The Electron-Pair wavefunction

Because of the correlation of the electron pairs, it is highly improbable that once the superconducting state has been entered, a pair will be scattered out of the flow (i.e. cause resistivity), since this would imply that all the electron pairs have to change their states. The wavefunctions of the electron-pairs are correlated by with the other pairs by a phase difference, meaning that the knowledge of the wavefunction at one place determines it at another. We will now examine this phase correlation.
The electron-pairs form a single state that can be described by

$$\Phi_{Pair} = \Phi e^{i \left( \frac{\mathbf{P} \cdot \mathbf{r}}{\hbar} \right)}$$ (1)

where $\mathbf{P}$ is the net momentum which is the same for all pairs and $\mathbf{r}$ the centre of mass.
Assuming a homogeneous current density, all electron-pairs in a superconductor have the same momentum and therefore have waves of the same wavelength. The superposition of a number of coherent waves of equal wavelength can be described by a single wave of a form

$$\Psi_P = \Psi e^{i \left( \frac{\mathbf{P} \cdot \mathbf{r}}{\hbar} \right)}$$ (2)

where $\Psi=\sum_n\Phi$, $\mid \Psi_P \mid^2$ is the density of electron-pairs and P is the momemtum per pair.This wavefunction will theoretically stay in phase forever.
Phase changes occur due to the non-zero net momentum of the Cooper pairs, i.e. a non-zero wave vector $\vec{k}$, where $\vec{P}=\hbar\vec{k}$. Inserting this into the expression for the wave, we can derive the difference in phase due to a supercurrent, flowing between X and Y, as

$$\left( \Delta \phi \right)_{XY} = \frac{4 \pi m_e}{h n_s e} \int^Y_X \mathbf{J_s} \cdot d\mathbf{l}$$ (3)

where $\frac{1}{2}n_s$ is the density of the electron pairs.
Applying an magnetic field changes the expression to

$$\left( \Delta \phi \right)_{XY} = \frac{4 \pi m}{h n_s e} \i... ...f{l} + \frac{4 \pi e}{h} \int^Y_X \mathbf{A} \cdot d\mathbf{l}$$ (4)

where $\mathbf{A}$ is the vector potential $\nabla\times\mathbf{A}=\mathbf{B}$.
This can be easily seen, since the canonical momentum $\mathbf{P_0}$ changes to $\mathbf{P_0}+q\mathbf{A}$.