Double Potential Well (Hubbard Model)

Double well potential
I think the double well potential should read $$ V(x) = V_0left(frac{x^2 - q^2}{q^2}right)^2 = V_0frac{(x - q)^2(x+q)^2}{q^4}, $$ i.e., it must be a polynomial of power $4$, that is approximately a parabolic potential near points $x = pm q$: $$ V(x)|_{xapprox pm q} = V_pm(x) = V_0frac{4(x mp q)^2}{q^2}. $$

Second quantization
The second quantization is formally done by expanding the field operator in terms of the preferred system of eigenstates $$ hat{psi}(x) = sum_k hat{a}_kphi_k(x), hat{psi}^dagger(x) = sum_k hat{a}_k^daggerphi_k^*(x), $$ and the evaluating the second quantized Hamiltonian as $$ hat{mathcal{H}} = int dx hat{psi}^dagger(x)Hhat{psi}(x) = sum_{k,q}H_{kq}hat{a}_k^daggerhat{a}_q. $$

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: $$ hat{psi}(x) = hat{a}_Lphi_-(x) + hat{a}_Rphi_+(x), $$ and defining $H_{--} = H_{++} = 0$, and $H_{+-} = H_{++} = -J,$ which results in $$ hat{mathcal{H}} = -Jleft(hat{a}_L^daggerhat{a}_R + hat{a}_R^daggerhat{a}_Lright). $$ 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: $$ hat{mathcal{H_C}} = frac{1}{2}int dx dx' hat{psi}^dagger(x)hat{psi}^dagger(x')Hhat{psi}(x')hat{psi}(x), $$ 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.