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NBODY in 2nd order IFCs #179

Closed Answered by ttadano
RicardoCecconello asked this question in Q&A
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The $n$ th order force constant is defined as

$$\Phi_{\mu_1\dots \mu_n} (\ell_1\kappa_1;\dots;\ell_n\kappa_n) = \frac{\partial^{n} U}{\partial u_{\mu_1}(\ell_1\kappa_1)\cdots \partial u_{\mu_n}(\ell_n\kappa_n)}$$

where $u_{\mu}(\ell\kappa)$ is the atomic displacement and $U$ is the Born-Oppenheimer potentail energy surface. https://alamode.readthedocs.io/en/latest/almdir/formalism_alm.html#interatomic-force-constants-ifcs

With this definition, the second-order IFCs involve only two atom indices, $(\ell_1\kappa1, \ell_2\kappa_2)$, so there are no multibody (3-, 4-body, etc.) interactions.

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