There are many good questions in the site about the microscopic derivation of the Josephson equations in a superconductor-insulator-superconductor system, see for instance this question and links therein.
I wonder if this effect also exists in the absence of electron spin.
Background
To a good approximation, the Hamiltonian $H_\text{JJ}$ for a Josephson junction is \begin{equation} H_\text{JJ} = H_\text{A} + H_\text{B} + H_\text{AB} \end{equation} with $H_\text{A}$ the BCS Hamiltonian, \begin{equation} H_\text{A} = \sum_{k,\sigma} \epsilon_k c_{k\sigma}^\dagger c_{k\sigma} - \sum_{k}(\Delta e^{-i\phi_c}c^\dagger_{k\uparrow}c^\dagger_{-k\downarrow}+\Delta e^{i\phi_c} c_{-k\downarrow}c_{k\uparrow}) \end{equation} and similarly for $H_\text{B}$ together with \begin{equation} H_\text{AB} = \sum_{k,p,\sigma} (tc^\dagger_{k\sigma} f_{p\sigma}+t^* f^\dagger_{p\sigma} c_{k\sigma}), \end{equation} where $c^\dagger_{k,\sigma}$ creates an electronic excitation with momentum $k$ and spin $\sigma$ in wire A and $f^\dagger_{k,\sigma}$ creates an electronic excitation with momentum $k$ and spin $\sigma$ in wire B.
This model should lead to the Josephson equations as usual.
Question: What if we remove spin?
That is, would the above Hamiltonian without spin indexes lead to the Josephson equations too?
If not: Why not? Is there a modification to the spinless Hamiltonian that leads to them?
