2}
\end{equation}
Here, $d_i$ destroys an electron in dot $i$, with $i=\{1,2\}$. Here $U$ is the interdot Coulomb interaction.
Then, this fact allows us to define the pseudospin variable as
\begin{equation}
T_{z}=\frac{1}{2}(n_1-n_2)
\end{equation}
where $n=1$ and $n=2$ are the quantum dot occupations for the dot $1$ and for the dot $2$. Similarly we can define
\begin{equation}
T_{+}=d_{1}^\dagger d_2, \quad T_{-}=d_{2}^\dagger d_1
\end{equation}
and the respective $T_x=1/sqrt{2}(T_{+}+iT_{-})$ and $T_y=1/sqrt{2}(T_{+}+iT_{-})$.
For the leads we consider one-dimensional contacts modeled by a tight-biding chain. Again, we employ the fact that the spin degeneracy is shifted by
the presence of a magnetic field and we just keep the lead index $i=1,2$. We allow that at certain distance $L$ the two leads can be coupled. Then the model for the two leads reads
\begin{equation}
H_{W}=\sum_{i=1,2}\sum_{j=1}^{j=\infty} \epsilon_0 c_{ij}^\dagger c_{ij} + t \sum_{i}\sum_{j=1} c_{ij}^\dagger c_{ij+1} + \sum_{j=N}^{j=\infty}\Delta (c_{1j} c_{2j} + h.c)
\emd{equation}
