What is the link between the BCS ground state and superconductivity? The link between the BCS ground state
$$
\left|\Psi_\mathrm{BCS}\right\rangle = \prod_k \left( u_k - v_ke^{i \phi} c_{k\uparrow}^{\dagger} c_{-k\downarrow}^{\dagger}\right) \left|0\right\rangle
$$
and the property of superconductivity (carrying current with no resistance) does not seem obvious to me, but I can't find a satisfying explanation anywhere. 
According to some authors, the gap explains superconductivity, but I am not convinced by this, since (1) insulators are gapped too, and (2) some superconductors have a vanishing gap for some values in $k$-space.
Isn't it possible to just calculate the linear response under DC field and compare with the normal state expectation value to see how this works? Does anyone know how to perform such a calculation?
 A: An electron-excitation-gap is necessary helpful for superconductivity. (Normally electrons scatter, but if there's a gap then they don't scatter, because there's no state to scatter into, as long as the temperature is low enough that they cannot jump the gap.**) But it is not sufficient. The filled states also have to be able to carry a current! The electrons in a filled semiconductor valence band have an electron-excitation-gap, like you say, but they do not carry a current. (If you think about it, in intrinsic GaAs near absolute zero, there are no electron scattering events!) On the other hand, the filled states in a superconductor CAN carry a current because...well I wasn't sure but I read the old BCS paper and they have a pretty basic explanation:

Our theory also accounts in a qualitative way for those aspects of
superconductivity associated with infinite conductivity...the paired
states $(k_{1\uparrow}, k_{2\downarrow})$ have a net momentum
$k_1+k_2=q$, where $q$ is the same for all virtual pairs. For each value
of $q$, there is a metastable state with a minimum in free energy and a
unique current density. Scattering of individual electrons will not
change the value of $q$ common to virtual pair states, and so can only
produce fluctuations about the current determined by $q$." And these
scattering events raise the free energy, unless all of the electrons
scatter simultaneously in exactly the right way to make a new
metastable state centered at a different q, which is extremely
unlikely.

Well that makes sense to me.... In your question you wrote the BCS ground state with $q=0$, but that's just one of the family of BCS meta-stable (ground-ish) states with different $q$, with different $q$s corresponding to different current flows.
In other words, BCS theory explains how the electrons pair up and then there is an energy gap for single-particle-excitations. Slightly changing the $q$ requires little or no energy (or in some cases even lowers the energy), but will not happen spontaneously because it requires trillions of electrons to change their state simultaneously in a coordinated way. (An electric field can cause this kind of coordinated change, but it can't just happen spontaneously. Generally, only single-particle-excitations happen spontaneously, and these are gapped.) So it is metastable. And the fact that you can have a metastable state which carries current is just another way of saying that current can keep flowing and flowing even with no electric field pushing it.
**Update: OK, yes there is such a thing as "gapless superconductivity". My mistake was conflating "superconductor" with "superconductor with no dissipation whatsoever". The latter doesn't exist—even with a full proper superconducting energy gap, remember the superconducting transition is above absolute zero, so there is certain to be some small but nonzero electron scattering rate that is tolerable without destroying the superconducting order. So by that logic, it's not surprising that a partial or nonexistent gap is compatible with superconductivity at a very low transition temperature.
A: The BCS wave function you write down depends on a parameter $\phi$, but the ground state energy is independent of it. This implies that $\phi$ is the (would be) Goldstone mode that governs the low energy dynamics of the system. The gradient of $\phi$
is the conserved $U(1)$ current $\vec\jmath \sim \vec\nabla\phi$, and the ordinary charged current is $\vec\jmath_s =n_s e \vec\nabla\phi /m$, where $n_s$ is the superfluid density of electrons. 
Because $\phi$ is a Goldstone mode the effective low energy action can only depend on gradients of $\phi$. By gauge invariance the effective action
is of the form $S[A_\mu-e\nabla_\mu\phi]$. The explicit form of $S$ can be computed from the BCS wave function, or more easily determined using diagrammatic methods. For our purposes the only important point is that $S$ has at least a local minimum if the field vanishes. This means that solutions of the classical equation of motion are of the form $A_\mu=e\nabla_\mu \phi$ (This is the London equation). Let's consider an applied electric field $\vec{E}=-\vec\nabla A_0$. I find
$$
\vec{E}=-e\vec\nabla \dot\phi=\frac{m}{n_s}\frac{d\vec\jmath}{dt}
$$ 
which shows that a static current corresponds to zero field, and the resisitivity is zero.
The effective action also governs other properties of the system, such as the Meissner effect, the critical current, and fluctuations of the current in a thermal ensemble. 
Postscript: A commenter argues that I really need to show that $S$ has a minimum
$$
 S \sim \gamma (A-e\nabla\phi)^2 + \ldots
$$
First, note that $\gamma$ determines the Meissner mass, so even without a calculation I have shown that the Meissner effect implies superconductivity. Beyond that I do indeed have to do a calculation of $\gamma$ based on the BCS wave function (I could appeal to the Landau-Ginzburg functional, but this only shifts the question to the gradient term in the LG functional). Fortunately, the calculation is straightforward, and can be found in many text books. For people with more of a particle physics interest there is a beautiful explanation in Vol II of Weinberg's QFT book. There is Anderson's famous paper on gauge invariance and the Higgs effect. I provided a version of the calculation in Sect. 3.4 of these lecture notes  https://arxiv.org/abs/nucl-th/0609075
Post-Postscript: How is this different from a weakly interacting electron gas? In the electron gas I have a low energy description in terms of electrons and phonons (and other degrees of freedom). For simplicity consider the high temperature limit, where a classical description applies (as explained by Landau Fermi liquid theory, this generalizes to low T). The equation of motion for a single electron is just $m\dot v=e E$, which superficially looks like the London equation. However, this is not a macroscopic current. When I pass from microscopic to macroscopic equations there is no symmetry that forbids the appearance of dissipative terms, so the conductivity is non-zero. There is indeed a subtlety in the coupling of electrons and phonons, because without the umklapp process momentum conservation would force the conductivity to vanish.     
In a superconductor the gradient of the Goldstone boson automatically describes a macroscopic current ($\gamma$ is proportional to the density of electrons). S is a quantum effective action, and dissipative terms are automatically forbidden. At finite temperature things do get a little more complicated because the total current is in general the sum of a non-dissipative supercurrent, governed by $S$, and a dissipative normal current. However, below $T_c$ part of the response is carried by a supercurrent.
A: As you rightly mention, the presence of a gap explains nothing of the superconducting phenomenology, except its DC (and quasi-DC) behaviour. This is quite natural : same causes, same consequences. So a superconducting behaves as a semi-conductor because it has a gap. This gap being quite small, conventional superconductors are not really interesting semi-conductors.
So, what are the crucial aspects of superconductivity hidden in the BCS Ansatz you wrote ? Well, many many, for instance


*

*this is an example of a coherent fermionic state

*it comes from the microscopic picture of the Cooper pairing (as explained in the first paragraph of Joshuah Heath answer (the rest of his answer is non-sense)

*it gives a phase (in the sense of $\varphi$ in the writing $e^{i\varphi}$, not in the sense of phase of matter) to the condensate, hence you got a macroscopic wave function associated to the condensate

*it explains the temperature dependency of the gap

*it makes the condensate a perfect diamagnetic material

*... 


Perfect diamagnetism goes hand to hand with no resistance. The demonstration that there is no current / perfect diamagnetism / no resistance associated to the BCS Ansatz is explained in great details in the historical account presenting the microscopic theory, namely

Bardeen, J., Cooper, L. N., & Schrieffer, J. R. (1957). Theory of Superconductivity. Physical Review, 108, 1175–1204.

Calculation details can also be found in 

Tinkham, M. (1996). Introduction to superconductivity (second edition). Dover Publications, Inc.

See also this answer of mine, about a related question.
As far as I remember, Leggett also provide many different calculations of this effect, as 

Leggett, A. J. (1975). A theoretical description of the new phases of liquid He3. Reviews of Modern Physics, 47, 331–414.

The calculation is nevertheless a bit cumbersome, so I'm reluctant to try it on this platform. Feel free to ask about unclear details in the linked references.
A: I am not sure about the second part of your question, but I think I can give you an answer to the first portion. The BCS trial wave function proposes a linear combination of a filled Fermi sea state (with probability $|u_k|^2$) and a state with a Cooper pair (with probability $|v_k|^2$). If one calculates the expectation value of the pairing Hamiltonian and minimizes the energy, one will find that the system favors being in the Cooper paired state. What this tells us is that, as long as we have some attractive potential (no matter how small), the system will prefer a Cooper paired state. In some lattice structure, phonon-electron interactions provide this attractive potential.
It is this inevitable formation of Cooper pairs that gives us zero resistivity. Small scale physics (like scattering) gets absorbed into the macroscopic quantum mechanics described by this system, as described in this answer. Hence, scattering processes won't effect the current. As suggested in your question and in the discussion in the link above, zero resistance is therefore not dependent upon the existence of a gap.
A good discussion of what causes superconductivity can be found here. For an excellent discussion of all things superconducting, see Tinkham's book.
A: The thing is that you can't compare an insulator's band structure and the one of the BCS superconductor. For the insulator, the band gap is an energy gap with zero density of states in electron space. Note that "electron" is not the right word as we have electron-like quasiparticles in a lattice because of interaction, but they are still very similar to electrons. Hence, "electrons" can't gain additional energy from an external voltage.
Now let us consider the BCS hamiltonian. We see a constant condensation energy and a Hamiltonian that looks famililar as it is written in k-space. However, the operators are not normal electron-like operators. They are Bogolibuov-quasiparticles, which are superpositions of electrons and holes. Hence, only for a sufficiently large energy, I can generate these particles. For energies (external voltages) below this threshold energy (at T=0), my ground state remains unchanged. Considering the ground state, we see a coherent state which has an associated phase with it, the superconducting phase. From that phase, one can compute many transport experiments and see supercurrents for example.
But what is the intuition here? We have to look again at a normal metal and see what resisitivity means. It translates to scattering. Electrons changing states or decaying into lower energy states and giving energy to an environment. Particles in a superconductor BCS state can't do that, for the same reasons as electrons in an insulator: There are no available states left. However, the electrons in a supercondcutor are tied to their coherent state. Thus, the gap actually protects the coherent state of electrons and thus the superconducting properties. Note, that the gap is not necessary for a SC. The only thing we need is the electron coherent state which can also form without a gap. There are unconventional superconductors that don't have a gap for certain directions, where cooper-pairs are easily broken down.
The question remains on why the coherent states grants the effects of superconductivity, namely perfect diamagnetism and zero resistivity. The answer lies in QUANTUM. What I mean by that is that there is no real intuitive explanation anymore. One can compute transport between two coherent states and see that there is current flowing without external influence. The dc-Josephson effect is a solely quantum mechanical effect. Same goes for the ac-Josephson effect. I don't even want to try to formulate an intuitive explanation onto why a constant voltage induces an alternating current between two superconductors. Of course, the math gives us an explanation (gauge invariant derivatives, etc.) however there is no "picture" associated to it. We have to use our "physics intuition".
