Does the lifetime of a molecular excited electronic state, for example the $S_1$ state, suffice for thermal equilibration before spontaneous emission occurs?

For example the fluorescence of coumarine, squaraine or cyanine dyes. Does the vibronically resolved emission spectrum of the $S_1$ state follow the distribution expected for emission from a canonical vibrational density operator of the emissive electronic state if the spectrum is measured at temperatures around 300 Kelvin, lets say in solution.

It is well known that in most cases electronic states relax quickly after excitation using for example a laser, before emission occurs from the $S_1$ state, following Kasha's rule. But it doesn't address the population of the vibrational state of the $S_1$ from which emission occurs. I assume that the vibrational ground of the $S_1$ state dominates in most cases, but what happens at higher temperatures? Is a change of the emission spectrum observable or is the emission spectrum generally insensitive to temperature, which would indicate that the fluorescence lifetime is not sufficient for thermal equilibration to occur.

My main concern is how the "initial conditions" for the simulation of vibronic emission spectra should be chosen. If thermal equilibrium can be assumed, then the density operator for a canonical ensemble is appropriate. I would like to know if this is justified by the rate for thermal equilibration vs fluorescence lifetime.


1 Answer 1


In solution generally yes, measurements show that vibrational relaxation takes less than 5-10 ps and excited state lifetimes are usually longer than this so a thermal population exists after this time.

In isolated molecules in the gas phase it depends on which vibrational level is excited. If $v=0,1,2$ then and transfer to other levels can be very slow, tens of nanoseconds, but as $v$ increases this transfer rate constant to other levels in the same or other states rapidly increases. Fermi Golden rule type of behaviour. There is a huge literature on this from the 1970's, non-radiative transitions. See Heller, Freed, Gelbart J. Chem. Phys. 56, 2309–2328 (1972)

Edit. If $v$ is small in the excited state and there are no collisions then emission comes from this level in the excited state. This is clearly not a Boltzmann distribution, but adding inert gas causes transitions between vibrational levels until Boltzmann dist'n is reached. This can take nanoseconds. See Proc. Roy. Soc. London. Series A, Mathematical and Physical Sciences, Vol. 340, No. 1623, 1974.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.