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In Jensen's Introduction to Computational Chemistry it says that the total non-relativistic Hamiltonian operator, transformed to the center of mass system, can be written, in atomic units, as $$ \hat{H}=\hat{T}_N+\hat{H}_e+\hat{H}_{mp} $$ where $\hat{T}_N$ is the kinetic energy of the nuclei, $\hat{H}_e$ is the electronic Hamiltonian, and $\hat{H}_{mp}$ is the mass polarization term, written as $$ \hat{H}_{mp}=-\frac{1}{2M_{tot}}\left(\sum_{i}^{N_{elec}}\nabla_i\right)^2$$ It happens that I did the transformation step-by-step and found that the mass polarization term should be $$ \hat{H}_{mp}=-\frac{1}{2M_{tot}}\sum_{i}^{N_{elec}}\sum_{j\neq i}^{N_{elec}}\nabla_i\cdot\nabla_j $$ So, my first question is, why the textbook consider the case where $j=i$ to be the only one? This is also strange, because in my derivation I only found mixed $j\neq i$ terms. To be sure about this I tried with a three particle Hamiltonian and got the same. No terms with $j=i$.

The other question is, if this term arises because it is not possible to separate the center of mass motion from the internal motion, then, how important is the contribution of mass polarization relative to non-adiabatic coupling elements (beyond Born-Oppenheimer approximation)?

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You may find derivation of eqs.120-124 in these notes Normal modes. The true story. helpful in revising your derivation.

Answering the other question: centre-of-mass can always be separated by the choice of centre-of-mass coordinates (cf. eq. 109-118) in Normal modes. The true story.. In almost all nuclear motion calculations the mass-polarisation term can be neglected: it is generally no bigger than finite-nuclei size effects or lowest order QED corrections.

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  • $\begingroup$ This is an interesting article, definitely on my to read list! There are a few other questions on vibrations which you may find interesting here and here $\endgroup$ – Cody Aldaz May 28 at 22:30

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