What is the quark magnetic moment?

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Does anyone know the magnetic moments of the up and down quarks? I am interested only in the best experimental data, not theoretical estimates.
 

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  • #2
tom.stoer
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I guess identical for all elementary spin 1/2 particles; why do you expect a difference? QCD corrections?
 
  • #3
NLB
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Have you ever hear of the Neutron Spin Crisis? It started in 1980's and has yet to be resolved. The Bjorken sum rule predicted that the spin of the proton is equal to the sum of spin of the valance quarks. This has been shown experimentally to be false, and that the sum of the spin of the quarks is very much less that of the proton. The spin 1/2 proton and spin 1/2 neutron do not have identical magnetic moments. All spin 1/2 particles do not have the same magnetic moment.
 
  • #4
tom.stoer
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I am perfectly aware of thre nucleon spin crisis ;-)

But you should read my post carefully

I guess identical for all elementary spin 1/2 particles; why do you expect a difference? QCD corrections?
btw.: I checked the PDG and found ... nothing
 
  • #5
NLB
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I know that, experimentally, the magnetic moment of the quarks is not what was expected. Do you know what the best experimental value is?
 
  • #6
NLB
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Thank you for checking the PDG. I too have looked in numerous places and have not yet found anything. Thanks for your help.
 
  • #7
tom.stoer
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I don't understand; the magnetic moment is defined as μ = g (q/2m) s = g q/4m with s = 1/2; so either there is a problem with the current quark mass m or with the g-factor; but the latter one is sensitive to QED corrections only, and these do not depend on QCD effects (OK, in leading order).

Sorry, you were asking for experimental bounds and I have to admit that I have no idea
 
  • #8
NLB
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I just recently read that the mass of the three quarks is much smaller than the mass of the proton (but I have not investigated the peer-review literature about this.) This is why certain theories suggest that there are more quarks in nucleons than just the three valance quarks, to make up for this mass difference. I also might expect the g-factor to be problematic for quarks, just as it is for atomic nuclei. Thank you again for your help.
 
  • #9
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One can define the form factors of effective dressed quarks in non-perturbative QCD. Some people who do that, recently
Dynamical chiral symmetry breaking and the fermion--gauge-boson vertex
Those are rather advanced questions however. If you just learnt about the mass-gap, I would suggest you postpone until a little bit later. There is no way to measure an experimental anomalous magnetic moment for quarks without some theoretical approximation. That is why PDG does not mention anything yet.
 
  • #10
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Does anyone know the magnetic moments of the up and down quarks? I am interested only in the best experimental data, not theoretical estimates.
Do you mean like an electron's magnetic moment? It's not possible to get an up or a down quark into that state, or any other quark into that state, as far as anyone knows. That's quark confinement. It takes too much energy to separate them by more than 10-15 m.

This means that up and down quarks are always strongly relativistic, with gamma ~ 100 in nucleons.

But one can easily estimate these quarks' magnetic moments when bound in nucleons. One uses some simple quantum mechanics. Quarks, being fermions, have an overall antisymmetric wavefunction. But they are in a "colorless" color-singlet state, which makes them color-state antisymmetric. This means that their flavor and spin states together must be symmetric.

For a proton with upward spin, one finds 1/sqrt(18) *
(2|u+u+d-> - |u+u-d+> - |u-u+d+> + 2|d-u+u+> - |d+u+u-> - |d+u-u+> + 2|u+d-u+> - |u-d+u+> - |u+d+u->)
with similar expansions for downward spin and neutrons.

One can easily find the magnetic moment:
mup = (4*muu - mud)/3
mun = (4*mud - muu)/3
With
mup = 2.79
mun = -1.81
in nuclear magnetons,
muu = 1.85
mud = -0.97
Dividing by the quark charges,
muu/qu = 2.78
mud/qd = 2.92
A 5% discrepancy.

I spent some time looking for bag-model and lattice-QCD estimates, but I couldn't find very much.
 
  • #11
tom.stoer
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I just recently read that the mass of the three quarks is much smaller than the mass of the proton ... This is why certain theories suggest that there are more quarks in nucleons than just the three valance quarks
But one can easily estimate these quarks' magnetic moments when bound in nucleons.

... 1/sqrt(18) *
(2|u+u+d-> - |u+u-d+> - |u-u+d+> + 2|d-u+u+> - |d+u+u-> - |d+u-u+> + 2|u+d-u+> - |u-d+u+> - |u+d+u->)

One can easily find the magnetic moment:
muu/qu = 2.78
mud/qd = 2.92
You guys should be careful when talking about a 'quark'.

What is a 'quark'?

My understanding of the question was that a quark is an elementary, undressed, current quark with nearly zero mass; it's magnetic moment is that of an elementary spin 1/2 particle; the only difference to the electron is its charge |q| = 1/3, 2/3, but its g-factor should be g=2 + QED corrections. You can 'see' those quarks rather diretly in hard processes like deep inelastic scattering where quarks are nearly free due to asymptotic freedom.

"This is why certain theories suggest that there are more quarks in nucleons than just the three valance quarks"; yes, QCD does this - b/c QCD does not talk about valence or constituent but undressed current quarks. The quark content of the nucleon is parameterized via the so-called structure functions.

You don't need any recent peer-reviewed articles to start with but simply a standard textbook on QCD.

When you talk about higher quark masses (approx. 1/3 of the nucleon mass), valence or constituent quarks, and when you are constructing constituent quark wave functions (w/o sea quarks and w/o gluons!) you are not talking about elementary particles in terms of QCD but about effective degrees of freedom. It's clear that you then get corrections to the current quark masses and their magnetic moment. But this is due to the fact that you now have 'dressed' quarks which are not elementary. You are no longer talking about QCD (and the deviations from g=2 are now model-dependent).

So you can do all this, but you have to be very clear what you mean by 'quark', which regime you are talking about and which model you are using.
 
  • #12
Vanadium 50
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Tom is right - this depends on what you call a "quark", and because the number you get depends on the definition you use, it's not very meaningful.

First, up to quantum corrections, the magnetic moment of a quark - like any spin-1/2 Dirac fermion - is q/2m.

Second, you can measure the magnetic moment of hadrons. The SU(3) flavor symmetry of hadrons enforces relationships upon them, such as the ratio of the neutron to proton magnetic moment is -2/3. If you take two from experiment (the proton and Lambda usually), you can calculate all the others, and as said before, this is usually good to a couple percent.

Textbooks usually report this as a triumph of the quark model. It is not. Any model that gives you SU(3) flavor will do this.

Third, one can look at the numerical value of the baryon magnetic moments (mesons are fairly useless for this) and from the quark model predictions use this to infer the quark mass. What you find is that the numbers that come out are about 100 times larger than the quark current masses.

Explaining this in detail is difficult in QCD. These calculations fall into the category of "higher-twist" and there are few people out there with the necessary facility in these kinds of calculations. If you would like a handwavy explanation, the mass of a quark as determined by the magnetic moment is a statement of how difficult it is to flip the spin of the quark. If I try and flip the spin of a quark in a hadron, I can't do it without dragging a lot of the glue field with the quark. So I see a very high mass: the thing that responds is not just the quark, but the quark and a whole mass of glue.

If you like, it's like calculating the mass of a bucket from the amount of material used in making the bucket, but only experimentally discovering that it's filled with water.

Anyway, I only have access to two things - the inferred quark magnetic moment from measurements of baryons, which has almost nothing to do with the quark itself and everything to do with the surrounding glue field, or the calculated quark magnetic moment from the current mass, which is experimentally inaccessible. The thing you would really like, "what would be the quark's magnetic moment if it weren't surrounded by glue" doesn't exist.
 
  • #13
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I remember seeing a "bag model" calculation somewhere, but I've been unable to find it again. A bag model makes each valence quark's magnetic moment equal to q/(2E), where E is a quark's total energy.

I'll now consider what would happen if the proton's angular momentum is shared with the "sea" -- gluons and sea quarks.

|total:1/2> = cos(a)*|sea:0,valence:1/2> + sin(a)*|sea:1,valence:1/2>

The sea having J = 0 is simple.

For J = 1, we find for the upward component:

1/sqrt(3)*(sqrt(2)*|sea:+1,valence:-1/2> - |sea:0,valence:+1/2>)

musea:1 = (2*musea - muvalence)/3
giving
mutotal = muvalence*cos(a)2 + (2*musea - muvalence)/3*sin(a)2

and if the sea has zero magnetic moment,
mutotal = (1 - (4/3)*sin(a)2)*muvalence

An early bag-model article: Phys. Rev. D 10, 2599 (1974): Baryon structure in the bag theory

Proton magnetic moment = 2.63 nuclear magnetons
Neutron magnetic moment = - 1.75 nuclear magnetons
Nucleon axial charge = 1.09
Proton rms charge radius = 1.04 fm
 
  • #14
NLB
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Thank you, everyone for your help. There is lots of very interesting information there that I will review further. Regarding this comment:

One can easily find the magnetic moment:
mup = (4*muu - mud)/3
mun = (4*mud - muu)/3
With
mup = 2.79
mun = -1.81
in nuclear magnetons,
muu = 1.85
mud = -0.97​

This above has been the theoretical predicted value for years, but it is only a theoretical estimate. I have read that the experimental data is roughly between 0% and 20% of expected theoretical value, but...I've had a very hard time finding the actual experimental numbers.

Since we know that the experimental data is anomalous when compared to the theoretical data (hence the cause of the Neutron Spin Crisis) then there must be some experimental data.
 
  • #15
tom.stoer
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I still don't see how the quark magnetic moment and the nucleon spin are related in QCD
 
  • #16
Vanadium 50
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NLB, what we are trying to tell you is there is no "expected theoretical value".

I don't know what your "mud"'s are, but the coefficients in the sort of relations you write come from SU(3). The numeric values come from experiment. All that theory lets you do is to take one or two measurements and use that to make a prediction for a third; it doesn't let you calculate the whole set ab initio.

For example, given the magnetic moment of the proton and lambda, one calculate the magnetic moment of the neutron is -1.86 (3% low), and the Omega- is -1.82 (11% low).
 
  • #17
NLB
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|
| I still don't see how the quark magnetic moment and the nucleon spin are related in QCD
|

They are related by the Bjorken Sum rule, which says that the sum of the spin of the quarks is equal to the spin of the nucleon. Experiment proved this to be not true. I assume the spin crisis relates to the magnetic moment of the quarks being different from what theory predicted, which are the numbers Ipetrich posted.

If the spin of a quark is 1/2 and the spin of a nucleon is 1/2, then there is no "Nucleon Spin Crisis". Are you saying the Nucleon Spin Crisis occurred because the experimental spin of the quarks was found to be something other than 1/2?

|
| what we are trying to tell you is there is no "expected theoretical value".
|

Please let me clarify again, I am not interested in any "expected theoretical value", QCD or otherwise, but thank you for all information nonetheless--it is very informative. I want to know the experimental data that caused the Spin Crisis. I really don't care what theory predicted, I want to know what experiment found.
 
  • #18
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They are related by the Bjorken Sum rule, which says that the sum of the spin of the quarks is equal to the spin of the nucleon. Experiment proved this to be not true.
False. The sum of the spins of the quarks is 1/2. The spin of the nucleon is 1/2.

I assume the spin crisis relates to the magnetic moment of the quarks being different from what theory predicted, which are the numbers Ipetrich posted.
False again.

The magnetic moments agree with the naive quark model. See my message for details.

The so-called Spin Crisis is the surprising fact that if I have a beam of protons than is 100% polarized along one direction, the summed spins of the quarks actually has a quite low polarization along that axis. This might not have matched some people's expectations, but it was hardly a crisis - and hardly unexpected. As my message points out, you can't change the spin of a quark without interacting with the glue, so it should have surprised no one that a good fraction of the spin is carried by the glue and orbital angular momentum.

If the spin of a quark is 1/2 and the spin of a nucleon is 1/2, then there is no "Nucleon Spin Crisis".
Well, I don't think there was a spin crisis, just muddy thinking. But the people who went around advertising it would not agree with your description of it. It's just wrong.
 
  • #19
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The Nucleon Spin Crisis:
HERMES searches for the nucleon's missing spin - CERN Courier
[hep-ph/9306288] The Nucleon Spin Crisis Bible
Nucleon Spin Structure, Experimental Review
http://wwwold.jlab.org/hugs/HUGS2011/Friday_June_3/Deshpande%20Lecture%206.pdf [Broken]

It's about the results of deep-inelastic-scattering experiments. At typical DIS-experiment momenta, only about 30% of a nucleon's spin is carried in the valence quarks.


I can't find much on nucleon magnetic-moment calculations in arxiv.org, but I could find some papers that discuss lattice calculations of its "axial charge", something which appears in weak interactions. This is the strength of its axial current, a relative of its vector current. While the vector current is constrained by conservation of current, the axial current isn't.

Experimental value: gA = 1.2695(29) (gV = 1, arxiv 1106.3163)
Bag-model value from earlier post = 1.09

[1111.5960] Hadron Structure in Lattice QCD
[1106.3163] Hadron Structure on the Lattice
gA ~ 1.15

[1106.1554] Form factors in lattice QCD
gA ~ 1.15 to 1.2 (mpion > 300 MeV)
gA ~ 1.2 to 1.3 (mpion <~ 300 MeV)

Nothing on magnetic moments, however.

In lattice-QCD calculations, the pion masses used are typically well above their physical ones (neutral: 135 MeV, charged: 140 MeV). This is because using their physical masses would require an excessively large lattice to get both pions and nucleons at reasonable resolution.
 
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  • #20
tom.stoer
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I said that "I still don't see how the quark magnetic moment and the nucleon spin are related in QCD", but in your response

They are related by the Bjorken Sum rule, which says that the sum of the spin of the quarks is equal to the spin of the nucleon. Experiment proved this to be not true. I assume the spin crisis relates to the magnetic moment of the quarks being different from what theory predicted, which are the numbers Ipetrich posted.
you talk about an assumption only. What is this assumption exactly? Where does it come from?

I still don't see any relation between "summing over spins" and "summing over magnetic moments".
 
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  • #21
NLB
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So here is what you've told me so far:
* There's no experimental data.
* There's no theoretical estimate either. But if there were, it is controversial depending on the definition of a quark and what specific theory we are using.
* The Spin Crisis isn't real, but just someone's muddy thinking.
* Spins and magnetic moments have no relation to each other.
* There is, however, an equation that relates them directly: "magnetic moment is defined as μ = g (q/2m) s = g q/4m with s = 1/2"
* In regards to the above equations, g and m are unknowns for quarks. Hence, there is no relation between spin and magnetic moment.

I think the hardest part here for me to understand is that the spin crisis is just "muddy thinking". How can it be even after over 25 years, that no one noticed that this is just muddy thinking?

Anyway, thank you all for your help and your efforts in helping to answer this question. Thank you Ipetrich for your web links. Even though I am frustrated by the lack of a real experimental number, I do appreciate the forum and all your help.
 
  • #23
tom.stoer
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* There's no experimental data.
No; we did not give you experimental data but just commented on wrong ideas; there are Terrabytes of experimental data

* There's no theoretical estimate either.
There are theoretical estimates; all what we were saying ways that quark or bag model estimates are too naive and are not suitable for an explanation (a theory that explains 50% of the data but fails to explain the other 50% is not really helpful - even if your interest is in the first 50%)

... it is controversial depending on the definition of a quark and what specific theory we are using
There is only one theory today - QCD - and in QCD there are no such ambiguities. Explaining the spin structures based on (non-rel.) quark models is like explaining black holes based on Newtonian gravity

* Spins and magnetic moments have no relation to each other.
There is a clear relation and I just quoted the correct formula using the g-factor; what I was saying is that I can't see how the quark magnetic moment and the nucleon spin are related; it was your claim that this should be the case, and therefore expected some explanation, reference, whatever

* There is, however, an equation that relates them directly: "magnetic moment is defined as μ = g (q/2m) s = g q/4m with s = 1/2"
Yes, I think I used that formula, but I can't see how it may help; you either explain the nucleon magnetic moment in terms of its spin (which you can't as long as you can't explain the spin), or you explain the quark magnetic moment in terms of its spin which is trivial and not the subject of the 'spin crisis'.

* In regards to the above equations, g and m are unknowns for quarks.
No, g = 2 + QED corrections, m for 'up'and 'down' is a few MeV; and you can find a lot of data for quark masses at the particle data group; but this is (afaik) irrelevant for the nuclon spin

How can it be even after over 25 years, that no one noticed that this is just muddy thinking?
I don't think it's muddy thinking. The problem is that all naive models fail to explain the spin, and that even in QCD not everything is well-understood.

A general remark: you should try to understand deep inelastic scattering, (polarized) nucleon structure functions, and perhaps Altarelli-Parisi / DGLAP equations in order to make progress.

A hint: if you are interested in a topic like DIS and spin structure it's not a good idea to focus on a singular question like "what is the quark magnetic moment?"

Another reference: http://arxiv.org/abs/0812.3535v2
 
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  • #24
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And let me clarify "muddy".

  • At the time, it was known that the quarks carried only 50% of the linear momentum of the proton. Why should carrying only 30% of the angular momentum be a "crisis"?
  • People were thinking that the baryon magnetic moments agreement with what you got starting with two and calculating the rest was evidence for a model where the quarks carried all the spin. As I mentioned, that's not true - this is enforced by the SU(3) flavor symmetry (and SU(2) isospin).
 
  • #25
edguy99
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... One can easily find the magnetic moment:
mup = (4*muu - mud)/3
mun = (4*mud - muu)/3
With
mup = 2.79
mun = -1.81
in nuclear magnetons,
muu = 1.85
mud = -0.97
Dividing by the quark charges,
muu/qu = 2.78
mud/qd = 2.92
A 5% discrepancy.

I spent some time looking for bag-model and lattice-QCD estimates, but I couldn't find very much.
... It's about the results of deep-inelastic-scattering experiments. At typical DIS-experiment momenta, only about 30% of a nucleon's spin is carried in the valence quarks.
If you cannot detect quarks by themselves, what logic is used to experimentally detect an amount of spin or a magnetic moment to the quarks?
 

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