It is possible to calculate it but the problem is far from trivial, and there are quite a lot of 'traps' in the process...

But first, this question has been done before - just follow the posted link.

Let's go step-by-step, then...

So, first approximation: since we're calculating this in an isolated molecule, we can discard collisional desactivation. So, this leave us with the following deactivation processes:

1) radiative transition, subdivided into:

 - spin- and dipole-allowed processes (very fast lifetime)

 - spin- and dipole-forbidden processes (low lifetime)

2) Non-radiative transitions. This can be due to internal conversion (interaction between electronic states of different or same spin or symmetry) as well as couplings with the kinetic energy operator, or even tunneling.

So, as you can see until now, a lot of assumptions have to be made. But we haven't yet scratched the surface...

So, two things are important to perform an ab initio calculation: the AO basis set, and of course, the theoretical approach.

Point one: You can't use a weak basis set. You need to add additional functions, specially if you're going to have a dissociation in excited atomic states. So you need a good basis set. The heavy kind. Bu the 'good news' don't stop here. Most likely, you're going to have some Rydberg states involved, and as such, you need to included also the corresponding diffuse functions that represent such states in the AO basis. Data on the AO's is widely available, so, you won't have a real problem with that - just on digesting this...

So, for a very feeble calculation, you need at least a double-zeta basis set with polarization included, and you need to include something to take into account electron correlation - which you obviously need to describe the actual excited orbitals.

Ok, that settles it for AO's. Now off to the method. So, the SCF (generic HF) doesn't work here. Reason? the movement of the particles is not completely independent from each other - in other words, there's correlation in their motion, something you don't take into consideration in the independent particle approach. This will give off errors in energy. So, what is missing here is the correlation energy. This correlation energy results from the pairwise interaction of electrons, obviously changing when this pair is broken in either dissociation or electronic excitation. Thus, this energy value differs from electronic state to electronic state, and as such, in order to get an accurate picture of it, you need to take into consideration electron correlation - specially the difference in electron correlation in the various states. So, forget pure HF.

So, ok, let's skip the Hamiltonian formalism, and assume you use a CI type calculation (sorry, I'm not going for the TD-DFT approach...)

So, you need to calculate the probability of spontaneous emission from the initial state ψn to the final state ψm. For that, you use good old Einstein transition probability, Anm:

Anm=(64π4/3h)νnm3 |Rnm|2 ; Rnm< ψn|Ô| ψm>

And since we have an exponential decrease in the number of systems in the state n,

Nn=N0n exp [-Anm]t , where t is time,

You'll get that after a certain amount of time, you'll have

τ= 1/Anm,

which means the number of systems left in the state n is reduced to 1/e of the initial number - which is also considered the lifetime of the state n.

And this is just skimming the surface!!! It's the most basic assumption you can make.

Now, literature on this: good and plenty, in the pic below.

Also, Petter Persson from the University of Lund is quite an expert on this kind of things. He's done some work envolving Iron (10.1021/jz500829w). You can find him here on ResearchGate and ask him for advice on how to tackle your problem. His RG profile is listed in the links below!

PS: RG, please, let one use HTML compliant code in answers... I hate to copy paste symbols when I can directly enter their html codes...

PPS: I've only now seen that this question has been answered by many people before, so I've been kind of repeating things you prolly know already...

https://www.researchgate.net/post/how_can_we_perform_excited_state_life_time_calculations

https://www.researchgate.net/profile/Petter_Persson

You must consider all decay channels, Wojciech.

OK, but still don't know how  :(

To my mind there is no standard procedure in software to do this. Formally, You must calculate probablities (matrix elements) for all possible transitions.

Yes, I have done it. Actually Gaussian gives you coefficients that allow you to calculate this. But still no idea how to calculate lifetime from that.

Inverse sum of probabilites is life time.

Thank you Sir :)

You can look into this article (http://pubs.rsc.org/en/Content/ArticleLanding/2014/RA/c4ra02473g#!divAbstract) where they have calculated excited state life time.

 Thank you very much. It helped as well. I could avoid getting matrix data from calculation. It was horrible ;)

Perhaps my answer come too late...

A good starting point is the work reported by the group of Nadia Rega:

http://pubs.acs.org/doi/abs/10.1021/jp3021485

What you need for estimating the RADIATIVE part of the excited state lifetime is to calculate the vertical emission energy and the transition dipole moment. For getting these date you need to optimize the excited state geometry and a SP (ground state) calculation on that optimized geometry.

Hope it helps!

Vasile Chis, Thanks for sharing an article on fluorescence lifetimes.

Is it possible to calculate non-radiative rates theoretically?

At least quantum mechanically I think it is not possible. However, you can estimate the non-radiative rate once you have the calculated radiative rate, the measured experimental fluorescence time and the measured quantum yield.

It is possible to calculate it but the problem is far from trivial, and there are quite a lot of 'traps' in the process...

But first, this question has been done before - just follow the posted link.

Let's go step-by-step, then...

So, first approximation: since we're calculating this in an isolated molecule, we can discard collisional desactivation. So, this leave us with the following deactivation processes:

1) radiative transition, subdivided into:

 - spin- and dipole-allowed processes (very fast lifetime)

 - spin- and dipole-forbidden processes (low lifetime)

2) Non-radiative transitions. This can be due to internal conversion (interaction between electronic states of different or same spin or symmetry) as well as couplings with the kinetic energy operator, or even tunneling.

So, as you can see until now, a lot of assumptions have to be made. But we haven't yet scratched the surface...

So, two things are important to perform an ab initio calculation: the AO basis set, and of course, the theoretical approach.

Point one: You can't use a weak basis set. You need to add additional functions, specially if you're going to have a dissociation in excited atomic states. So you need a good basis set. The heavy kind. Bu the 'good news' don't stop here. Most likely, you're going to have some Rydberg states involved, and as such, you need to included also the corresponding diffuse functions that represent such states in the AO basis. Data on the AO's is widely available, so, you won't have a real problem with that - just on digesting this...

So, for a very feeble calculation, you need at least a double-zeta basis set with polarization included, and you need to include something to take into account electron correlation - which you obviously need to describe the actual excited orbitals.

Ok, that settles it for AO's. Now off to the method. So, the SCF (generic HF) doesn't work here. Reason? the movement of the particles is not completely independent from each other - in other words, there's correlation in their motion, something you don't take into consideration in the independent particle approach. This will give off errors in energy. So, what is missing here is the correlation energy. This correlation energy results from the pairwise interaction of electrons, obviously changing when this pair is broken in either dissociation or electronic excitation. Thus, this energy value differs from electronic state to electronic state, and as such, in order to get an accurate picture of it, you need to take into consideration electron correlation - specially the difference in electron correlation in the various states. So, forget pure HF.

So, ok, let's skip the Hamiltonian formalism, and assume you use a CI type calculation (sorry, I'm not going for the TD-DFT approach...)

So, you need to calculate the probability of spontaneous emission from the initial state ψn to the final state ψm. For that, you use good old Einstein transition probability, Anm:

Anm=(64π4/3h)νnm3 |Rnm|2 ; Rnm< ψn|Ô| ψm>

And since we have an exponential decrease in the number of systems in the state n,

Nn=N0n exp [-Anm]t , where t is time,

You'll get that after a certain amount of time, you'll have

τ= 1/Anm,

which means the number of systems left in the state n is reduced to 1/e of the initial number - which is also considered the lifetime of the state n.

And this is just skimming the surface!!! It's the most basic assumption you can make.

Now, literature on this: good and plenty, in the pic below.

Also, Petter Persson from the University of Lund is quite an expert on this kind of things. He's done some work envolving Iron (10.1021/jz500829w). You can find him here on ResearchGate and ask him for advice on how to tackle your problem. His RG profile is listed in the links below!

PS: RG, please, let one use HTML compliant code in answers... I hate to copy paste symbols when I can directly enter their html codes...

PPS: I've only now seen that this question has been answered by many people before, so I've been kind of repeating things you prolly know already...

https://www.researchgate.net/post/how_can_we_perform_excited_state_life_time_calculations

https://www.researchgate.net/profile/Petter_Persson

Alcides, my comment on the top of this page was an answer to the last question on the previous page, related to the possibility to calculate the non-radiative rates.

Yes, Vasily, I've just realized that there were two pages in this thread. My bad! I missunderstood it. Sorry!

If you use semiempirical methods, Alcides, then correlations are included in empirical parameters.