flux quantization in superconducting ring I am trying to understand SQUID microscopy from the ground up, so I am starting with flux quantization in superconducting rings. I found a nice presentation that covers some of the details, but I am having a hard time getting my head around it. The presentation is here, and my questions here related to slide 4 as labelled in the presentation.
The equation for the wavefunction is given as $\psi=\rho_s^{\frac{1}{2}}\exp(i\phi(r))$. I am guessing that $\rho_s$ is the average charge density(?) over the loop and $\phi$ is just $\frac{\vec{p}\cdotp \vec{r}}{\hbar}$ is the phase, so this looks basically like free particlesto me. 
First question: where does this wavefunction form come from? Is this the electronic wavefunction, or charge density of cooper pairs, what? The factor of 2 in the momentum equations everywhere suggests this is for cooper pairs, so I guess the idea here is that a superconducting loops is basically a zero-potential loop where we can approximate particles as free?
My second question is for the phase difference equation:
$\Delta \phi =\oint\frac{\vec{p}}{\hbar}\cdotp d\vec{l}$
Where does that equation come from? Why is the phase difference equal to the line integral of the momentum? I think I've seen this before but it's been a while since I did any QM. 
My last question for the moment is a bit more general: I can see that flux is quantized for a superconducting loop. However, in a SQUID device, you don't really have a loop anymore - you have two half-loops separated by Josephson junctions. So why does the quantization continue to hold for the SQUID?
 A: All phenomenology of superconductivity is described by solid state theory with spontaneously broken electromagnetic $U_{EM}(1)$ symmetry, so let's answer on your questions by using this idea. 

First question: where does this wavefunction form come from?

So, EM symmetry group is spontaneously broken bu such VEV. Which VEV breaks this symmetry? It is the electron bilinear form $\psi_{\sigma}(\mathbf p)\psi_{-\sigma}(-\mathbf p)$, which describes two-electron state, which has with opposite spins and momentums - so-called Cooper pair. Corresponding field $\psi$ is expressed through classical grassmannian electron field $\Psi_{\sigma}(\mathbf r)$ as
$$
\tag 0 \psi (\mathbf r) \simeq \sum_{\sigma , \sigma{'}}\Psi_{\sigma}(\mathbf r)\Psi_{\sigma {'}}(\mathbf r) + h.c. \equiv \Psi_{\sigma}(\mathbf r)\Psi_{-\sigma}(\mathbf r) + h.c.,
$$ 
So that the final symmetry group is discrete $Z_{2}$ symmetry group, i.e., there is invariance of theory under transformations with gauge field
$$
\tag 1 \Lambda = 0, \quad \Lambda = \frac{\pi}{e} 
$$
We have to construct gauge invariant effective field theory which describes such breaking. There exist the theorem that for each spontaneously broken symmetry generator there is corresponding massless state - the Goldstone degree of freedom (for case of broken local it is unphysical) $\varphi (\mathbf r)$. In the case of EM symmetry group $U_{EM}(1)$ there is only one generator, and it becomes broken; we may extract corresponding goldstone phase from electron field $\Psi_{\sigma}(\mathbf r)$ in the next way:
$$
\tag 2 \Psi_{\sigma}(\mathbf r) = e^{i\varphi (\mathbf r)}\tilde{\Psi}_{\sigma}(\mathbf r)
$$
It parametrizes factor-space $U_{EM}(1)/Z_{2}$, so that $\varphi (\mathbf r) $ and $\varphi (\mathbf r) + \frac{\pi}{e}$ are equal, as it must be (see $(1)$).
Your parametrization conveniently follows from Eqs. $(0)$ and $(2)$. You see that corresponding wave function has nothing similar (formally and physically) to free particle wave function.

My second question is for the phase difference equation

Let's continue thinking in the direction of the answer on your first question. We need to construct effective gauge invariant field theory containing $\varphi , A_{\mu}$. After integrating out electrons corresponding lagrangian takes the form
$$
\tag 3 L = -\frac{1}{4}\int F_{\mu \nu}F^{\mu \nu} + L_{s}(A_{\mu} - \partial_{\mu}\psi (\mathbf r)),
$$
where in most cases $L_{s}$ has true minimum for $A_{\mu} = \partial_{\mu}\varphi$ inside the superconductor, so that energy of superconductor is minimal for $A_{\mu} = \partial_{\mu}\varphi$. Suppose now your superconductor ring. Let's assume circuit $C$ inside the ring. We know from previous sentences that for this circuit 
$$
\tag 4 |A_{\mu} - \partial_{\mu}\varphi| = 0,
$$ 
and that due to equivalence of $\varphi$ and $\varphi + \frac{\pi}{e}$ field $\varphi$ may be changed only up to $\frac{\pi n}{e}$, where $n$ is integer number. So that integral of $\nabla \varphi $ over such circuit is quantized:
$$
\Delta \varphi \equiv \oint_{C} \nabla \varphi \cdot d\mathbf l = |(4)| = \oint_{C} \mathbf A \cdot d \mathbf l = \int_{S_{C}}\mathbf B \cdot d\mathbf S = \frac{\pi n}{e}
$$

So why does the quantization continue to hold for the SQUID?

Let's complete the story.
Suppose now you want to discuss behaviour of system of two superconducting pieces which are separated by the gap. From gauge invariance follows that $L_{s}$ from $(3)$ depends on the difference $\Delta \varphi $ between the fields of Goldstone phases in these two pieces:
$$
L_{S} = AF(\Delta \varphi )
$$
where $A$ is the square of the gap. Moreover, if we shift $\Delta \varphi$ in each direction on the quantity of $\frac{\pi}{e}$, nothing will changed, so that $F$ must be periodical:
$$
F(\Delta \varphi) = F\left(\Delta \varphi + \frac{\pi n}{e}\right)
$$
