BCS state and its superconductivity I've learned in BCS theory about its ground state by applying Bogoliubov annihilation operator on it to be zero; however, in the textbook the total momentum of electrons is set to be zero. It's okay to me for this state to be a ground state for the effective Hamiltonian; however, I cannot understand why this state exhibits superconductivity. I was considering yo apply perturbation say a constant electric field $E=U/L$ to the system and calculate some kind of linear response. However, I'm not sure about the results I derived so far. 
 A: bcs state is superconducting because excitation spectrum has a gap. which mean to create a quasi particle on ground state you need non zero energy. 
creating a qp can be interperted as exciting an cooper pair. cooper pairs can be excited by scattering in lattice. 
so lattice scattering of your charge carriers needs energy. but this scattering what causes the resistance. hence having a resistence causes an energy. thats why you don't have resistance in bcs.
A: Now even though I haven't derive a concrete solution to what happens when applying external electric field to BCS superconductor, I eventually get an explanation of why gapped BCS states has relation with superconducting. 
Considering the excitation energy spectrum as illustrated (gapped): 

And then considering an impurity with some velocity to scatter with the superconductor. Because it is superconductor, no quasi-particle should be excited to consume energy or effectively enforce some kind of friction into the impurity. If the dispersion of impurity is classical, i.e., $E_{\text{imp}} = \dfrac{1}{2}m_{\text{imp}}{\bf v}^2$, and in/out with velocity ${\bf v}_{\text{in}}, {\bf v}_{\text{out}}$. For the conservation of energy & momentum, we have
$$\dfrac{1}{2}m_{\text{imp}}{\bf v}_{\text{in}}^2 = \dfrac{1}{2}m_{\text{imp}}{\bf v}_{\text{out}}^2 + E({\bf k}) $$
$$m_{\text{imp}}{\bf v}_{\text{out}} = m_{\text{imp}}{\bf v}_{\text{in}} - \hbar{\bf k}$$
square the second equation and divided my $2m_{\text{imp}}$, we have
$$\dfrac{1}{2}m_{\text{imp}}{\bf v}_{\text{out}}^2 = \dfrac{1}{2}m_{\text{imp}}{\bf v}_{\text{in}}^2 - {\bf v}_{\text{in}}\cdot\hbar{\bf k} + \dfrac{\hbar^2{\bf k}^2}{2m_{\text{imp}}}$$
compare with the first equation we have
$$ E({\bf k}) = {\bf v}_{\text{in}}\cdot\hbar{\bf k} - \dfrac{\hbar^2{\bf k}^2}{2m_{\text{imp}}}\le \hbar|{\bf v}_{\text{in}}||{\bf k}| $$
which set the lower bound for the incident scatter velocity, as follow:

therefore, at low energy range, as long as the system is gapped, the superconducting property remains at this level. The linear response calculation of current is somehow not correct, for it's completely a non-perturbative phenomenon. 
