14

There's two facets to this question: What methods can be used for excited states in crystals? (the title, and final sentence) What methods can be used for excited states with plane waves? (paragraphs 1 and 3) I will begin by answering the second question, which is in some sense grounded by this assumption: "it is not possible, as far as I know, to do ...


12

As you mentioned in the question, excitons are indeed bound electron-hole pairs. They are often considered to be the signature of optical spectra in insulating solids. Adding onto the comment by tmph, there are two types of excitons: opposite spins of electrons will lead to a dark exciton with S=0 (since it doesn't allow for momentum conservation). Same spin ...


11

tldr; it's related to the excitation energy The lifetime of an excited molecule assuming there are no non-radiative pathways, is related to the Einstein A coefficient: $\left(\frac{dn_2}{dt}\right)_\mathrm{spontaneous} = -A_{21}n_2$ where $n_2$ is the population of the excited-state. I used this reference to get the following equations: J. Braz. Chem. Soc., ...


11

$\Delta$SCF This method generates excited states by changing the occupancy of a ground state determinant and then carrying out a new SCF with that initial guess, with some restriction throughout to prevent variational collapse back to the ground state [1]. The most common approach to stay out of the ground state is the Maximum Overlap Method (MOM), which ...


9

GW+BSE: Excited states in the framework of many-body Green's function comprise charged excitations, where the number of electrons in the system changes from $N$ to $N-1$ or $N + 1$, and natural excitations, where the number of electrons remains constant. In the $|N\rangle \rightarrow |N-1\rangle$ case, an electron in the valence band (occupied orbital) is ...


9

The Einstein A coefficient (as mentioned in Cody's answer) tells you the probability per unit time of spontaneous radiative decay from an excited state to a lower state. The Fermi golden rule (as mentioned in sleepy's answer) also gives a probability per unit time to transition from a specific state, but the destination is usually a group of states in a ...


8

Fermi golden rule should give you the transition probability (per time), from which you should be able to estimate the lifetime. You should know of be able to estimate the density of states in the final (ground) state, of course.


7

The question cites a 2008 paper about fairly large molecules (e.g. bithiophene N-succinimidyl esters), in which TD-DFT gave significantly wrong results, so the authors recommended RI-CC2. The question then asks if there's any example where TD-DFT was insufficient for a smaller system. After searching the literature, I have found an example of a small ...


6

I cannot see your output file, but I am pretty sure that the program is running into an error when starting the optimization i.e. when it is trying to generate the guess hessian. You are using Cartesian coordinates for optimization, and GAMESS by default will attempt to eliminate the rotational and translational modes from the guess hessian, before the ...


6

First of all, if you are only interested in molecules with about the same size as the ones in the article you cited, then TDDFT is probably not the best method for this purpose. ORCA supports many methods that are much more reliable than TDDFT for singlet-triplet gaps, yet are still affordable for molecules of this size, for example DLPNO-STEOM-CCSD, NEVPT2, ...


5

This is not a full answer, because it does not solve the problem. But I hope to shed some light on why you are not getting the correct frequencies. Short answer: I suspect there is a bug in GAMESS Long answer: First of all I would advise you to go through the manual of GAMESS. There are lots of problems in the input file, and most of them can be easily ...


5

I can see two major problems: CCD != CC2 and I'm not certain if CC2 is available in GAMESS. CC2 is an approximation of CCSD, which is available in GAMESS. You may be able to use this since $\ce{NH3}$ is a fairly small molecule, though I don't know how time consuming this will be relative to your initial CCD calculation. TZV != TZVPP. The def2-TZVPP basis ...


5

This is a complicated question for which I do not have a full answer. However, here are some thoughts: Excited state relaxation. In solids, the calculation of excited states typically requires the many-body GW approximation for quasi-particle energies and the solution of the Bethe-Salpeter equation for excitonic properties. The calculation of forces, a ...


4

TD-DFT works best for valence-valence excitation energies, but does not very well for charge-transfer excitation (unless xc potentials are constructed to lower delocalization error). Core-excitation energies are also not described will with TD-DFT.


3

As specified in the GAMESS Documentation: nspace=n means the active space will be split into $n$ groups. mstart sets the starting orbital for each of the $n$ groups. In your case, the first group extends from orbital 3 to orbital 9, while the second is from orbital 10 to NCORE+NACT (the number of core and active space orbitals). They mention that in this ...


2

While I'm not familiar with surfaces, I can provide the theory for photoredox molecules. Presumably, this can be extended to surfaces with the correct level of theory, which perhaps others can elaborate on. I'm quoting from reference 1 with some commentary in brackets The reduction potentials associated with the excited states cannot be directly measured [...


Only top voted, non community-wiki answers of a minimum length are eligible