Is there any theoretical way to know if a material emits a fluorescent light or a phosphorescent one? Is there any relationship between this process and the singlet-triplet energy splitting $\Delta E = E_s - E_t$ (which is the energy difference between singlet and triplet excited states)?
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Yes, approximately.
Mostly one simply calculates transition rate according to the Fermi Golden Rule:
$$\Gamma_{1\,2}\propto \left|\langle \psi_1\mid \vec{x}\mid\psi_2\rangle\right|^2\tag{1}$$
so that the rate will be nought (theoretically no transition!) when $\langle \psi_1\mid \vec{x}\mid\psi_2\rangle=0$. The rate is nonzero only if the beginning electron state $\psi_1$ and the end ground state $\psi_2$ are of opposite parity.
Here, naturally, $\vec{x}$ is the position observable. I discuss all this further, together with the vagueness inherent in the words "fluorescence / phosphorescence" in my answer to this physics SE question.
If the transition rate is nonzero, the decay can only happen as a many stage process, and one would need to write down detailed rate equations to calculate the correction to the rate if the rate in (1) turns out to be zero. The corrected rates naturally tend to be much longer, hence one usually says that fluorescent materials are ones where the Fermi Golden Rule rate is nonzero, and phosphorescent materials the others.
But the difference can be vague: some phosphorescent processes (by this definition) are still remarkably swift.
Selene Routley
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Thanks for the useful reply. Do you have any idea about the reason why we need to calculate the singlet-triplet energy splitting? – Ou Hala Feb 23 '18 at 17:42