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I am trying to get the molecular orbital for arbitrarily placed nuclear centres (identical). In other words , I am trying to solve the following one electron hamiltonian. $$h=-\dfrac{1}{2}\nabla^{2}-\sum_{i=1}^{N}\dfrac{1}{|\vec{r}-\vec{R_{i}}|}$$

I assumed the charge on each nuclear centre is one and to solve the problem I took the following linear combination as the trial function.

$$\psi(\vec{r})=\sum_{i=1}^{N}\alpha_{i}1s(\vec{r}-\vec{R_{i}})+\sum_{j=1}^{N}\beta_{j}2p_{x}(\vec{r}-\vec{R_{j}})+\sum_{k=1}^{N}\gamma_{k}2p_{y}(\vec{r}-\vec{R_{k}})+\sum_{m=1}^{N}\delta_{m}2p_{z}(\vec{r}-\vec{R_{m}})$$

of course, I start with s orbitals and then I proceeded with the above linear combination. My question is that if there are some symmetries in the structure. Does variational principle take care of that and gives the right $$ \alpha, \beta,\gamma,\delta$$, the coefficients of the trial wave functions? Do I need to impose the symmetry of the molecule (if there is any) on the trial wavefunctions? Does it give totally wrong answer if I take the above trial wave functions without worrying about the symmetry of the molecule?Can someone throw some light on it? Also, I am curious to know how do I apply MO theory for polymer like chain molecules? Please suggest some good reference to understand that

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    $\begingroup$ Sinmetry always simplify things so I would say use it. With long molecule it will be translational. Not an answer at all but you definitely profit of symmetry for things like molecular hydrogen ion. So impose it. $\endgroup$ – Alchimista Feb 11 at 15:28
  • $\begingroup$ Thanks a lot for the answer .if the molecule doesn't have symmetry at all, somewhat like protein. Does MO theory work for that? Can I use the above-mentioned trial functions to solve it? $\endgroup$ – user135580 Feb 11 at 15:31
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Running the calculation without imposing symmetry constraints should give the same ground energy as one with them because the true electronic ground state of the system (within the Born-Oppenheimer approximation) always has the same symmetry as the nuclei. The variational principle always puts an upper bound on your ground state energy, so in principle it will work just as well. Using symmetry, however, could speed up your calculation significantly because the symmetry operators commute with you Hamiltonian. This gives you an additional set of constraints on the coefficients. It is worth mentioning that single reference methods like Hartree-Fock and DFT will sometimes give lower ground state energies without symmetry constraints because the true ground state is a symmetric superposition of several configurations, but the single reference methods can only examine a single one.

For polymer molecules and other 1D infinite systems you would use periodic boundary conditions. Just make the box really large in the two dimensions you don't want to be periodic (so the system doesn't 'see' itself) and only evaluate non-gamma k-points along the periodic vector.

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