Are neutrons gauge neutral to all gauge interactions?

Neutron has mass, so it does couple to gravity.

However, if we focus on the strong, electromagnetic EM, and weak forces,

are there gauge interactions that can act on neutrons?

  • Neutron must be a SU(3) color singlet, which means it is in the trivial representation 1 of the SU(3) color gauge group.

  • Neutron must be a neutral carrying charge 0 under U(1) electromagnetism EM.

  • Question 1: Can a neutron carry the gauge charge of the (acted by) U(1) hypercharge gauge field?

  • Question 2: Can a neutron carry the gauge charge of the (acted by) SU(2) weak gauge field?

It looks like question 1 and question 2 depend on whether the neutrons are formed by the left-handed quarks (SU(2) weak doublet) or the right-handed quarks (SU(2) weak singlet).

If a neutron is formed by three right-handed quarks, then it looks like question 1 and 2 have answers yes, because three SU(2) weak singlets form a singlet.

If a neutron is formed by three left-handed quarks, then it looks like question 1 and 2 have answers no, because three SU(2) weak doublets cannot form a singlet.

But do my interpretations seem too weird?

Thanks for comments and answers!

PS: Of course, a neutron $n$ can $\beta$ decay to a proton $p^+$ under weak interaction. But this decay is via the internal weak interaction inside a neutron. The above question I am asking the net gauge charge of the neutron being acted by external gauge forces.

  • $\begingroup$ Left and right quarks interconvert to each other constantly via their mass terms. $\endgroup$ – Cosmas Zachos Apr 1 at 2:50
  • $\begingroup$ so? Can a net Neutron carry the gauge charge of (acted by) SU(2) weak gauge field? thanks for comments! $\endgroup$ – annie marie heart Apr 1 at 2:57
  • $\begingroup$ Since a neutron is perforce composed by booth these species, it cannot be weak-neutral. Current algebra estimates the charge, in serious textbooks, such as Li & Cheng. The same must hold for weak hypercharge, since the sum with t3 vanishes! $\endgroup$ – Cosmas Zachos Apr 1 at 11:23
  • $\begingroup$ @CosmasZachos That seems like an answer, rather than a comment. $\endgroup$ – rob Apr 1 at 11:45
  • 1
    $\begingroup$ Or is it possessive (for the neutron)? - "Are neutrons' gauge..." Can a physicist clear it up for us non-physicists? From Gauge theory: "In physics, a gauge theory is a type of field theory in which the Lagrangian does not change (is invariant) under local transformations from certain Lie groups." $\endgroup$ – Peter Mortensen Apr 1 at 16:34

Are neutrons gauge neutral to all gauge forces?

No. Neutrons have a substantial magnetic moment and thus feel a magnetic field.

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  • $\begingroup$ Based on the examples given in the question, the question is using "neutral" as a synonym for "has zero net charge", not a synonym for "inert." $\endgroup$ – Chiral Anomaly Apr 2 at 0:18
  • $\begingroup$ @ChiralAnomaly Yes, I should have read to the end of the question. $\endgroup$ – G. Smith Apr 2 at 1:00

There's a nice complementarity between the constants which describe couplings been the various weak isospin states and the photon (which we refer to as "electric charge") and a corresponding set of couplings between those flavor states and the weak neutral current, carried by the Z boson. It turns out that, in a system of units where the neutrino and the neutron have approximately unit "weak charge" (with opposite signs), the electron and proton weak charges are "small" in a way that's sensitive to the Weinberg angle.

Here's a more detailed answer about the proton's weak charge, with some links to the literature.

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The neutron is composed of quarks (and gadzillions of antiquarks) both left and right-chiral: the quark mass terms connect them to each other and both species have a role to play.

In addition, you readily see in the PDG review that the weak hypercharge connects to both species in a magnificently lopsided way (Feynman used to call it "cockeyed"). So, by necessity, the neutron cannot be weak neutral.

For instance, the quark effective Lagrangian in the SM has a term like $$ W^+_\mu J^{\mu ~+}= W^+_\mu \bar u \gamma^\mu P_L d , $$ and, as you indicated, this elicits the decay of the free neutron, since $$ \langle p| J^+ |n\rangle \neq 0 , $$ as you may read up on in standard texts, like the one by M. Schwartz.

But the range of the weak interaction represented by this is 0.1% the size of the neutron, one fermi, so it all happens deep inside it, if you wanted a dream metaphor. It's hard to see what an "external weak force" would be like. (It might be a virtual pseudoscalar coupling, like $K^-$, but pay no mind...) So the weak isospin charge of the neutron is a marshmallow mess computable in current algebra, and it is non vanishing. (Cf. Erler & Su Progress in Particle and Nuclear Physics Volume 71, July 2013, Pages 119-149.)

With similar arguments, you may see the neutral current coupling of the neutron is nonzero, and it probes both L and R quarks inside it, because the weak hypercharge U(1) couples nontrivially to both. But the clever Weinberg combination of electromagnetism remains unbroken, and a Ward identity ensures that the very long distance interactions of a photon with the neutron vanish: zero charge. (On shorter distances, there are magnetic interactions, as G.Smith's answer points out.)

  • Comment on extra comment

    Suppose we do not consider the quark sea. Just the representation theory of a fermion bound state. Does the right-handed neutron as a bound state count as gauge-neutral to all gauge forces? 𝑒𝑅𝑑𝑅𝑑𝑅, suppose that the Higgs condensate is zero (so, right-handed [SU(2) singlet] quarks cannot be paired to the left handed [SU(2) doublet] via the mass term.

OK, in this hypothetical science-fiction scenario, that operator would be a singlet. In other words, IF you managed to abscond with all left-handed fermions to another universe, leaving only R fields behind, then, indeed, your operator would be gauge-neutral. (I'll pretend the binding is not done by color, since color confinement is linked in mysterious ways to dynamical chiral symmetry breaking, that actually generates quark masses in reality: constituent quarks get a ~ 300 MeV mass even for massless current quarks.) There is also the issue of the hypercharge, now unbroken, which thus need not mix with $T_3$ of $SU(2)_L$ anymore (!); zero Weinberg angle, so it would then be identifiable with EM. Beyond shorter-distance magnetic Pauli-moment interactions, at a long distance, that notional operator would be neutral. (The black hole that ate it would provide no memorable trace of it.)

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  • $\begingroup$ thanks so much also physics.stackexchange.com/q/540856/42982 $\endgroup$ – annie marie heart Apr 1 at 14:47
  • $\begingroup$ Mysterious undetectable (except gravitationally) dark matter? $\endgroup$ – Cosmas Zachos Apr 1 at 14:50
  • $\begingroup$ arent there fermion bound states within standard model obeying this? $\endgroup$ – annie marie heart Apr 1 at 14:52
  • $\begingroup$ Of course not. . . $\endgroup$ – Cosmas Zachos Apr 1 at 14:55
  • $\begingroup$ If "there are NO fermion bound states within standard model being gauge neutral to all gauge forces." Can one prove that any fermion bound states canNOT carry a trivial representation for all gauge groups? (for all U(1), SU(2), SU(3).) It must be a nontrivial representation for some gauge group? Any literature/refs on this? $\endgroup$ – annie marie heart Apr 1 at 15:11

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