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There is no sacred difference between the way a gauge boson and a Goldstone boson couple to a matter field. Goldstone bosons are just ordinary bosons that are massless due to the breaking of a global symmetry. To avoid Lorentz violating these have to be scalar fields so that does turn out to restrict the possible coupling that they could have, but the fact ...

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You can use either potential for either purpose; it's just that some potentials are better for the different purposes. The reason is that these are empirical potentials; their constants are tweaked to work for a certain purpose. For example, if you're looking at the phonon band structure, you want $\omega\left(\vec{q}\right)$ to be as accurate as possible. ...

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Phonons are collective modes in solids and a general derivation is needed independent of particular lattice constants to first order. Lattice constants define individual solids. Are you aware of the harmonic oscillator approximation? All symmetric potentials have as a first term in their expansion the quadratic, thus the harmonic oscillator ...

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(I will write $\ell$ instead of $l$, because it confuses me less frequently.) Assuming that $\delta$ definition is a typo, and it is supposed to be $$\delta=(\ell+1)[2(\ell-1)j_{\ell-1}(kR)-kRj_\ell(kR)]$$ Your matrix's determinant is nonzero... \begin{align} \alpha\delta-\beta\gamma &= \ell \left[R j_{\ell}(kR) k - 2 j_{\ell+1}(kR) \left(\ell + ...

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