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Hadrons have electrical moments since they are made up of both positive and negative charges. Water molecules have dipole moments for the same reason even though they are electrically neutral.

Since hadrons have three different colors then are there color moments? The quarks are electrically distinct and spatially separated enough to be detectable in scattering experiments, so the color charges should also be spatially separated enough to have a moment. Some kind of tri-pole instead of a dipole.

If not then why? Does color confinement generically and nonlinearly negate color charge seen outside of a hadron?

If so then is there any effect or is the moment so weak to have no effect at all?

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No, stable hadrons exist in color singlet states which are color neutral and have color distributions which are perfectly spherically symmetric. Thus, they have precisely zero color moment of any order. Fundamentally, the origin of this behavior is the confinement property of QCD.

The common picture of having three quarks in protons or neutrons is a bit misleading in the sense that the quarks are not spatially arranged in a triangle. I should note, however, that excited states, which are unstable under most conditions, in principle may have color moments, but I have not seen any experimental suggestion of how that might be possible.

You can understand this lack of color moment by taking a look at the analogous spin singlet states $$\frac{\left\vert \uparrow \downarrow \right\rangle - \left\vert \downarrow \uparrow\right \rangle}{\sqrt2},$$ which also have no magnetic moment of any order (quadrupole, octupole, etc.) despite conceptually being formed by two opposing dipoles.

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    $\begingroup$ In other words, your saying that color is a quantum number, not a field, right? $\endgroup$
    – Jason
    Commented Jan 26, 2021 at 12:23
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    $\begingroup$ I should clarify, I mean their color distribution is spherically symmetric (unless you know of an experiment that shows otherwise?). Certainly the spin is not spherically symmetry. $\endgroup$
    – KF Gauss
    Commented Jan 26, 2021 at 18:31
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    $\begingroup$ Note: the assumption that the electric charge distribution is spherically symmetrical is closely related to the assumption that $CP$ violation is small in the QCD vacuum. The neutron has no electric monopole moment, but the $CP$ violation evidenced by our low-antimatter universe predicts some nonzero electric dipole moment. Whether the same arguments apply to a color dipole with no net color charge seems like a nontrivial question, confinement or no. I've written an answer that's basically orthogonal to this one. $\endgroup$
    – rob
    Commented Jan 26, 2021 at 22:55
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"Color moment" could mean two different things. One of them is certainly not possible, but I'm not sure about the other one:

  • One is something like "mostly green on one side and mostly blue on the other side." That type of color moment is not possible.

  • The other is "mostly regular color (red/green/blue) on one side and mostly anticolor (antired/antigreen/antiblue) on the other side." This one is more closely analogous to the elctromagnetic version, and I'm not sure if this type of color moment occurs.

To explain, I'll need to use a little math.

Colors

Let $q_c$ denote a quark with color $c$. The $c$ is an index that takes three different values. We could call those values $1,2,3$, but calling them colors $r,g,b$ is more fun.

Similarly, let $\bar q_{\bar c}$ denote an antiquark with anticolor $\bar c$. Once again, $\bar c$ is an index that ranges over three values, which we call $\bar r,\bar g,\bar b$ (antired, antigreen, antiblue).

The relationship between $c$ and $\bar c$ is analogous to the relationship between positive and negative electric charges.

To construct color-neutral combinations (more precisely, gauge-invariant combinations), we need to sum over these indices as shown below.

Meson-like combinations

The simplest color-neutral combination is the meson-like combination involving one quark and one antiquark matheamtically connected to each other by a color-field matrix $U$ called a Wilson line: $$ \sum_{c,\bar c} q_c U_{ c \bar c} \bar q_{\bar c}. \tag{1} $$ The sums over $c$ and $\bar c$ are required by gauge invariance.

The meson-like combination has an analog in electromagnetism: it's analogous to a bound state of two charges with opposite signs. The difference is that for electromagnetism, the indices only take one value each (only one color), so the sums are not needed.

Baryon-like combinations

Another color-neutral combination is the baryon-like combination $$ \sum_{a,b,c,\bar a,\bar b,\bar c} q_a q_b q_c U_{a\bar a}U_{b\bar b}U_{c\bar c} \epsilon^{\bar a \bar b\bar c}. \tag{2} $$ Each of the letters $a,b,c$ is a color index that runs over all three colors. The quantity $\epsilon^{\bar a \bar b\bar c}$ is completely antisymmetric in its three indices, which ensures gauge invariance because $U$ is a $3\times 3$ unitary matrix with determinant equal to $1$. Once again, the sums over the color indices are required by gauge invariance.

The baryon-like combination does not have a perfect analog in electromagnetism. The analog would be $q_a U_{a\bar a}$ with only one factor of $q$, one factor of $U$, and no sums (because there's only one color), but the analogy is imperfect because the $U$ in electromagnetism does not have determinant restricted to $1$. If it's determinant were restricted to $1$, then $U$ itself would be equal to $1$, so there wouldn't be any charged matter at all.

Is either type of color moment possible?

Thanks to the sums over the color indices, we cannot have a color moment in the sense of something like "mostly green on one side and mostly blue on the other side."

The other possibility is something like "mostly regular color (red/green/blue) on one side and mostly anticolor (antired/antigreen/antiblue) on the other side." This is more closely analogous to the electromagnetic case, and I'm not sure it's impossible. The baryon-like case (2) does seem to have this type of asymmetry, but (2) doesn't account for the geometric configuration, and the expressions (1) and (2) are not meant to be complete descriptions of real mesons or baryons anyway. They are just the simplest color-neutral combinations that are meson-like and baryon-like. Real mesons and baryons are more complicated, so I'm not sure.

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  • $\begingroup$ Im thinking the answer is thats its a quantum number not a field, like digital vs analog, or like spin. Spin for electrons is related to magnetic properties but is a quantum number vs a field, so I think this might be the same case ignoring if Im even right about spin. That is what you seem to be implying as well but not 100% $\endgroup$
    – Jason
    Commented Jan 26, 2021 at 17:03
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    $\begingroup$ It seems to me that a dipole with more particles at its north pole than at its south pole would be $CP$-odd, and so would only be present in a $CP$-even particle like a nucleon at the same scale as a permanent electric dipole moment. But sneaky sign business like this makes me not-confident in that assertion. $\endgroup$
    – rob
    Commented Jan 26, 2021 at 23:07
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(This post is right on the line between a comment and an answer, because it's some things that you might enjoy digging into the literature for, rather than a proper explanation. But it's closer to an answer than to an attempt to improve your question, and plus it doesn't fit in the comment box.)

In the context of unconfined QCD, people talk about "color magnetism," in analogy with magnetic interactions between electric charges. However, one description of the phase transition to color confinement is that the low-temperature vacuum is a "color superconductor." You are probably aware that an electrical superconductor expels magnetic fields; one interpretation of color confinement is that the low-temperature vacuum expels all of the color fields. I don't know how robust this analogy is.

Also relevant to you (and hinted at in another answer, which mentions spherical symmetry): the Wigner-Eckart theorem limits the nonzero multipole moments which are available to a system, depending on its total angular momentum. This is a big deal in the community of people who are trying to measure the electric dipole moments in the neutron, proton, and electron. Those particles have spin $\hbar/2$ and can therefore have nonzero monopole moment and nonzero dipole moment, but not higher. It's common to explain the search for a permanent electric dipole moment as a measurement of a fundamental particle's non-spherical shape.

Having thought about this for all of ten minutes, I think your intuition about a "tri-pole moment" and the Wigner-Eckart restriction means that the spin-half baryons don't have enough degrees of freedom to participate in any residual-color interaction. The smallest "tri-pole" moment is the octupole, and the simplest system which can support an octupole moment has unit spin. That suggests the low-energy experimental avenue for exploring color moments would have to start with polarized vector mesons or high-spin baryon excited states. A mess.

Of course many people describe the entire nuclear force as a "residual color interaction," and would be perturbed if I said nucleons (and pions, which are pseudoscalars) couldn't participate. But even the lowest-energy two-baryon system, the deuteron, has quite a bit of $d$-wave contribution to its wavefunction; this evidence that the low-energy nuclear interaction is a tensor force was an important result in the early days of nuclear physics.

One of the things I learned trying to puzzle out the allowed quantum numbers in hadronic parity violation was that, even at zero energy, nucleons interacting by exchanging mesons are not a very parsimonious model for describing what goes on inside of a nucleus. The boundary between $\rho$ meson exchange and color non-confinement is blurrier than my strong-interaction colleagues like to believe. I would be deeply skeptical about a nonzero "color moment" assigned to any strong-unstable baryon or meson, due to medium effects, and I have already suggested above that I think all the "color moments" of the proton and neutron should be exactly zero for symmetry reasons.

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    $\begingroup$ +1 your answer and that of Chiral Anomaly are making me doubt my own answer. I'll have to think more deeply about this issue, not as trivial as I assumed in the beginning. I think the moments are still zero, but my answer needs more rigor $\endgroup$
    – KF Gauss
    Commented Jan 26, 2021 at 23:01
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    $\begingroup$ I only skimmed the other answers before I wrote this one, but reading both and thinking for an hour I'm wondering whether a color-anticolor dipole might also be allowed at the same $CP$-violating scale as the electric dipole moment. I'm not in that field any more, but I might put a bug in somebody's ear about it. $\endgroup$
    – rob
    Commented Jan 26, 2021 at 23:19

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