# Do Baryons have Quantized Vibrational and Rotational States?

Given that baryons are not point particles but are composed of three quarks, are there quantized vibrational and rotational states analogous to those of molecules?

If not why not, and if so are there spectroscopically observable transitions between such states?

What do you think a $\Delta^+$ or $\Delta^0$ is, if not an excited nucleon? (To be sure the $^{++}$ and $^{-}$ states do not correspond directly to a nucleon, because there is no allowed lower spin state with that valence content.)

The thing I am not sure about is how closely these excitation match the ones you are envisioning. They match nuclear excitation pretty well.

There are also excited states of mesons, such as the $\eta' (958)$.

A couple of people have mentioned (in the comments and in another answer) that the first excitation energy is comparable to the binding energy, which is rather different from the molecular case where the excitation energies are often very much less than the binding energies.

This is roughly true.

That said, these systems are bound by the strong force, so they exhibit a feature that may be surprising to people used to atomic of molecular physics: they remain bound at all separation (until pair creation occurs).

The proton has a mass of $938 \,\mathrm{MeV}$ and the Deltas have a mass of $1232 \,\mathrm{MeV}$. Higher resonances of the same valence content have masses of $1600$, $1630$, $1700$, $1905$, $1910$, $1920$, $1930$, $1950$, $2420 \,\mathrm{MeV}$ ... Which we see can be several times the original mass of the system!

• ok, thanks, so $\Delta^+$ is an excited proton and $\Delta^0$ is an excited neutron. I'll have to read more about this. I'm from more of chemistry background and never knew that. – DavePhD May 8 '14 at 1:44
• For various valence quark contents ($uud$, $udd$, $uds$ and so on) you can find a lowest energy example and some higher energy examples. The higher energy example represent excitation of the lower energy system, but they will be more like excitations of individual atoms then the geometrically complicated excitations of non-trivial molecules (no flexing where the benzene rings come together type of thing). – dmckee --- ex-moderator kitten May 8 '14 at 1:51
• One might want to add that for systems bound by the strong force, the energy gap between the ground state and the first excited state is of the same order of magnitude as the total mass of the ground state system. This is very different in electromagnetically bound systems! – Neuneck May 8 '14 at 10:13
• Beware that valence quark content isn't all that matters: the $\Delta$ baryons have a different strong isospin than the $N$ baryons. This is why all the $\Delta$ have four charge states, instead of two. – rob May 9 '14 at 13:50

Not 100% sure about this, but exciting a rotational energy level requires a photon with energy of h^2/(m*r^2). This is true for any object of mass m and size scale r. For molecules, m and r are are in a sweet spot where the photons are low enough energy that spectroscopy is possible. (For rotational states of molecules, this is typically in the microwave-to-submillimeter regime.)

But for a baryon, m and r are absolutely tiny. I think that the single-quantum of rotational energy may be so large that it's comparable to the baryon rest mass itself. So a photon coming in with that energy will in effect cause it to 'spin apart', classically speaking.

Sort of like for a hydrogen atom: m and r are already getting pretty small, and the 'rotational' energy h^2/mr2 is simply the Bohr scale which is close to the binding energy. If you start to spin the electron in a hydrogen atom, it goes from n=1 to 2 etc. and then pretty quickly ionization occurs.