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I apologize, but QFT is not my domain. What I ask is connected with the question Do the fields exist without electric charges? .

By analogy with the electron and proton, that carry the electric charges of the electrostatic field, could a neutron and anti-neutron represent the positive, respectively negative, charges of some field? I am sure that, if the answer is positive, the supposed field is not the gravitational one. In the gravitational field all the charges (masses) are positive.

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    $\begingroup$ Sofia, neutrons are bound states of three quarks and they have no net electrical charge because the quark charges balance out. It's like asking if hydrogen atoms could be the source of some field because they have no net charge either. $\endgroup$ – John Rennie Dec 11 '14 at 12:02
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The neutron is not a fundamental particle. It carries no electric charge, yet it can interact with photons as its components - the quarks - carry electric charge and thus couple to photons. Macroscopically/classical, these interactions cancel out since its net charge is zero, but quantumly, there is a very big difference between objects with charged components and objects with no charged components (or without components at all).

Additionally, quantumly, one would say "to be charged under some field" means to transform non-trivially under some gauge group. The Standard Model has the electroweak $\mathrm{SU}(2)_R \times \mathrm{U}(1)_Y$ and the strong $\mathrm{SU}(3)$ as gauge groups. The neutron is color-neutral and electrically neutral. A left-handed neutron composed of three left-handed quarks has an overall weak isospin of $\frac{1}{2}$, though, and is thus also overall "charged" under the weak $\mathrm{SU}(2)_L$, a right-handed neutron has weak isospin $0$ (Note that the actual states of a neutron may well be a superposition of the two). Since anti-particles have opposite quantum numbers (transform in the conjugate representations), a anti-neutron has either $-\frac{1}{2}$ or $0$ as the weak isospin of its chiral states. Therefore, you might say the neutron is (or can be) "charged under the weak interaction".

Note that non-vanishing weak isospin $T_3$ also implies non-vanishing hypercharge $Y$, since vanishing electric charge means $T_3 + \frac{Y}{2} = 0$. Yet the boson of the $\mathrm{U}(1)$ combines with one of the three bosons of the $\mathrm{SU}(2)_L$ to form the photon, which couples to electric charge. Non-vanishing hypercharge is, in and of itself, not very interesting.

Note also that the above looks very unlike any statement "the neutron is the source for a field". That is because, in the quantum view of interactions, all such classical statements only emerge in some non-relativistic, macroscopic limit. The quantum description of force understands under "being charged" really nothing else than "transforming non-trivially under some gauge group". Do not think of the neutron now somehow generating a "weak field" around it like one classically thinks of charges generating electric fields. This view is wholly classical, and not justified from the quantum picture alone, and, for such non-Abelian gauge groups, a bad idea anyway, since their "field strength" is not observable.

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  • $\begingroup$ Why is weak hypercharge called Y? $\endgroup$ – Harry Wilson Dec 12 '14 at 13:05
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    $\begingroup$ @HarryWilson: I haven't the slightest idea. $\endgroup$ – ACuriousMind Dec 12 '14 at 13:24

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