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Quark Model Spectroscopic Notation 
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#1
Aug2613, 08:18 AM

P: 14

I'm a little confused about what is actually meant by the J^{P} = 1/2 ^{+} octet.
I'm pretty sure it means the particles must have spin 1/2 in the ground state (l=0) and an even parity but what I'm confused over is the absence of an sss, uuu or ddd state. I understand that this would give I_{3} = 3/2 but why couldn't we have this but maintain spin 1/2? I'm sure I'm just missing something obvious here, thanks a lot in advance for any help :D 


#2
Aug2613, 09:33 AM

Mentor
P: 16,321

It's a fermion, so the total wavefunction has to be antisymmetric. uuu is antisymmetric in color, so it needs to be symmetric in spin: e.g. 3/2.



#3
Aug2613, 10:34 AM

Sci Advisor
Thanks
P: 4,160

It's a product (space)(spin)(flavor)(color). space is ℓ = 0, ℓ' = 0, which is totally symmetric color is colorless, which is totally antisymmetric if the flavor is uuu, that is totally symmetric Together these imply that the remaining part, the spin part, must be totally symmetric, which requires it to be s = 3/2. 


#4
Aug2613, 11:26 AM

P: 14

Quark Model Spectroscopic Notation
Thanks! That makes sense, I didn't think about the wavefunction symmetry at all.



#5
Sep1013, 01:06 PM

P: 530

What's what in J^{P}:
J = total spin P = parity As to the lightbaryon octet, it's a rather long story. Quarks have color, flavor, spin, and orbit states. Since they have spin 1/2, they follow FermiDirac statistics, meaning that their combined wavefunction is antisymmetric. They are antisymmetric in color: (rgb + gbr + brg  rbg  bgr  grb)/sqrt(6) making them symmetric in the other states combined together. Orbit states we can ignore for the longerlived baryons; the quarks are all in groundstate orbits. That leaves flavor and spin. I'll consider only the three light flavors of quark: u, d, s, because those were the first ones discovered. At first thought, one might want to have a state that's symmetric in flavor and spin separately. For flavor, one gets uuu, uud, udd, ddd, uus, uds, dds, uss, dss, sss  10 flavor states. The spin is all the quarks' spins parallel, giving spin 3/2. Thus, one ends up with 10 spinflavor states. This is the lightbaryon decuplet. But we want an octet, so we must consider other possibilities. We must consider a system where spin and flavor separately have mixed symmetry, but combining into full symmetry. The simplest case to examine is the case of all three quarks having different flavors: uds Let's see how the u and d spins combine. Parallel, symmetric: spin 1 Antiparallel, antisymmetric: spin 0 Add the s spin, and use angularmomentum addition. One gets three states: Spin 3/2  we already did that Spin 1/2, u and d parallel Spin 1/2, u and d antiparallel So we get not one, but two uds states. If two of the quarks have the same flavor, then those quarks must have symmetric, parallel spins, giving total spin 1. That means that there is only one spin1/2 state. If all three quarks have the same flavor, then no spin1/2 state is possible. We thus get these states: uud, udd, uus, 2(uds), dds, uss, dss These 8 states are the lightbaryon octet that we were looking for. This includes two uds baryons: the Lambda0 and the Sigma0. The Sigma0 is close in mass to the Sigma+ (uus) and the Sigma (dds), meaning that the u and d quarks are symmetric in all three particles. The Lambda0's mass is more difference, because the QCD equivalent of the magnetic moment is somewhat less for the strange quark than for the up and down quarks. 


#6
Sep1013, 01:44 PM

P: 14

Thanks a lot for such a detailed reply! That really helps my understanding of wave function symmetry, especially in the 3 particle case.
Is there a reason the baryons are antisymmetric in color? Couldn't we have the wave function symmetric in color and antisymmetric in flavor and spin? 


#7
Sep1013, 02:28 PM

Sci Advisor
Thanks
P: 4,160

A quark by itself has any one of three colors, in fact a quark state transforms under SU(3) as a 3vector V_{a} in the color space. What you need to do is couple three quarks together to form an invariant. Obviously the only way to do this is to use the antisymmetric symbol ε_{abc}. Thus the baryon wavefunction is like ε_{abc}U_{a}V_{b}W_{c}, which is totally antisymmetric under interchange of the quarks. 


#8
Sep1013, 03:13 PM

P: 530

One can certainly imagine such states, but we don't observe evidence of them. When we try to make color nonsinglets at high energies, nonsinglets like quarks or gluons, they make jets of hadrons as they separate.
Calculation of the QCD "charge" or couplingconstant strength g reveals that it has this behavior when it is relatively small: g^{2} ~ b/log(p/p_{0}) ~ b/log(r_{0}/r) where b is somewhere around 1, p is the interaction momentum, and r is the interaction distance. Note that p*r ~ 1 (uncertainty principle). This is from oneloop calculations; for better results, one needs more loops, of course. But when one does such calculations, one has good agreement with experiment for interaction momenta greater than a few GeV. g becomes large at around p ~ p_{0} ~ a few hundred MeV and r ~ r_{0} ~ a fermi (10^{15} m), and one must do lattice calculations. One finds there also that the effective coupling constant increases with distance. Back to energies much greater than a GeV, these correspond to distances much less than a fermi. When quarks and gluons start getting about a fermi apart, they start making quark and gluon pair production and combination of quarks and gluons into hadrons. Thus, jets of hadrons. All this has led to a theoretical concept called "color confinement". Looking at size scales greater than about 1 fermi, a state must be a color singlet, even if it must be composite in order to do so. 


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