# Photon Spin in Fluorescence and Phosphorescence

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}