ChrisVer said:
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?
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
, and ii) many peopel do think it is simple 
