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If I have a open shell system (10 alpha electrons and 9 beta electrons) and running TDDFT calculations, is it possible to have a 9a→10b transition? Or does it have to be 9b→10b? Because both exciation won't change the multiplicity of the system, ending with one un paired electrons.

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If by alpha and beta you mean spin up and spin down, you would need spin flip TDDFT.

Standard TDDFT is for the density - density response function $\chi_{\rho\rho}$. It describes the response of the system to a scalar potential $V$ which couples with the density. This captures the effect of an external longitudinal electric field. The density operator is composed by a pair of operators with the same spin, and moreover the spin index is summed. $\rho=\sum_\sigma \rho_\sigma=\sum_\sigma \psi^\dagger_\sigma\psi_\sigma$. This is why you cannot flip spin.

First you need to move to TDSpinDFT. A theory in terms of $\rho_\sigma$ and $V_\sigma$. This captures both a longitudinal electric field and the z component of a magnetic field.

Then you need to generalize this to include the spin flip channel, using $\rho_{\alpha\beta}= \psi^\dagger_\alpha\psi_\beta$. With that you also capture the response to the x and y components of an external magnetic field. This is what can flip the spin.

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  • $\begingroup$ Welcome! It is not necessary to post your answer again. For future reference, if you cannot undelete your post, raise a custom flag on it, so moderators can do it for you. $\endgroup$ Jul 5, 2022 at 20:42

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