I am very much familiar with Atomic spectroscopy, not much with nuclear spectroscopy.

In atoms (electronic cloud), we have electromagnetic interaction that plays the leading role, whose exchange boson is the photon, and that's why we have photon spectroscopy.

Now, we have the nucleus, where allegedly strong force plays a crucial role, whose exchange boson is the pion (Yukaway, Nobel prize 1949), although later on it's been understood that gluons are responsible on a more microscopic level, but we can still use pions in the low energy regime. Having this in mind, I would expect that nuclear spectroscopy be done with pion beams. But that's not generally the case! "... studies of $\gamma$-ray emission have become the standard technique of nuclear rspectroscopy", Krane, Introduction to Nuclear Physics. Why is so?

To conclude, we of course have Coulomb interaction in nuclei, since they contain a bunch of protons. But I expect Coulomb interaction to be of secondary importance, being of much less strength. Furthermore, I would expect the excitations caused by ELM interaction to be independent from the ones caused by Strong interaction, thus leading to two spectroscopy patterns. Do we have two spectroscopy patterns in nuclear physics? If not, why?


To make myself clearer, I add something taken from the discussion within comments below, since I think it makes better the point: "Suppose to have a universe with only He nucleus and strong interaction. You would not even know what a photon is, since ELM interaction is switched off. Then, why would you need a photon to excite the He nucleus?"

  • $\begingroup$ Because transitions between states in the nucleus can emit gamma rays? There are tables and tables of nuclear energy levels in all kinds of different nuclei based on measurements of gamma emissions. $\endgroup$ – Jon Custer Oct 24 '18 at 22:43

1) In QCD without the electro-weak interaction every nucleus has many excited states (states with energies above the ground state energy) that are absolutely stable. This is because both the total baryon number and the total charge (the number of protons and the number of neutrons) are conserved, and because angular momentum and parity are conserved. This means that any state that differs from the ground state in terms of angular momentum, parity, or isospin, cannot decay to the ground state. If the state has an excitation energy above the pion mass then it could in principle decay by pion emission, but this is irrelevant in practice because nuclear level spacing are much lower than the pion mass.

2) These states become unstable when coupled to the weak and EM interactions, and they can decay by photon (or neutrino etc) emission. This is very useful precisely because the photon is weakly coupled. It implies that the width of the state is small, and that it can be resolved with high precision.

3) You can try to to resolve excited states with hadronic (pion or proton) beams, but this is much more complicated because the probe is now strongly coupled to the nucleus, and the scattering reaction is not only sensitive to the nuclear levels, but also to initial and final state interactions.

  • $\begingroup$ Thanks. This helps me understand a little, even though it looks all strange to me. Also in atomic physics the charge, angular momentum, parity, energy are conserved. Yet we get a decay via the interaction boson, the photon, that carries away the difference in angular momentum and else. It looks to me that this brings into the game an isospin asymmetry in nuclear strong realm: the protons' part of the nuclear WF that is due to the strong force is weakly coupled to ELM, while neutrons' is not. I was hoping to have somewhat more independence between strong-elm. $\endgroup$ – Wizzerad Oct 29 '18 at 18:45
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    $\begingroup$ Yes, a nucleus without EM is like an atom with only the Coulomb interaction, and no coupling to the radiation field (all levels are now stable, so you can't do spectroscopy!). Also, isospin breaking is indeed a complicated business: 1) Isospin is already broken by the strong interaction, 2) Isospin selection rules are crucial for understanding EM transitions, but $\Delta I=0$ is allowed, even for an $E1$ transition (except in a $N=Z$ nucleus), 3) Neutrons don't directly couple to the charge (they couple indirectly, via the strong coupling to protons), but they have a large magnetic moment. $\endgroup$ – Thomas Oct 29 '18 at 19:46
  • $\begingroup$ Thanks. Everything's approx clear now. Could you suggest me something to read about it (nuclear wavefunctions, ELM coupling, gamma ray transition matrix elements, ...)? $\endgroup$ – Wizzerad Oct 30 '18 at 21:48
  • $\begingroup$ This is discussed in standard text books on nuclear physics, eg. chapter 10 of Krane, "Introductory Nuclear Physics" $\endgroup$ – Thomas Oct 31 '18 at 4:18
  • $\begingroup$ I could not find much in Krane. I found some more details here www.umich.edu/~ners311/CourseLibrary/bookchapter16.pdf $\endgroup$ – Wizzerad Nov 1 '18 at 5:41

In both atomic and nuclear spectroscopy, we can gain information about the spacing of the energy levels inside an object by exciting the object and then looking at the energy emitted when it transitions back to the ground state. In atoms, this energy is in the form of photons in the infrared/visible/UV range. In nuclear spectroscopy, this energy is emitted in the form of photons in the gamma-ray range (of course, some nuclei also tend to emit beta or alpha radiation, but this emission fundamentally changes the ground state of the nucleus, so at that point we're no longer studying nuclear structure anymore). The photons emitted from the transition of a nucleus back to the ground state are gamma rays because the spacing between energy levels in the nucleus is large.

The reason that photons are emitted during these transitions instead of gluons is because, though gluons are massless, they are themselves color-charged and hence tend to self-interact before they propagate very far. The practical consequences of this is that the strong force has a very short range, and therefore does not have much of an ability to transfer energy over macroscopic distances, unlike the photon, which is electrically-neutral and does not self-interact. This short "range" of the strong force is reflected in the effective field theory that describes the residual strong force via its use of massive force carriers. Pions have a mass of 135 MeV. In order to create an on-shell pion (i.e. a "real" pion, as opposed to a virtual pion, and importantly, one that is able to propagate long distances) during a transition, the transition must release at least that much energy. But 135 MeV is far more energy than almost any nuclear transition would release. In contrast, photons are massless, and can therefore be on-shell at any energy.

As for the use of a pion beam in scattering experiments, there are several reasons why this is not typically done. For instance, it's difficult to make a pion beam, certainly more difficult than making an electron beam. It's also more difficult to analyze the resulting collisions; even if the pion remains intact throughout the collision (which is not guaranteed), neutral pions do not show up in electromagnetic calorimeters, which easily detect electrons (electrons don't disappear during collisions, either, due to conservation of lepton number). In addition, due to the pion mass being much closer to the proton and neutron mass, it is more difficult to distinguish pions and protons/neutrons than it is to separate protons/neutrons from electrons. Pions also may decay on their way to the detector, in which case their trajectory must be reconstructed by inference from their decay products, whereas electrons do not. Finally, there's usually no real advantage to using a pion beam, as opposed to an electron beam, since what you're usually looking for in a spectroscopy experiment is how the intact nucleus reacts to an input of energy, whether that energy was deposited in the nucleus with a photon, a weak boson, or a gluon. That's not to say that people aren't using pion beams to probe interesting aspects of nuclear structure (neutron scattering is another useful tool that probes mainly the residual strong and weak interactions within the nucleus), but these are several reasons why they aren't the go-to experimental tool.

  • $\begingroup$ Thanks. I agree that "gluons are colored and hence tend to self-interact before they propagate very far", but I do not see it as a reason for which photons should be used instead. As I see it, a quantum mechanical theory that aims to describe something with the Yukawa model where the exchange boson is 'X' (different from the photon) needs to conceive transition amplitudes performed by the particle 'X'. Saying it differently, a particle that interacts strongly and not charged (say the neutron, low energy), must not be (and is not) sensible to photons. Again differently: isospin symmetry. $\endgroup$ – Wizzerad Oct 25 '18 at 21:16
  • $\begingroup$ Similarly, I agree that pions, having a strong mass, cannot be used as transition quanta, unless the transition energies are close to 135MeV. But again I do not see it as a valid reason. If your theory conceives pions as transition-bosons, and then you discover that they are too heavy, it's not ok to say "yeah, let's keep the theory, but for transition amplitudes just use photons instead". $\endgroup$ – Wizzerad Oct 25 '18 at 21:20
  • $\begingroup$ @Wizzerad The neutron is sensitive to photons, though; it has a nonzero magnetic dipole moment due to its internal structure. $\endgroup$ – probably_someone Oct 25 '18 at 22:04
  • $\begingroup$ I said low energy. The B field from a photon is a factor c suppressed compared to E field. We see this from the planewave or, for ex, in the Rosenbluth formula, where $G_M$ is multiplied by Q^2/m^2 with respect to electric $G_E$. Magnetic dipole is in general a 2nd order coupling. What I mean is very simple. We are physicists, we love to suppose ideal scenarios. Suppose to have a universe with only He nucleus and strong interaction. You would not even know what a photon is, since ELM interaction is switched off. Then, why would you need a photon to excite the He nucleus? $\endgroup$ – Wizzerad Oct 26 '18 at 17:23

A better way to look at this might be to compare spectroscopy done with electromagnetic interactions (gamma-ray spectroscopy) to spectroscopy done with strong interactions. People certainly do spectroscopy with strong interactions. For example, you can send a beam of deuterons (2H) at a target and observe reactions in which a neutron is absorbed from the deuteron into the target nucleus, and the proton flies off and has its energy measured.

But such techniques have limitations. For example, a light beam particle like 2H can't bring in very much angular momentum, so you can't study high-spin states. Also, a nucleus can have a very large number of excited states, and to disentangle them, you need to do more than just measuring a spectrum. The standard technique with gamma-ray spectroscopy is to measure correlations between different gamma ray energies.

Emission of pions would occur only at very high excitation energies, and at such energies, you're not really finding out the structure of the nucleus (nuclear structure physics), you're blowing the nucleus apart and studying things like the transition to quark-gluon plasma.

I would expect that nuclear spectroscopy be done with pion beams.

Usually when people say "beam," they mean an accelerator beam. Here, that doesn't seem to fit the context of the rest of your question. Did you just mean detecting pions emitted by nuclei?

  • $\begingroup$ Thanks. Your reasoning is similar to @probably_someone, so I progressed the discussion there, sorry. I edited the question to make myself clearer. $\endgroup$ – Wizzerad Oct 26 '18 at 18:57

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