# What does phonon exchange do in the theory of superconductivity?

In the theory of superconductivity, the phonon-mediated electron-electron scattering leads to an effective interaction, the BCS hamiltonian,$$\hat{H}_{\rm BCS}=\sum\limits_{\vec k,\sigma}\epsilon_{\vec k}c^\dagger_{\vec k,\sigma}c_{\vec k, \sigma}-\kappa^2\sum\limits_{\vec k, \vec k^\prime}c^\dagger_{\vec k,\uparrow} c^\dagger_{-\vec k,\downarrow} c_{-\vec {k'},\uparrow} c_{\vec {k'}, \downarrow}$$ which represents an attractive interaction between pairs of electrons (or Cooper pairs) rather than between two electrons.

So my question is, what does phonon exchange do? Does it result in an attractive interaction between two electrons of equal and opposite momenta and spins OR does it cause an attractive interaction between two Cooper pairs $$|\vec k,\uparrow;-\vec k,\downarrow\rangle$$ and $$|\vec k',\uparrow;-\vec k',\downarrow\rangle$$ OR both?

In your denotation, you have already replaced phonon exchange by effective interaction. You should start from the electron-phonon interaction, $$H=\gamma\int d^3x\,n(x)\nabla\cdot u(x)=\gamma\sum_{q,\lambda}\frac{iq_{\lambda}}{\sqrt{2m\omega_q}}(a_{q\lambda}+a^{\dagger}_{-q\lambda})n_q,n_q=\sum_qc^{\dagger}_kc_k,$$ where $$u(r)$$ is lattice distortion, $$a$$ is phonon operator and $$c$$ is fermion operator. Electron feels effectiv charge $$\rho\sim\nabla\cdot P$$, where $$P\sim u$$ is polarization. Then, you should add free phonons & free electrons terms and integrate out phonon fields. The resulting effective action seems like $$S_{\text{eff}}=S_{0}-\frac{\gamma}{2m}\sum_n\int\frac{d^3q}{(2\pi)^3}\frac{q^2}{\omega_n^2+q^2}\rho(i\omega_n,q)\rho(-i\omega_n,q),$$ where $$\rho$$ is electron density, the summation is performed over Matsubara frequincies. After analytic continuation into real frequency, you should see that for small enough momenta $$q$$, you can just neglect this term and consider effective local 4-fermion attractive interaction.