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Using only 1s, 2s, and 2p orbitals, which of the matrix elements $H_{kl} = \int \chi_k \hat{H} \chi_l d\tau$ (where H is the full molecular Hamiltonian and $\chi_i$ are the atomic orbitals eg. 1s AO of H-atom) vanish based only on symmetry arguments without needing a calculation?

Will the integral vanish if you have for example a $p_x$ and a $p_y$ orbital where the z-axis is the molecular axis, because you have minimal overlap of the AOs?

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  • $\begingroup$ I see you're new to the site. If you find any answer helpful to you, please upvote or accept them. $\endgroup$ – pentavalentcarbon Nov 26 '16 at 16:22
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The hydrogen molecule and its various ions fall into the $D_{\infty\mathrm{h}}$ point group.

For an integral to be non-zero, the product of its components must contain the totally symmetric irreducible representation (irrep) under said point group, which here is $\mathrm{A_{1g}}$. $s$-orbitals are perfectly spherical, so they always transform as the totally symmetric irrep. $p_x$ and $p_y$ both transform as $\mathrm{E_{1u}}$, and $p_z$ transforms as $\mathrm{A_{1u}}$. The non-relativistic molecular Hamiltonian is invariant to all these symmetry operations, so it too transforms as $\mathrm{A_{1g}}$.

To determine the final product, refer to a direct product table. The product of any irrep with itself will always give the totally symmetric irrep. So, for example, if your $\{\chi\}$ are both $p_x$ orbitals, the integral will not vanish due to symmetry arguments. In general, if your $\{\chi\}$ are different, because the product of different irreps never contains the fully symmetric irrep, those integrals will be zero.

However, due to $p_x$ and $p_y$ transforming identically, matrix elements between them don't necessarily vanish. In the matrix representation of the Hamiltonian in the atomic orbital basis, these rules will lead to a diagonal matrix, except for a block corresponding to $p_x$ and $p_y$.

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