The Hamiltonian for Josephson Junctions

In the physics of nanostructures one often uses something like: $$hat{H} = -E_J coshat{varphi},$$ where $hat{varphi}=hat{varphi}_1 - hat{varphi}_2$ is the superconductor phase difference, related to the number of particles via the commutation relation $$[hat{varphi}_j, hat{N}_j] = -i.$$

In the same spirit as the tight-binding approach for atomic orbitals in a lattice, you assume that the spatial wavefunctions (let's call them $phi_{1,2}(mathbf{r})$ here) on either site of the barrier are approximately orthogonal,

$$ int phi_1^* phi_2 , dmathbf{r} simeq 0,$$

whereas the barrier allows particle to tunnel through and lifts this orthogonality

$$int phi_1^* V(mathbf{r}) phi_2 = K.$$

Integrating out the spatial coordinates from the Schrödinger equation,

$$i hbar dot{Psi} = [hat{H}_0 + V(mathbf{r})] Psi,$$

where the total wavefunction of the system is factorized as $Psi = psi_1 phi_1(mathbf{r}) + psi_1 phi_2(mathbf{r})$ and $psi_{1,2}$ are the probability coefficients (or particle coefficients if one considers superconductors or condensates). You then have,

$$ i hbar begin{pmatrix} dot{psi}_1 \ dot{psi}_2 end{pmatrix} = begin{pmatrix} U_1 & K \ K & U_2 end{pmatrix} begin{pmatrix} psi_1 \ psi_2 end{pmatrix}.$$

Here, $U_{1,2}$ are the bulk energies (or un-perturbed energies) of the wavefunction at each side.