Double Potential Well (Hubbard Model) The Hamiltonian from the Hubbard model for the double well potential $V(x) = V_0 \frac{x^2 - q^2}{q^2}$ is given by
\begin{equation}
H = -J(a_L^\dagger a_R +  a_R^\dagger a_L) + \frac{U}{2} (a_L^\dagger a_L^\dagger a_L a_L + a_R^\dagger a_R^\dagger a_R a_R)
\end{equation}
corresponding to the left side and right side respectively.
I am very new to second quantisation and some basic concepts arn't making sense to me. In this example, how would I find the ground state energy of a particle in either side of the wells?
 A: Double well potential
I think the double well potential should read
$$
V(x) = V_0\left(\frac{x^2 - q^2}{q^2}\right)^2 = 
V_0\frac{(x - q)^2(x+q)^2}{q^4},
$$
i.e., it must be a polynomial of power $4$, that is approximately a parabolic potential near points $x = \pm q$:
$$
V(x)|_{x\approx \pm q} = V_\pm(x) = V_0\frac{4(x \mp q)^2}{q^2}.
$$
Second quantization
The second quantization is formally done by expanding the field operator in terms of the preferred system of eigenstates
$$
\hat{\psi}(x) = \sum_k \hat{a}_k\phi_k(x), \hat{\psi}^\dagger(x) = \sum_k \hat{a}_k^\dagger\phi_k^*(x),
$$
and the evaluating the second quantized Hamiltonian as
$$
\hat{\mathcal{H}} = \int dx \hat{\psi}^\dagger(x)H\hat{\psi}(x) = \sum_{k,q}H_{kq}\hat{a}_k^\dagger\hat{a}_q.
$$
Transfer Hamiltonian
In this case however the expansion done in an approximate basis, taking as the basis the ground states in the two wells, i.e. in the approximate potentials $V_\pm(x)$, neglecting overlap of these states and all the other states:
$$
\hat{\psi}(x) = \hat{a}_L\phi_-(x) + \hat{a}_R\phi_+(x), 
$$
and defining $H_{--} = H_{++} = 0$, and $H_{+-} = H_{++} = -J,$ which results in
$$
\hat{\mathcal{H}} = -J\left(\hat{a}_L^\dagger\hat{a}_R + \hat{a}_R^\dagger\hat{a}_L\right).
$$
In fact, since we use a defective basis, the value of exchange integral $J$ estimated as simply the matrix element between $\phi_\pm$ is incorrect. Although there exist techniques to correct it, taking account for overlap, most of the time its value would be inferred empirically (if one actually needs to get down to the numbers, which is often not the case when using the Hubbard model).
Coulomb term
The Coulomb term is actually the sum of the one-site Coulomb terms, neglecting the interaction between the two wells. It is obtained using the prescription cited above adopted for two-particle operators:
$$
\hat{\mathcal{H_C}} = \frac{1}{2}\int dx dx' \hat{\psi}^\dagger(x)\hat{\psi}^\dagger(x')H\hat{\psi}(x')\hat{\psi}(x),
$$
and using the fact that $\hat{a}_{L,R}^2=0$ and, as mentioned above, neglecting the cross terms. Special attention has to be paid to the order of the operators (see the order of $x$ and $x'$).
I think in terms of second quantization this is pretty much a homework problem, but the transfer part is somewhat non-trivial.
