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I'd like to compute the decay rate using

$$ \Gamma = 2\pi \ | \langle ab|V|cd \rangle|^2 \delta\ (\epsilon_\mathrm{f}-\epsilon_\mathrm{i}), \tag{1}$$

where $a,$ $b,$ $c,$ $d$ are the MOs, $V$ is the Coulomb interaction, and $\epsilon_\mathrm{f}$ and $\epsilon_\mathrm{i}$ are the final and the initial energies of the system, respectfully.

I am interested in computing this value regarding a special atom (for example, atom $\ce{A}),$ which means MOs should be expanded in terms of a basis set:

$$ \Gamma = 2\pi \ \sum_{\mu, \nu,\lambda, \sigma \ \in \ce{A} }C_{\mu,a}\ C_{\nu,b}C_{\lambda,c}C_{\sigma,d}|\langle \mu\nu|V|\lambda \sigma \rangle|^2 \delta \ (\epsilon_{\mu}-\epsilon_{\nu}-\epsilon_{\lambda}+\epsilon_{\sigma}) \tag{2}$$

where $\mu,$ $\nu,$ $\lambda,$ $\sigma$ are the states in Gaussian basis sets, and $\sigma_\mu,$ $\sigma_\nu,$ $\sigma_\lambda,$ $\sigma_\sigma$ are the corresponding energies.

It seems like a straightforward calculation, but I am completely confused because I don't know how I can calculate the energy of a basis sets! I think the energy of the basis set is not meaningful.

Somebody suggested using a numerical atomic orbital basis set to compute this kind of a decay rate. For a numerical atomic orbital, the energy of states is meaningful, but I don't know how to compute this kind of a basis set either!

Are there any suggestions to solve this problem?

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    $\begingroup$ I'm not sure how the partitioning to atom $A$ will work. Would all AO basis functions $\mu, \nu, \lambda, \sigma$ have to be centered on $A$? Just the ones on one side of the integral? (The same is true for the answer by @jezzo) $\endgroup$
    – TAR86
    Commented Apr 15, 2020 at 19:18
  • $\begingroup$ In the summation, one just considers the terms which localized on atom A. $\endgroup$ Commented Apr 15, 2020 at 19:33

1 Answer 1

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First, these Coulombic interaction integrals can be simplified to $ \langle \mu\nu|\frac{1}{r_{12}}|\lambda \sigma \rangle $, which is the common two-electron integral in the AO basis (your atom-centered gaussian basis functions).

You are correct that energies are only well-defined for eigenfunctions of the Hamiltonian, and thus basis set functions (the choice of which is arbitrary) do not have a well-defined energy. Therefore, the most straightforward workaround for this is to calculate the two-electron integrals in the AO basis, forming the rank 4 tensor of two electron integrals, $g(\mu,\nu,\rho,\sigma) = \langle \mu\nu|\frac{1}{r_{12}}|\lambda \sigma \rangle $. Then, you must (approximately) solve the Schrödinger equation with some level of electronic structure theory (e.g. CCSD, MP2, etc.) to solve for the molecular orbital coefficient matrix, C. The a$^{th}$ column of the matrix C is the weightings of the AO basis (atom-centered Gaussian basis functions) to give the a$^{th}$ molecular orbital with well-defined energy $\epsilon_a$.

Finally, you should rotate $g(\mu,\nu,\rho,\sigma)$ with the coefficient matrix C to yield $g(a,b,c,d)$: $$g(a,b,c,d)=C_{\mu a}C_{\nu b}g(\mu,\nu,\rho,\sigma)C_{\rho c}C_{\sigma d}$$ Note that, in this form, this change of basis scales as $N^8$, where N is the number of basis functions used, but can be implemented efficiently (by doing one rotation at a time) to scale as $N^5$. See the Crawdad programming tutorial on MP2 (the site seems to be down right now) for more discussion on tensor rotations. Although it was done for MP2, it works identically for you.

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  • $\begingroup$ I am not wondering about rotating rank 4 tensor of two-electron integrals. The main problem is the delta function which makes sure energy conservation in the system. To implement energy conservation in the system, the energy of the basis set is required (because by using the molecular orbital coefficient matrix, the two-electron integral is transferred to Gaussian basis space) but as I mentioned it is not a meaningful concept. $\endgroup$ Commented Apr 15, 2020 at 19:30

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