# How do I know which type of atomic transition this is?

In general, if I observe an atomic transtion between two internal states, how do I know whether an atomic transition is due to the magnetic dipole moment, electric dipole moment, electric quadrupole transition etc?

These all have different selection rules and parity considerations, sure, but are there any other ways to tell?

For example I am looking at the ground state of $^{87}\rm{Rb}$ in high-magnetic field:

and I know that, by shining a Radio frequency of the correct frequency, I am exciting the transitions indicated by the red arrows.

How do I know which type of transition this?

It preserves parity, so I guess it cannot be electric dipole transitions... anything else?

The transitions you described are magnetic dipole transitions.

It is not an electric dipole transition because the orbital state is still S-orbital (you stay in the ground state manifold), and electric dipole transitions should change parity (e.g. S->P. Like the D1 and D2 optical transitions in alkali metals). Also, in the high field region as you have indicated, the electron spin and nuclear spin decouple (as it is already obvious in the Breit-Rabi diagram), so you are just flipping the nuclear spin. Spins only couple to magnetic field, unless they indirectly couple to electric field via spin-orbit coupling.

You may ask, what about magnetic quadrupole and higher order terms? Well, the nuclear spin possesses a dipole moment and no higher order moment, so it only interacts with the magnetic field via the dipole interaction * (see comments below).

I agree mostly with what @jgerber already said. I will add a few remarks.

1. In general, the higher order transitions are much weaker (see https://quantummechanics.ucsd.edu/ph130a/130_notes/node422.html. It says $\vec{k}\cdot \vec{r} \sim \alpha$, and I believe this scaling is obtained as $k = \frac{\hbar\omega}{\hbar c}$, $r = a_0 = \frac{\hbar}{mc\alpha}$, and $\hbar\omega_{\text{hydrogen}} \sim \alpha^2 mc^2$), unless you shine an intense, short pulse of light whose amplitude varies significantly over the size of the atom. But I don't think you can localize a RF wave (long wavelength) so tightly.

2. I don't think you can drive any electric quadrupole transition within the ground state hyperfine manifold. The ground state is an S-orbital, so it is isotropic and has no structure.

There is also a nice book consisting of AMO problems called "Atomic Physics: An Exploration Through Problems and Solutions" by Budker et al. There are several exercises on figuring out types of atomic transitions, and how "forbidden" transitions (see alkaline-earth elements like strontium for example) work.

Extra resources: See "Quantum Optics" lecture notes by Daniel Steck. Also consult any of the alkali metal data sheet (e.g. rubidium) by Steck.

• Thanks this is great. Two questions. About only flipping the nuclear spin: but how do you know if the nuclear angular momentum I only comprise spin and not also nuclear orbital angular momentum? Sep 1 '18 at 21:59
• And why does the nucleus not have anything higher than a dipole moment? Sep 1 '18 at 22:01
• @SuperCiocia , regarding the first question (I did not know this beforehand), we commonly refer to total angular momentum of the nucleus as the nuclear spin (see hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/nspin.html). This makes sense because in most atomic transitions we are not sensitive to the internal dynamics of the nucleons - we treat the nucleus as a single entity. Also see Bill N's answer in this StackExchange post (physics.stackexchange.com/questions/169342/…).
– wcc
Sep 2 '18 at 14:25
• Also, regarding higher magnetic moments of nucleus (again I admit I did not know this beforehand), permanent non-zero magnetic quadruple moment violates CP symmetry, and this is not something that is expected in the Standard model. See this research description (hutzlerlab.com/research/nuclear-mqm).
– wcc
Sep 2 '18 at 14:34
• Sorry where do they say specifically that the internal orbital angular momentum of the nucleus does not affect the lines? Sep 3 '18 at 11:16

Given no other information I don't knows if there's much more you can say about it. You can look at this selection rules table on Wikipedia and see which types of excitation can drive this transition. As you point out parity is conserved so that rules out odd multipole electric transitions and even multipole magnetic transitions.

This leaves for example magnetic dipole, electric quadrapole, magnetic octopole, etc.

The other piece of information you have given is that $\Delta m_J = -1$. This basically doesn't rule out anything. If instead it had been $\Delta m_j = \pm 2$ then that would have ruled out magnetic dipole transitions.

However, it is of course worth pointing out that which transition you are driving depends on the multipolar decomposition of the electric/magnetic field you are driving with. For example, if the field is uniform across the entire atom (which is of size ~$a_0$) then the field can basically be expressed entirely as a dipole field. This means the excitation photons have the mode function of a dipole field. In this case ONLY dipole transitions are driven. If the field has some gradient across the atoms then it now includes quadrapole components which drive quadrapole transitions. Once you determine whether the field configuration includes dipole and quadrapole elements you must now calculate the transition matrix elements for the different types of transitions you are considering. This is basically to say how well does my ground state, when I act the dipole, or quadrapole operators on it, overlap with the excited state. It is a geometric statement about the ground and excited states. It turns out (perhaps I can give a more thorough explanation why at a different time) that these matrix elements are increasingly suppressed by certain large factors as you move to higher multipole orders.

So to answer your question, it is most likely that you are driving a magnetic dipole transition here. This is because

1) It is allowed by the selection rules (and electric dipole isn't allowed)

2) It is the lower order multipole excitation which is allowed which means it has the largest matrix element of all multipole transitions

3) You are most likely driving the transition with a field which is predominantly dipolar from the perspective of the atom. I don't know how you would get an RF field which varies dramatically over a length scale of $a_0$.

However, if you were a bit crazy, you could try to build some RF emitter which could realize a significant field gradient and try to drive this transition by, for example, an electric quadrapole transition. Perhaps you could do it with some small atom chip with tiny wires layered on it. I think if you got the polarization and frequency right you could try to drive purely $m_F = \pm2$ transitions which could be evidence that you are excited it via an electric quadrapole transition and not the easy magnetic dipole transition.