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In high energy accelerator collisions, why does the De Broglie wavelength of the incident particle affect the type of interaction it has with the target nucleus?

E.g. In 280 MeV proton, "direct reactions", the De Broglie wavelength is comparable to nucleon-nucleon distances so proton interacts with a single nucleon not the whole nucleus. At higher energies wavelength becomes fraction of a proton; the interaction occurs in quark scale.

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There are a couple of different "hand-wavy" ways to think about this qualitatively. Neither of them are fully correct, and neither of them explain every phenomenon, but both are decent rules of thumb.


Binding energy

A nucleus has a particular binding energy, as does a proton. If the momentum transfer of the colliding thing is significantly smaller than the binding energy of the nucleus, then the force transferred to part of the nucleus will easily be absorbed by the nucleus without affecting it very much. In contrast, if the energy of the colliding thing approaches the binding energy of the nucleus, then the nucleus no longer has the ability to absorb the energy of the collision, and you begin to push individual nucleons around with impunity (in nuclear physics, this marks the transition between elastic scattering and quasielastic scattering).

But the binding energy of the nucleon is significantly higher than the binding energy of the nucleus, so the nucleon can still easily absorb the energy of the collision without it affecting the nucleon much. But at still higher scales, when the momentum transfer of the colliding thing approaches or exceeds the binding energy of the nucleon, then the nucleon can no longer completely absorb the momentum transfer of the collision, and so the colliding thing can start pushing quarks and gluons around.

This also works in atomic physics: for photons with energy significantly less than the binding energy of the electron cloud, the electron cloud can easily absorb the momentum transfer, so these photons tend to induce large-scale collective motion of the entire electron cloud. Photons of higher energy, ones that approach the binding energy of the electrons in the valence shell, begin to push around individual electrons, exciting electronic transitions instead. And photons with still higher energy, exceeding the binding energy of the electrons in the valence shell, will unbind the electron cloud, ionizing the atom.


Diffraction

Waves diffract around things that are significantly smaller than their wavelength. This is a fact of wave mechanics in general, classical and quantum alike. A colliding thing with a de Broglie wavelength longer than a nucleon will tend to diffract around individual nucleons, only being stopped by objects that are larger, like a nucleus. In contrast, a colliding thing with a de Broglie wavelength smaller than a nucleon will not diffract around (and hence will interact with) a nucleon, but will still diffract around individual quarks. A colliding thing with a tiny de Broglie wavelength, likewise, will begin to "see" (i.e. not diffract around) individual quarks.

This viewpoint is particularly useful for explaining the penetrating nature of gamma rays. Lower-energy photons diffract around the fine-grained details of a solid, seeing instead a relatively uniform cloud of charged material that it continuously interacts with. But higher-energy photons don't diffract nearly as much, which means that they see a solid as being mostly empty space populated with some nuclei and isolated electron wavepackets. As such, they travel much further before being absorbed.

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    $\begingroup$ thanks for this very clear and understandable exposition. $\endgroup$ May 23 '20 at 20:08
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This is to be read in parallel to the answer by probably_someone.

The de Broglie relation,(as also the Bohr planetary model of the atom), belongs to the time before the quantum mechanical theory was validated, and it is one of the experimental observations that led to the theory.

In the small dimensions of molecules, atoms, nuclei, nucleons and quarks there are no matter waves,as the de Broglie observation implies by modeling the data, with

debrog

where p is the momentum,λ the wavelength and h Planck's constant.

In the quantum mechanical framework which is the one validated for particles and nuclei, there is no spread of the mass of the elementary point particle, there is only the probability of finding the particle at a particular (x,y,z,t) . It is the probability that displays a wave nature. The particle appears whole at each (x,y,z,t). The de Broglie relation is an envelope of this mathematical description , in a similar way that the Heisenberg uncertainty principle is an envelope of the mathematics of commutators in the theory of QM.

For quantum mechanical particles composed of elementary particles, like nucleons, the premise that the probability holds the wave nature also applies. In molecules there are the molecular orbitals, probability loci. In nucleons complicated quantum mechanical models exist , and the basic premise is that any measurement at the quantum scale is a probability measurement.

In this light, of probability waves,

E.g. In 280 MeV proton, "direct reactions", the De Broglie wavelength is comparable to nucleon-nucleon distances so proton interacts with a single nucleon not the whole nucleus.

Because the probability of interacting is very large mathematically when the de Broglie wavelength is of the size of the nucleus, when calculating the scattering of a proton of that energy with the nucleus.

At higher energies wavelength becomes fraction of a proton; the interaction occurs in quark scale.

Again, in calculations it is a matter of QM probabilities. Rutherford scattering was first found experimentally, and it was then explained rigorously in quantum mechanics. The higher the energy of the projectile the smaller the wavelength for high probability of scattering. That is why higher and higher energy colliders are required to check for compositness of the present elementary particles .

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  • $\begingroup$ Thank you for your response, could you elaborate on "probability of interacting", is it meaning at the instance of the collision the wavefunction probability density is higher for that confined region; roughly interpreting as more dB wavelengths can fit in that region. $\endgroup$
    – Efe Utku
    May 24 '20 at 13:16
  • $\begingroup$ @EfeUtku If one calculates the scattering crossection , the higher the energy of the particles, the smaller the volume in phase space available with measurable interactions. The de broglie wavelegth is an envelope of what real calculations would give. $\endgroup$
    – anna v
    May 24 '20 at 15:43

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