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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 $\sqrt{\hslash /(2{\rho }_{\mathrm{ion}}V{\omega }_{\mathbf{q}})}$ for a single simple branch. Expanding the electronic field in bloch states then gives a band-projected Fröhlich HamiltonianHere $c,c^{†}$ obey canonical anticommutation relations and remove/create an electron with crystal momentum $\mathbf{k}$ and spin $\sigma$; $b,b^{†}$ 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, $M_{-\mathbf{q}}=M_{\mathbf{q}}^{\ast }$ 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 $A_{\mathbf{kq}\sigma }=c_{\mathbf{k}+\mathbf{q},\sigma }^{†}{c}_{\mathbf{k}\sigma }$ and ${\Delta }_{\mathbf{kq}}={ϵ}_{\mathbf{k}+\mathbf{q}}-{ϵ}_{\mathbf{k}}$, one possible generator isIts Hermitian conjugate is its negative after relabelling momenta. The Baker--Campbell--Hausdorff formula givesThe 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 ${ϵ}_{\mathbf{k}}$ and eliminate the explicit phonon sector. For elastic pair scattering the two virtual time orderings have the combined amplitudeUsing the usual factor one-half to count unordered pairs gives the coefficient in the printed Hamiltonian, with $\Delta ={ϵ}_{\mathbf{k}}-{ϵ}_{\mathbf{k}-\mathbf{q}}$. The two outgoing momenta are $\mathbf{k}-\mathbf{q}$ and ${\mathbf{k}}^{\prime }+\mathbf{q}$, 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: 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.^[ref]

For the pair calculation, make the model restrictions explicit. The filled Fermi sea is inert, and $|0⟩$ 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 ${\xi }_{\mathbf{k}}={\hslash }^2{k}^2/(2m)-E_F$ and $D=\hslash {\omega }_D$. 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 $E_{\mathrm{abs}}=2E_F+E$. 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 $|\mathbf{k}⟩=c_{-\mathbf{k}\downarrow }^{†}{c}_{\mathbf{k}\uparrow }^{†}|0⟩$. 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 asFor $|\Psi ⟩={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}|\mathbf{k}⟩$ and $C={\sum }_{⟨\mathbf{k}⟩}{\phi }_{\mathbf{k}}$, matching each basis coefficient givesThis 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, $|C{|}^{-2}=g^2{\sum }_{⟨\mathbf{k}⟩}(E-2{\xi }_{\mathbf{k}}{)}^{-2}$. It cannot be set to zero for the nontrivial bound solution.

Put $E=-B$ with $B>0$ and impose the sum defining $C$: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 yieldsThe 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. An arbitrarily weak net attraction binds a zero-momentum singlet below the free-pair threshold, because the near-Fermi-surface integral diverges logarithmically as $B\downarrow 0$. 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.^[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.^[ref]