Quantum mechanics experiment

So it appears that if a hydrogen atom emits a photon, it not only has to transition between two states whose energy difference matches the energy of the photon, but it is restricted in other ways as well, if its mode of radiation is to be dipole. For example, a hydrogen atom in its 3p state must drop to either the n=1 or n=2 energy level, to make the energy available to the photon. The n=2 energy level is 4-fold degenerate, and including the single n=1 state, the atom has five different states to which it can transition. But three of the states in the n=2 energy level have l=1 (the 2p states), so transitioning to these states does not involve a change in the angular momentum quantum number, and the dipole mode is not available.

So what's the big deal? Why doesn't the hydrogen atom just use a quadrupole or higher-order mode for this transition? It can, but the characteristic time for the dipole mode is so much shorter than that for the higher-order modes, that by the time the atom gets around to transitioning through a higher-order mode, it has usually already done so via dipole. All of this is statistical, of course, meaning that in a large collection of hydrogen atoms, many different modes of transitions will occur, but the vast majority of these will be dipole.

It turns out that examining details of these restrictions introduces a couple more. These come about from the conservation of angular momentum. It turns out that photons have an intrinsic angular momentum (spin) magnitude of $\hslash$, which means whenever a photon (emitted or absorbed) causes a transition in a hydrogen atom, the value of l must change (up or down) by exactly 1. This in turn restricts the changes that can occur to the magnetic quantum number: $m_l$ can change by no more than 1 (it can stay the same). We have dubbed these transition restrictions selection rules, which we summarize as:

$\Delta l=\pm 1,\Delta m_l=0,\pm 1$

(1)