In atomic physics we say that only transitions with $\Delta l = \pm 1$ are allowed. This is since photons are bosons.

But for example in nuclear physics we also consider higher order EM transitions ($\Delta l > 1$). There the photon-boson explanation isnt possible.

So why we use this explanation for the specific case of $\Delta l = \pm 1$ ? What am I mixing up?


2 Answers 2


In atomic physics we say that only transitions with $\Delta l = \pm 1$ are allowed.

Not really. The electronic transitions between different energy levels in atoms are typically restricted to $\Delta l = \pm 1$, but this is not a universal rule.

Instead, there is a whole hierarchy of selection rules: electric dipole ("E1") transitions are restricted to $\Delta l=\pm 1$, but there are also magnetic dipole ("M1") transitions with $\Delta l=0$ and $\Delta s=1$, electric quadrupole ("E2") transitions with $\Delta l=\pm 2$, and so on; Wikipedia has a nice summary table of the resulting selection rules.

As a general rule, each step up in this transition (i.e. going from E1 to M1 or E2, from there to M2 or E3, and so on) makes the transition less likely by a factor of $a/\lambda$, where $a$ is the characteristic size of the states involved (for electronic transitions in atoms, ${\sim}1\:\unicode{xC5}$) and $\lambda$ is the wavelength of the emitted radiation (${\sim}500\:\rm nm$ in the visible domain).

For electronic transitions in atoms, this ratio is rarely bigger than $1/1000$, so we can generally ignore (and call "forbidden") any transition higher than E1, as they will not show up in experiments unless we specifically go looking for them. (On the other hand, it is perfectly possible, and quite routine, to find these transitions if you do go looking for them. You just need a powerful laser and a high-resolution spectrometer.)

In nuclear physics, the characteristic size $a$ is typically smaller (on the order of 1 fm), but the wavelength is often also much shorter (and often approaching 1 fm in the hundred+ MeV range), so the ratio $a/\lambda$ is much shorter.

Finally, this explanation,

This is since photons are bosons.

is just dead wrong. The fact that photons are bosons has nothing to do with anything. What does matter is that photons have spin angular momentum $1$, which puts the lowest-order transitions at $\Delta l=\pm 1$; higher-order rungs (as well as anything with $\Delta J=0$) requires a photon with orbital angular momentum instead of just spin, and this is what drives the transition probability down.


The selection rule expresses angular momentum conservation and photons have spin S=1. For bosons with a different spin a different rule will apply. Higher order transitions will break this rule, as will spin-orbit coupling.


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