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A recent Scientific American article brought up an old issue, which is this: According to quantum chromodynamic models, the emergence of exactly 1/2 unit of spin in a proton (or a neutron, or any other 3-quark quark system) apparently is quite mysterious due to the ambiguity of how quarks, virtual quarks, and gluons.

That brings to mind a simple question for which I hope there is a simple answer: Since the 1/2 spin of a silver atom is the result of a system of electrons and protons bound by photons, why is that situation fundamentally different (or is it?) from a set of quarks bound together by gluons?

My first guess would be that the energy levels involved in quark binding are high enough that you lose the asymptotic simplicity of QED. So if it's just a matter of "QCD is messier doncha know," I'm OK with that, I guess. But even there it would seem that the question would be one of degree, not of absolutes, and that silver atoms should still suffer from a similar ambiguity at a much lower level.

My second guess would be that since gluons carry color (analogous to photons having charge), that tosses in some factor that makes it harder to ignore their contribution.

Insights, anyone? Hardly a critical issue, but just... interesting, at least to me.


Addendum - 2015-05-25.2307 EST Mon

I am exceedingly annoyed to have to make the observation that silver atoms, despite their rich history as part of the uncovering the strange half-spin of fermions... are not fermions.

Both isotopes of neutral silver, 107 and 109, are bosons. The para versions with net nuclear spin opposite to net electron spin are spin 0 bosons, and the ortho versions with net nuclear spin parallel to net electron spin are spin 1 bosons. That's because both Ag-107 and Ag-109 have a net nuclear spin of $\frac{1}{2}$.

As best I can figure it, what is really going on with silver atoms zipping through a Stern-Gerlach apparatus is a geometric effect of the outermost electron of silver interacting far more strongly with the magnetic field gradient than the enclosed nuclear-electron composite of +1 charge and net spin $\pm\frac{1}{2}$. The geometric enclosing the composite positive fermion within the zero-orbital-momentum outermost electron orbital apparently so eclipses the inner fermion that the result appears to be a fermionic silver atom, right down to its bizarre separation into two groups when passing through a sufficiently strong magnetic gradient.

I am annoyed by this because I've always assumed that the silver nucleus had an even number of nucleons, at least in the context of Stern-Gerlach. It literally never occurred to me that it might be odd, since that (as noted) would make silver atoms bosons.

So: Why do none of the prevalent physics descriptions -- at least the ones I've seen, including in particular Feynman's very detailed discussion -- bother to mention this rather critical little point?

Does this affect my question? Well, it certainly does if the issue is how you measure spin, since clearly the "visibility" of the spin if a composite charged fermion is a complicated issue that, in the case of silver, can approach zero in certain cases.

But I think more broadly there's still and interesting simple question in all of this: If QM quantizes spin, always, then QM regardless of the binding forces involved will necessarily require the components involved to "line up" appropriately whenever the particle is in a situation where its quantized spin is measurable. Unless that more generic "top down QM quantization coordination" issue is well understood and modeled, it's a bit hard to imagine how a bottom-up perspective on spin can ever be complete.

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    $\begingroup$ Could you link the article (or, better, some papers on that)? $\endgroup$
    – ACuriousMind
    Commented May 15, 2015 at 14:58
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    $\begingroup$ if you google "lattice qcd and the spin of the proton" you will see that it is an active area of research. $\endgroup$
    – anna v
    Commented May 15, 2015 at 15:03
  • $\begingroup$ Isn't this more about the fact that you don't have a clean definition of quarks in a bound state? Like, the proton is made up of a complicated set of parton distribution functions due to flavor mixing and the rest. To a very good approximation, the silver atom is made up of electrons and a nucleus. The analogous statement about protons and quarks isn't true--the binding energy is much larger than the masses of the constituent quarks. $\endgroup$ Commented May 15, 2015 at 17:34
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    $\begingroup$ look at equation 1 here int.washington.edu/talks/WorkShops/int_14_55W/People/Zhao_Y/… . The goal is to justify the sum rule and it seems still a goal. $\endgroup$
    – anna v
    Commented May 15, 2015 at 18:36
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    $\begingroup$ Regarding part II: a weak magnetic field couples to the total angular momentum along a given direction, not the total spin. For example, paired spin-up and spin-down electrons do cancel out. So you are certainly correct that the S-G experiment did not directly measure total spin; I'm not sure how one would in a simple fashion, but that isn't because total spin is somehow ill-defined. See also: physics.stackexchange.com/questions/33021/… $\endgroup$
    – Rococo
    Commented May 26, 2015 at 5:47

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The binding energy of the electrons in a silver atom is far less than the rest energy of an electron, so there is no ambiguity about the number of electrons in a silver atom. That makes adding up the spins a straightforward business.

By contrast, the combined mass of the two up and one down quarks in a proton is about 10MeV (it isn't precisely known) but the proton mass is around a GeV. So 99% of what's in a proton isn't the three quarks. The balance is made up of a seething mass of virtual particles, all of which contribute to the proton spin. Is it any wonder that the proton spin is hard to calculate?

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  • $\begingroup$ Thanks John, that also helps me realize why the problem bugs me: Aren't all the virtual particles in that seething mass also by definition paired in ways that ultimately exactly cancel all of each other's properties, including spin? If so... well, where's the problem exactly? Only the unbalanced real-particle spins will be left. I must be missing something subtle?... $\endgroup$ Commented May 15, 2015 at 19:38
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    $\begingroup$ @TerryBollinger: We think of electrons in atoms as falling into pairs in atomic orbitals. However this is an approximation that ignores electron correlation. The correlation mixes up the atomic orbitals, however in atoms this effect is relatively small because the electrons are relatively far apart. In dense strongly interacting systems the correlations are so strong there are no separate orbitals that you can populate in pairs. You cannot just pair up the particles and see what's left. $\endgroup$ Commented May 16, 2015 at 5:16
  • $\begingroup$ [1 of 3] Actually, even in silver atoms the strong force impact on spin cannot be totally ignored, since strong-force-bound nuclear protons and neutrons add more $\frac{1}{2}$ spins to the total than do the electrons. But your excellent point about the easier mutual isolation of QED-controlled electron spins is well taken. $\endgroup$ Commented May 17, 2015 at 6:16
  • $\begingroup$ [2 of 3] For baryons, if the positive spin vector is made coincident with a coordinate axis such as $z$, conservation of angular momentum combined with quantization of spin permits only two possible spin outcomes, e.g. $uud\rightarrow\{+\frac{1}{2}(p),+\frac{3}{2}(\Delta^+)\}$. To model the effect of this baryon-level spin quantization you would need to trace its coercive effect all the way down to the quarks, where it would necessarily cause the three quark spins to "line up" precisely (and only) at the exact moment when the whole-proton spin measurement is made. That's not QCD, that's QM. $\endgroup$ Commented May 17, 2015 at 6:17
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    $\begingroup$ @user253751 You are quite correct and I'm being careless with the terminology. Inside a hadron the interactions between the three valence quarks are so strong that the state of the quark quantum field cannot be simply decomposed into three distinct particles. Instead we approximate it as the sum of states of the valence quarks and many other off shell quarks and gluons. When I use the term virtual particles I am using shorthand for this approximation. The virtual particles do not actually exist. $\endgroup$ Commented Jan 7, 2023 at 5:09
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Let us make clear that the problem

If proton spin emergence from quarks and gluons is mysterious, why is silver atom spin not?

is a modelling problem. The spin both of the proton and the silver atom is measured and known to identify them.

John's answer covers it, the energy carried by the virtual quarks and gluons within the proton are much larger than the energies carried by the virtual protons neutrons within the silver nucleus , and more so of the virtual electrons and the photons they exchange between them and with the nucleus . This means that a central-potential-well approximation will work to first order for nucleons and atoms, and that makes modeling both the nucleus and the silver atom successful, i.e. fit the observations. This is false for the interactions within a proton, no central potential well can be devised to model it.

Modelling QCD interactions cannot be reduced to such a simplified model, and "mysterious" just reflects the difficulty of the mathematical tools necessary and the observations necessary to explain how within a QCD model the spin arises. Here is a talk of some time ago but it sets up the problems. It is a current matter of research both theoretically and experimentally, as can be seen here. .

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  • $\begingroup$ I like your emphasis on it being a modeling (modelling? US vs UK?) problem. $\endgroup$ Commented May 30, 2015 at 2:20
  • $\begingroup$ modelling , does not get red from dictionary check, nor modeling ; see double consonants en.wikipedia.org/wiki/… $\endgroup$
    – anna v
    Commented May 30, 2015 at 4:31

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