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This resource ended up basically explaining the answer to my question: https://wiki.physics.udel.edu/phys824/Discretization_of_1D_Hamiltonian


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Just to be clear, the two or more atoms do not have to be of different type. Optical phonons are related to the relative vibrations of atoms within the unit cell, while acoustic phonons describe the relative vibration of different unit cells. Optical phonons arise whenever the unit cell has at least one such degree of freedom, meaning at least two atoms in ...


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Crystal lattices are classified in a way that is not necessarily the most natural one. A first classification is by their lattice class, which is determined by its associated unit cell, which is not necessarily the same as a fundamental domain for the lattice. There can be different Bravais lattices in each lattice class, determined by possible additional ...


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In quantum treatment, phonons are quantum harmonic oscillators, which is studied in any quantum mechanics textbooks. The energy spectrum is readily studied there. https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator The $1/2\hbar\nu_E$ comes from uncertainty principle. (A quantum harmonic oscillator is confined in space so its momentum cannot be zero) ...


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The lack of the $e^{-i \omega t}$ term is just because we're using complex wave notation. If you've ever taken an electrical engineering course, it's the same sort of thing that is used there: We're using $A e^{i \omega t}$ to stand for $A \cos (\omega t - \delta)$, with the implicit assumption that we're only interested in the real part of the quantities ...


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The standard trick is partial bosonization, a.k.a. the "Hubbard-Stratonovich" trick. Consider ${\cal L}=g(\bar{\psi} \psi)^2$. Introduce a dummy field $\sigma$ with purely Gaussian lagrangian ${\cal L}_\sigma=-\frac{1}{g}\sigma^2$. You can always insert a factor 1 in the path integral $$ 1=\frac{1}{Z}\int D\sigma \exp(iS_\sigma). $$ Now shift the scalar ...


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Let't me give it a shot. One possible explanation is to imagine we soften the integer condition for $m$'s in $S_{2}$. Then, the cosine term in $S_{1}$ is what we want to add because when $K$ is large, it enforce the integer condition and go back to $S_{2}$.



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