= Solution
Start from <electrons> moving in the periodic equilibrium ionic potential. Displace ion $l$ by $\mathbf u_l$ and expand its contribution as $V(\mathbf r-\mathbf R_l-\mathbf u_l)\simeq V(\mathbf r-\mathbf R_l)-\mathbf u_l\cdot\nabla V(\mathbf r-\mathbf R_l)$. Quantizing the harmonic <normal modes> expresses $\mathbf u_l$ as a sum of creation and <annihilation operators> with displacement scale $\sqrt{\hbar/(2\rho_{\rm ion}V\omega_{\mathbf q})}$ for a single simple branch. Expanding the electronic field in <bloch states> then gives a band-projected <Fröhlich Hamiltonian>
$$
H=H_0+V_{e\text{-}ph},\qquad H_0=\sum_{\mathbf k\sigma}\epsilon_{\mathbf k}c_{\mathbf k\sigma}^{\dagger}c_{\mathbf k\sigma}+\sum_{\mathbf q}\Omega_{\mathbf q}b_{\mathbf q}^{\dagger}b_{\mathbf q},
$$
$$
V_{e\text{-}ph}=\sum_{\mathbf k\mathbf q\sigma}M_{\mathbf q}c_{\mathbf k+\mathbf q,\sigma}^{\dagger}c_{\mathbf k\sigma}(b_{\mathbf q}+b_{-\mathbf q}^{\dagger}),\qquad \Omega_{\mathbf q}=\hbar\omega_{\mathbf q}.
$$
Here $c,c^{\dagger}$ obey <canonical anticommutation relations> and remove/create an <electron> with <crystal momentum> $\mathbf k$ and <spin> $\sigma$; $b,b^{\dagger}$ are <boson> operators for lattice modes. The <matrix element> $M_{\mathbf q}$ contains the derivative of the ionic potential, <phonon polarization>, displacement normalization and electronic <wavefunction> overlaps. With phases chosen appropriately, $M_{-\mathbf q}=M_{\mathbf q}^*$ ensures a <Hermitian> Hamiltonian. The displayed one-band, one-branch notation suppresses additional branch and band labels. “<Fröhlich Hamiltonian>” here means the linear electron-phonon model used in the question; it is not a requirement to use specifically the polar-optical-phonon coupling of a polar <semiconductor>. A harmonic lattice, small displacements and a weak enough coupling to organize <perturbation theory> are the underlying approximations.
The <Fröhlich canonical transformation> removes real linear <phonon> operators from the electron-only effective problem to first order. Choose an <anti-Hermitian> $S$ with $[S,H_0]=-V_{e\text{-}ph}$. For $A_{\mathbf k\mathbf q\sigma}=c_{\mathbf k+\mathbf q,\sigma}^{\dagger}c_{\mathbf k\sigma}$ and $\Delta_{\mathbf k\mathbf q}=\epsilon_{\mathbf k+\mathbf q}-\epsilon_{\mathbf k}$, one possible generator is
$$
S=\sum_{\mathbf k\mathbf q\sigma}M_{\mathbf q}A_{\mathbf k\mathbf q\sigma}\left(\frac{b_{\mathbf q}}{\Delta_{\mathbf k\mathbf q}-\Omega_{\mathbf q}}+\frac{b_{-\mathbf q}^{\dagger}}{\Delta_{\mathbf k\mathbf q}+\Omega_{\mathbf q}}\right).
$$
Its <Hermitian> conjugate is its negative after relabelling momenta. The <Baker--Campbell--Hausdorff formula> gives
$$
e^SHe^{-S}=H_0+\frac12[S,V_{e\text{-}ph}]+O(M^3).
$$
The <commutator> describes virtual emission and absorption of a <phonon>. <normal ordering> yields an electron-electron term as well as one-electron <energy> renormalizations and <phonon> terms. Retain the former, absorb single-particle shifts into $\epsilon_{\mathbf k}$ and eliminate the explicit <phonon> sector. For elastic pair scattering the two virtual time orderings have the combined amplitude
$$
V_{\rm eff}(\mathbf q,\Delta)=|M_{\mathbf q}|^2\left(\frac1{\Delta-\Omega_{\mathbf q}}-\frac1{\Delta+\Omega_{\mathbf q}}\right)=\frac{2|M_{\mathbf q}|^2\Omega_{\mathbf q}}{\Delta^2-\Omega_{\mathbf q}^2}.
$$
Using the usual factor one-half to count unordered pairs gives the coefficient in the printed Hamiltonian, with $\Delta=\epsilon_{\mathbf k}-\epsilon_{\mathbf k-\mathbf q}$. The two outgoing momenta are $\mathbf k-\mathbf q$ and $\mathbf k'+\mathbf q$, so total <momentum> and the <spins> are conserved. A fully off-shell effective operator can instead use symmetrized denominators for both external legs; the displayed expression is the usual elastic-scattering reduction. Eliminating nearly resonant real <phonons> requires more care than this second-order construction.
For $|\Delta|<\Omega_{\mathbf q}$ the denominator is negative: \b[virtual <phonon> exchange is attractive in this low-energy channel]. Physically, an <electron> changes the ionic density and a second <electron> can gain <energy> by responding to the delayed distortion. This is a retarded interaction, not an unscreened instantaneous attraction at every <energy> transfer. At large transfer the sign changes, and screened Coulomb repulsion must also be included when deciding the net low-energy pairing strength. The connection between lattice dynamics and <superconductivity> is supported by the isotope effect, but the mathematical attraction alone does not determine a measured transition <temperature>.<Electron-Phonon Interactions and Superconductivity by John Bardeen>[pp. 55–57, lattice-coupling mechanism and isotope-effect motivation]{ref}
For the pair calculation, make the model restrictions explicit. The filled <Fermi sea> is inert, and $|0\rangle$ denotes that sea with no added particles, not an empty electronic vacuum. Project the interaction onto the previously empty shell $0\le\xi_{\mathbf k}\le D$, where $\xi_{\mathbf k}=\hbar^2k^2/(2m)-E_F$ and $D=\hbar\omega_D$. Measure the added-pair <energy> relative to $2E_F$; this is the negative $E$ in the final formula. If absolute one-electron energies are retained in the preceding <eigenvalue> equation, its <eigenvalue> must instead be $E_{\rm abs}=2E_F+E$. A shell-restricted state would not in general be an exact eigenstate of the completely unprojected, all-momentum contact Hamiltonian.
Write $g=\gamma^2>0$ and $|\mathbf k\rangle=c_{-\mathbf k\downarrow}^{\dagger}c_{\mathbf k\uparrow}^{\dagger}|0\rangle$. Within the zero-total-momentum pair subspace the two opposite-spin terms cancel the interaction's factor one-half, while same-spin s-wave terms cancel by fermionic antisymmetry. Thus the projected Hamiltonian acts as
$$
H'_{\rm pair}|\mathbf k\rangle=2\xi_{\mathbf k}|\mathbf k\rangle-g\sum_{\langle\mathbf p\rangle}|\mathbf p\rangle.
$$
For $|\Psi\rangle=\sum_{\langle\mathbf k\rangle}\varphi_{\mathbf k}|\mathbf k\rangle$ and $C=\sum_{\langle\mathbf k\rangle}\varphi_{\mathbf k}$, matching each basis coefficient gives
$$
(E-2\xi_{\mathbf k})\varphi_{\mathbf k}=-gC,\qquad \boxed{\varphi_{\mathbf k}=-\frac{\gamma^2C}{E-2\xi_{\mathbf k}}.}
$$
This is the stated coefficient formula in the consistent Fermi-level <energy> convention. Its even dependence on $\mathbf k$ produces an even orbital state with a <spin> singlet. The constant $C$ is fixed up to an overall phase by normalization; for a <bound state>, $|C|^{-2}=g^2\sum_{\langle\mathbf k\rangle}(E-2\xi_{\mathbf k})^{-2}$. It cannot be set to zero for the nontrivial bound solution.
Put $E=-B$ with $B>0$ and impose the sum defining $C$:
$$
1=g\sum_{\langle\mathbf k\rangle}\frac1{2\xi_{\mathbf k}+B}\simeq g\rho(E_F)\int_0^D\frac{d\xi}{2\xi+B}=\frac{g\rho(E_F)}2\log\frac{2D+B}{B}.
$$
Here $\rho(E_F)$ counts orbital states for one <spin> species, as appropriate to the unrestricted $\mathbf k$ sum in this pair basis, and is approximately constant over the thin shell. If a two-spin total <density of states> is used, it must be divided by two in this integral. State-counting and the volume convention in $g$ and $\rho$ must be consistent. With $\xi=g\rho(E_F)$ as the paper's dimensionless coupling, inversion yields
$$
\boxed{E=-\frac{2\hbar\omega_D}{e^{2/\xi}-1}\simeq-2\hbar\omega_D e^{-2/\xi}\quad(\xi\ll1).}
$$
The first approximation is the constant-density-of-states, restricted-shell model; the final exponential additionally assumes weak attraction. The binding <energy> is nonanalytic at zero coupling, so a finite-order power expansion in $g$ cannot produce it.
Cooper's problem asks whether two <electrons> added above an inert <Fermi sea> bind despite Pauli exclusion of lower states. \b[An arbitrarily weak net attraction binds a zero-momentum singlet below the free-pair threshold], because the near-Fermi-surface integral diverges logarithmically as $B\downarrow0$. Its <wavefunction> draws on many time-reversed <momentum> states and becomes dominated by a narrow region near the <Fermi surface> at weak coupling; the corresponding pair is spatially extended. This establishes a <Cooper instability> of the unpaired reference state.<Bound Electron Pairs in a Degenerate Fermi Gas by Leon Neil Cooper>[pp. 1189–1190, two added particles above the filled sea and the bound-state criterion]{ref} The full superconducting <ground state> needs a coherent many-electron treatment in which overlapping pairs and the reorganized sea are included. <BCS theory> supplies that step; the two-electron binding scale alone is neither a calculation of the many-body gap nor a derivation of zero resistance.<Theory of Superconductivity by John Bardeen and others>[pp. 1175–1177, paired many-electron ground state and excitation gap]{ref}
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