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For ordinary one-phonon infrared optical absorption, both an energy and a momentum selection rule must hold. The optical photon has negligible crystal momentum on the scale of the Brillouin zone, so crystal momentum conservation selects $\mathbf{q}\simeq 0$ modulo a genuine reciprocal lattice vector. In addition, the mode must change the macroscopic electric polarization, with a component along the incident polarization of an electromagnetic wave.

In terms of atomic displacements ${\mathbf{u}}_j$ and Born effective charge tensors $Z_j^{\ast }$, the relevant derivative is proportional to ${\sum }_j{Z}_j^{\ast }{\mathbf{u}}_j$ per primitive unit cell. Thus an optical phonon at $\mathbf{q}=0$ is not automatically infrared active: its dipole contributions may cancel. Disorder-assisted and multiphonon processes relax these simple first-order selection rules.

The periodic boundary condition requires $e^{ikM}=1$, so there are exactly $M$ independent wavevectors in a Brillouin zone:The inverse discrete Fourier transform, with the sign convention of the paper, isFourier orthogonality gives the tight binding dispersion relation ${ϵ}_k=-2t\cos k$. For the interaction, first use the canonical anticommutation relations at distinct sites to write $n_{\sigma ,j}{n}_{\sigma ,j+1}=c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}^{†}{c}_{\sigma ,j+1}{c}_{\sigma ,j}$. Summing its discrete Fourier transform over $j$ enforces crystal momentum conservation, givingEvery wavevector is reduced modulo $2\pi$. The phase $e^{iq}$ comes from the displacement of the second site by one lattice spacing. Equivalently replace $e^{iq}$ by $\cos q$: exchanging the two incoming and the two outgoing fermions changes $q$ to $-q$ with two cancelling fermionic signs. Thus the usual interaction vertex is $2V\cos q$ accompanied by the factor $1/(2M)$.