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So in a similar way to electrons moving in atoms, causing induced dipole-dipole interactions, can neutrons momentarily attract or repel?

Qmechanic
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Joshua Farrell
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1 Answers1

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It seems that within the standard model of particle physics

A permanent electric dipole moment of a fundamental particle violates both parity (P) and time reversal symmetry (T). These violations can be understood by examining the neutron's magnetic dipole moment and hypothetical electric dipole moment. Under time reversal, the magnetic dipole moment changes its direction, whereas the electric dipole moment stays unchanged. Under parity, the electric dipole moment changes its direction but not the magnetic dipole moment. As the resulting system under P and T is not symmetric with respect to the initial system, these symmetries are violated in the case of the existence of an EDM. Having also CPT symmetry, the combined symmetry CP is violated as well.

Thus experiments trying for new physics beyond the standard model have tried to measure it for the neutron:

The neutron electric dipole moment (nEDM) is a measure for the distribution of positive and negative charge inside the neutron. A finite electric dipole moment can only exist if the centers of the negative and positive charge distribution inside the particle do not coincide. So far, no neutron EDM has been found. The current best upper limit amounts to |dn| $< 3.0×10^−26 e⋅cm.$

As CP violation has been observed in particle data, to a very small amount, the value due to CP violation for the electric dipole moment of the neutron is expected to be around $ 10^−31 e⋅cm$, way smaller than the present limit.

can neutrons momentarily attract or repel?

Do not forget that the neutrons are in the realm of strong interactions , and will attract other neutrons or protons, in orders of magnitude stronger reaction. The technique for the electric dipole measurement is described here.

anna v
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    but OP is asking about non-permanent dipole interactions, like London dispersion force – DavePhD Apr 06 '18 at 16:10
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    @DavePhD the standard model does not allow either permanent or instantaneous violation. Beyond the standard model , as the CP violation number shows, the effect should be equally very small, afaik. It is a moot point since quarks are a soup of strongly interacting gluons and quark antiquark pairs within the neutron , cannot be compared with orbitals in molecules and atoms. Will need lattice QCD for calculations. – anna v Apr 06 '18 at 17:32
  • Stationary states can not have electric dipole moments, but time dependent states can. http://robotics.cs.tamu.edu/dshell/cs689/papers/anderson72more_is_different.pdf – DavePhD Apr 06 '18 at 23:43
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    Since a neutron is not a fundamental particle in the Standard Model, is the first quote on point? – hmakholm left over Monica Apr 06 '18 at 23:44
  • @HenningMakholm , it should say "quanrum mechanical entitites" I guess. CP violation in hadrons , it is not simple : https://www.ncnr.nist.gov/summerschool/ss09/pdf/Filippone_FP09.pdf – anna v Apr 07 '18 at 03:39
  • @DavePhD The analysis I have seen need time or CP violation to do this, i.e. outside the standard model. Do you have a link that disputes this? After all the standard model is not a la cart, instantaneous or averaged? It should hold for all (x,y,z,t)? The London dispersion forces do not appear for a single atom, but for a quantum mechanical system of two atoms where there are large distances within the charge densities of the individual atoms. This is not true for two neutrons with their complicated quark content . ( form factor analysis does show radial distributions of charges). – anna v Apr 07 '18 at 03:56
  • See https://arxiv.org/pdf/0705.2409.pdf – anna v Apr 07 '18 at 03:58
  • In your link, the argument is based on laws of motion, the hadronic stationary states are not governed by laws of motion, in this case complicated QCD cannot be approximated as far as instantaneous differences go, the way it can be done with atoms. – anna v Apr 07 '18 at 04:17