Multiwfn forum / Emission energy

Re: Emission energy

dummy@example.com (solan) — Tue, 02 Dec 2025 10:18:57 +0000

Thank you so much for your answers. I was mixing up some concepts. Thank you so much!

Re: Emission energy

dummy@example.com (sobereva) — Fri, 28 Nov 2025 03:50:17 +0000

I don't understand your question.
Excitation energies are dependent of geometry. For example, the excitation energies calculated at S0 minimum and S1 minimum geometries are different.

Re: Emission energy

dummy@example.com (solan) — Wed, 26 Nov 2025 16:37:24 +0000

I have a new question: if I calculate for a specific molecule (as a single point) without optimization its absorption spectrum (S0->Sn) and its emission spectrum (S1->S0), given that it starts from the same geometry and there is no optimization process, is the same expected for the first Excited State 1?

Re: Emission energy

dummy@example.com (sobereva) — Mon, 24 Nov 2025 18:17:52 +0000

This is not necessarily emission wavelength.
It is emission wavelength only if the two conditions are satisfied:
(1) The current geometry is the minimum of potential energy surface of actual emission state.
(2) The calculated 1st excited state indeed corresponds to the actual emission state.
For fluorescene, usually the emission state corresponds to S1.

Emission energy

dummy@example.com (solan) — Mon, 24 Nov 2025 18:07:54 +0000

Dear Tian Lu:
First of all, thank you very much for your versatile program and for this very useful and enriching forum.
My question today concerns the calculation of the emission spectrum. Many reviews of TD-DFT show equations for calculating absorption and emission vertical energies. Gaussian's TD-DFT calculation provides the following:

Excited State 1: Singlet-A 4.2390 eV 292.48 nm f=0.0316 =0.00.

Does this value in nm directly indicate the emission wavelength? Or should this value be used to calculate the emission energy?

Thank you very much for everything and best regards.