This is all well and good, but I'm not sure it explains why a free proton can't radiate. After all, we have already determined that the proton is, in fact, a composite particle (made of quarks and gluons).
It seems that (at length scale ~0.1fm) the constituent quarks do not get excited (or bothered) by each others em-fields. So no radiation is expected from their bound state (the proton).
One can also say;
since the quarks "inside" the proton are almost "free", therefore they can not emit photons.
It is still an open question in the physics community as to whether or not a proton can decay (producing other particles that can decay into photons),
One should distinguish between decay processes and radiation processes, specially when the decay is an
em-process.
It is better to agree on the following:
a process that changes the identity of the system is (called) a decay process. And call radiation process (emiting photons), a peocess that keeps the system intact.
For example;
\pi^{0} \longrightarrow 2 \gamma
is an em-decay process ( the pion is gone),
but
e^{-} + (Z,A) \longrightarrow e^{-} + (Z,A) + \gamma
is a radiation process ( the electron and the nucleus are still there after radiation).
I don't think I've ever heard anyone discuss "energy states" for the proton. Why is this?
The problem of bound states in QCD is still unsolved.
If one knows the exact form of the
q-q potential, then (with great patience) one could find the energy levels (states) of proton and all other hadrons by solving Dirac (or even Schrodinger) equation.
People have "experimented" with many potentials, the frequently used potentials (adjusted to fit the observed states of the charmonium and bottomium) are
V(r) = -\frac{1}{2r} + 0.2 r
V(r) = 0.7 ln(r/2)
However, even the exact dependence on the quark separation (
r) does not solve the 2-body(meson) or 3-body(baryon) problem completely. as well as (
r) the potential should also depend on the ordinary and the unitary spins of the quarks. Well, the story does not end here. The exact potential should account for the fact that the proton "lives" as 5-body system for a considerable length of time;
p = uudq\bar{q}
So, I do not think we will live to see the
{}^{2S+1}L_{J}
(energy) states of the proton.
There is another model for the
q-q potential. Feynman et al.(1971) considered a relativistic hadron consisting of 2 or 3 quarks bound together by a
covariant harmonic oscillator potential. It was shown (Kim & Nos 1977) that the Lorentz-squeezed wave function of the covariant oscillator can be used to resolve a paradox in the
parton model of hadrons;(
here the OP can see what the proton looks like)
When the hadron (meson or baryon) is at rest, it appears as a "
bound state" of
2 or 3 quarks. When it moves with its velocity close to that of light, the hadronic matter becomes concentrated along one of the light-cones, with wide-spread distributions in both space-time and momentum-energy. As a consequence, the hadron appears as a collection of a large number of "
free"
partons.
So for a stationary observer, the
proton appears as a
bound state of
3quarks. But, for an observer who is moving very fast, the
same proton appears as a
plasma of
free partons
Obviously, the two observers will quarrel over the structure of the proton.
The Lorentz-squeezed wave function of the covariant harmonic oscillator (model of potential) settles this quarrel (the paradox).
Feynman R.A., Kislinger M and Ravndal F.(1971) Phys.Rev. D3, 2706-2732.
Kim Y.S. and Noz M.E.(1977), Phys.Rev. D15, 335-358.
regards
sam