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solan

Member

Registered: 2024-04-02

Posts: 14

Emission energy

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 <S**2>=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.

sobereva

Tian Lu (Multiwfn developer)

From: Athens, Greece

Registered: 2017-09-11

Posts: 2,153

Website

Re: Emission energy

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.

solan

Member

Registered: 2024-04-02

Posts: 14

Re: Emission energy

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?

sobereva

Tian Lu (Multiwfn developer)

From: Athens, Greece

Registered: 2017-09-11

Posts: 2,153

Website

Re: Emission energy

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.