# Proton spin

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1. Mar 25, 2014

### ChrisVer

I am currently confused... I read that we don't know yet where does the proton spin come from.
But I wonder...
1) Doesn't the proton have effectively 3 quarks of spin 1/2 (I said effectively to leave out the quark-gluon sea within the proton)? In that case, 3 spin 1/2 particles can't be added up to spin 1/2? Where's the problem in that?

I also learned they tried to calculate the spin of the interactive gluons within the proton, and it was not enough to give its spin. Now the idea that is being proposed (as I know it'll be tested in CERN LCH-COMPASS experiment) is that the needed spin comes from the angular momentum of the rotating quarks.... Isn't that a weird idea? I mean there is a reason (spacetime) we make the distinguishing between angular momentum and spin...I think that the one can in fact be chosen zero in the rest frame, while the other still exists.... In that case, how could angular momentum produce spin?

2. Mar 25, 2014

Staff Emeritus
If that were the whole story, the magnetic moment of the proton would be 100x larger than it is measured to be.

3. Mar 25, 2014

### Staff: Mentor

4. Mar 26, 2014

### samalkhaiat

Do you think that the proton is a box with 3 marbles spinning inside it? If the story were as naïve as you imagine it to be, I would sleep at night without having nightmares about it. To explain the complications involved here, let me start by simple example and consider the 3-momentum and 3-angular momentum of the free photon in EM. From Noether’s theorem, we find the following CANONICAL expressions
$$\vec{ P } = \int d^{ 3 } x \ E^{ i } \vec{ \nabla } A^{ i } , \ \ \ (1)$$
$$\vec{ J } = \int d^{ 3 } x \ \vec{ E } \times \vec{ A } + \int d^{ 3 } x \ E^{ i } \ ( \vec{ x } \times \vec{ \nabla } ) A^{ i } . \ \ (2)$$
Notice that the angular momentum is nicely decomposed into spin and orbital angular momentum parts: $\vec{ J } = \vec{ S } + \vec{ L }$. However, there is problem with these CANONICAL expressions. They are NOT gauge invariant and therefore cannot be measurable quantities: experimentally measured quantities CANNOT be gauge dependent. However, since the theory is Poincare’ invariant, we can construct equivalent expressions for $\vec{ P }$ and $\vec{ J }$ that are gauge invariant and therefore measurable:
$$\vec{ P } = \int d^{ 3 } \ \vec{ E } \times \vec{ B } , \ \ \ (3)$$
$$\vec{ J } = \int d^{ 3 } \ \vec{ x } \times ( \vec{ E } \times \vec{ B } ) . \ \ \ (4)$$

It looks as if the problem is solved! But no, far from it, because what we measure in experiments is the photon SPIN and Eq(4) has NO SPIN term. So, we gained gauge invariance and lost the very useful decomposition $J = S + L$. This is the story in the FREE field theory, but if we go to QED or QCD, the problem gets even more complicated. Again, the CANONICAL expressions are not gauge invariant and therefore not measurable but the total angular momentum is nicely decomposed into spin and orbital parts for photon (gluon) and electron (quark). But more importantly, these canonical expressions satisfy the Poincare’ algebra and generate the correct translations and rotations on the fields involved. These last two features are the corner stone of any QFT and one should not mess with it. But this is exactly what happens when we try to make $P$ and $J$ gauge invariant. So, in QED and QCD, the problem is not just
$$\vec{ J }_{ \mbox{tot} } \ne ( \vec{ S } + \vec{ L } )_{e} + ( \vec{ S } + \vec{ L } )_{ \gamma } , \ \ (5)$$

but the most important equations in QFT’s get screwed: (1) The operators $P$ and $J$ are no longer the generators of translations and rotations
$$[ i \vec{ P } , \phi ( x ) ] \ne \vec{ \nabla } \phi ( x ) , \ \ (6)$$
$$[ i \vec{ J } , \phi ( x ) ] \ne ( \vec{ L } + \vec{ S } ) \phi ( x ) . \ (7)$$
(2) The following Poincare’ subalgebra get messed up
$$[ P^{ i } , P^{ j } ] \ne 0 , \ \ [ J^{ i } , P^{ j } ] \ne i \epsilon^{ i j k } P^{ k } , \ \ (8)$$
and
$$[ J^{ i } , J^{ j } ] \ne i \epsilon^{ i j k } J^{ k } . \ \ \ \ (9)$$
So, to solve the Proton Spin Problem, we in the last 20 years have been trying to write EQ(5) as equality with each separate term in it represents a gauge-invariant operator, and at the same time Eq(6) to Eq(9) are changed to equalities. Simple is it not?

Sam

5. Mar 26, 2014

### Bill_K

Here's a comprehensive review of the proton spin.

6. Mar 26, 2014

### ChrisVer

Well, just to clarify something in the name of my honor... I was thinking of the proton as:
$p=\frac{1}{\sqrt{18}}(2u_{+}u_{+}d_{-}+2u_{+}d_{-}u_{+}+2d_{-}u_{+}u_{+}-u_{+}u_{-}d_{+}-u_{-}d_{+}u_{+}-u_{+}d_{+}u_{-}-d_{+}u_{-}u_{+}-d_{+}u_{+}u_{-}-u_{-}u_{+}d_{+})$
notation $q_{s_{z}=\pm}$
and that seaquarks (as gluons do) contributing nothing to the spin, since they would appear in pairs of different orientation. So that's why I gave the mistaken impression that I was speaking of marble-like things in the proton.

Although my question is answered I think from the above post #4....
Nice review too Bill, thanks

Last edited: Mar 26, 2014
7. Mar 26, 2014

Staff Emeritus
The magnetic moment of a Dirac particle goes as 1/m. Since the u and d quarks have a mass of a few MeV, they have magnetic moments of a few hundred nuclear magnetons. The proton has a magnetic moment of something under 3. (2.79, if I remember right)

The explanation of this is as follows: if I want to measure the mass of a quark in a proton, I take my finger, find a quark, push it with a known force, and see how fast it accelerates. If I do this, I find the quark appears to be about 100x heavier, because to move it, I need to also move all the gluons and sea quarks that are surrounding it. This glob of glue makes the effective quark mass higher.

So the fact that we see this large effective mass means that the gluons and sea quarks are participating in the motion of the quarks. This includes the angular momentum, and needs to be accounted for.

8. Mar 26, 2014

### strangerep

Ha! You should have put a smiley face or a wink after that.

More seriously, thanks for the insightful post highlighting some of the subtleties that arise when dealing with interacting reps of the Poincare group.

9. Mar 26, 2014

### samalkhaiat

Well, I didn't because i) I don't want to get told off :tongue:, and ii) many peopel do think it is simple

You welcome, it is pleasure to talk about my baby.

Sam

10. Mar 26, 2014

### samalkhaiat

Even if you add colour and iso-spin labels to the quarks and make the "wavefunction" two page long, you are still dealing with 3 marbles in a box. The dynamical properties of the 3-body bound state (the proton) can only be investigated by the QCD framework.

Sam

11. Mar 27, 2014

### strangerep

That's better.

Oh,... is this (one of) your main lines of research? Anything on the arxiv about it?

12. Mar 30, 2014

### lpetrich

Proton: +2.79, neutron: -1.81

From a simple constituent quark model, where the quarks' spin is 100% of the total nucleon spin,
μp = (4μu - μd)/3
μn = (4μd - μu)/3
giving
μu = (4μp + μn)/5
μd = (4μn + μp)/5

Up: +1.87, down: -0.89

The down one is close to -(1/2) of the up one, as one would expect of the quark model.

Lattice-QCD calculations apparently agree reasonably well: [1403.4686] Review of Hadron Structure Calculations on a Lattice

A similar approach comes from a bag model, where the quark is assumed confined in a spherical cavity about the size of a hadron. This is much less than the Compton wavelength of an up or a down quark, and those quarks are thus highly relativistic with γ ~ 100. From a bag model, magnetic moments are ~ 1/E, where E is a quark's total energy, even if it is much greater than the rest mass. When one calculates the up and down quarks' magnetic moments from a bag model, one finds approximate agreement with experiment.

13. Mar 31, 2014

### ChrisVer

I am not sure for the magnetic moments approach either. For the magnetic moments you also have to take in consideration the orbital angular momentum... an illustration of it is by trying to calculate the magnetic moment of the Deuterium (D)...
In that case, taking only the ground state $L=0 (S)$ gives inconsistent result. The result becomes better when you take some contribution from $L=2 (D)$ states too.. (you can also try to take states with L=1, but 1st you will break parity symmetry, and you will also get negative probabilities -if you take only L=1 to keep the parity)

In maths you'll have:
$\psi_{d}= \psi_{^{3}S_{1}} +\psi_{^{3}D_{1}}$
giving (MJ=J=1):
$\mu_{d}= 0.8796 p _{^{3}S_{1}}+0.3102 p_{^{3}D_{1}}$
and the solution for the probabilities are
$0.8796 p _{^{3}S_{1}}+0.3102 p_{^{3}D_{1}}=0.8574$ (the 0.8574 is the deuterium magnetic moment)
$p _{^{3}S_{1}}+p_{^{3}D_{1}}=1$
with solutions $p _{^{3}S_{1}}=0.96$ and $p_{^{3}D_{1}}=0.04$
(or in other words, Deuterium's ground state consists of mostly an S state, mixed with a small contribution of D state, in order to give the correct results for its magnetic moment)

On the contrary taking L=1 only eigenstates
$\psi_{d}= \psi_{^{1}P_{1}} +\psi_{^{3}P_{1}}$
giving (MJ=J=1):
$\mu_{d}= 0.5 p _{^{1}P_{1}}+0.6898 p_{^{3}P_{1}}$
and the solution for the probabilities are
$0.5 p _{^{1}P_{1}}+0.6898 p_{^{3}P_{1}}=0.8574$ (the 0.8574 is the deuterium magnetic moment)
$p _{^{1}P_{1}}+p_{^{3}P_{1}}=1$
with negative solutions for the probabilities p

Last edited: Mar 31, 2014
14. Apr 1, 2014

### lpetrich

That is certainly correct.

Let's see what happens when the sea quarks and the gluons mix with the valence quarks.

Total angular momentum: 1/2

jval = 1/2
jsea = 0
|1/2> = |1/2,0>
jsea = 1
|1/2> = sqrt(1/3)*|1/2,0> - sqrt(2/3)*|-1/2,1>

jval = 3/2
jsea = 1
|1/2> = sqrt(1/2)*|3/2,-1> - sqrt(1/3)*|1/2,0> + sqrt(1/6)*|-1/2,1>
jsea = 2
|1/2> = sqrt(1/10)*|3/2,-1> - sqrt(1/5)*|1/2,0> + sqrt(3/10)*|-1/2,1> - sqrt(2/5)*|-3/2,2>

μ(jval = 1/2, jsea = 0) = μ(jval = 1/2)
μ(jval = 1/2, jsea = 1) = - (1/3) * μ(jval = 1/2)

μ(jval = 3/2, jsea = 1) = (5/9)*μ(jval = 3/2)
μ(jval = 3/2, jsea = 2) = - (1/3)*μ(jval = 3/2)

For the proton,
μ(jval = 1/2) = (4μu - μd)/3
μ(jval = 3/2) = 2μu + μd

The neutron has u and d reversed.

Treating the up and down quarks as dynamically identical, we can set μu = (2/3)*μq, μd = - (1/3)*μq, and we get for the proton and neutron
μ(jval = 1/2) = μq, - (2/3)*μq
μ(jval = 3/2) = μq, 0

So the nucleons may have a little bit of admixture of jval = 3/2. Sea effects may also mean that my calculated magnetic moments are underestimates.

One can do a similar analysis for the delta baryons, though it's hard to get magnetic-moment measurements for them because of their very short lifetime.

15. Apr 2, 2014