I think i still lack basic understanding of how absorption and emission of light actually works. Like fluorescence is the radiative transition from an excited singlet state to the ground (singlet) state. This is allowed, because there is no spin flip of the electron, i.e. (1) is allowed, (2) is forbidden, because of the Pauli principle:

$$ \begin{align} (1) \quad |\uparrow_\text{groud}, \downarrow_\text{excited} > \quad \rightarrow \quad |\uparrow_\text{groud}, \downarrow_\text{ground} >\\ (2) \quad|\uparrow_\text{groud}, \uparrow_\text{excited} > \quad \rightarrow \quad|\uparrow_\text{groud}, \uparrow_\text{ground} >\end{align} $$

But ... why? The emitted photon has a spin of 1 which corresponds to real angular momentum that has to come from somewhere. If there is no spinflip, how is angular momentum conserved? By this logic phosphorescence would be the allowed-by-default transition and not fluorescence. The spin flips from +1/2 to -1/2 and the difference goes to the emitted photon:

$$ \begin{align} |\uparrow_\text{groud}, \uparrow_\text{excited} > \quad \rightarrow \quad|\uparrow_\text{groud}, \downarrow_\text{ground} >\end{align} $$

I know this is wrong, but I don't know why. Please help me understand.

  • $\begingroup$ The answer by @LonelyProf (below) is nice and clear. The post chemistry.stackexchange.com/a/61501 doesn't answer your question quite so clearly, but it does indicate how the same answer can be expressed mathematically. $\endgroup$ – Chiral Anomaly Nov 30 '18 at 2:56

You may well get a more comprehensive answer from someone else, but the short answer is that the electron spin is not the only thing carrying angular momentum. Most fluorescent systems are molecular, and an electronic transition is usually accompanied by a change in the molecular rotational state (and the vibrational state too). Generally, the one unit of angular momentum carried by the photon can easily be cancelled out by a corresponding change in total angular momentum of the molecule, satisfying the law of conservation.


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