Register to reply

Quark Model Spectroscopic Notation

by SteDolan
Tags: model, notation, quark, spectroscopic
Share this thread:
SteDolan
#1
Aug26-13, 08:18 AM
P: 14
I'm a little confused about what is actually meant by the JP = 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 I3 = 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
Phys.Org News Partner Physics news on Phys.org
'Squid skin' metamaterials project yields vivid color display
Team finds elusive quantum transformations near absolute zero
Scientists control surface tension to manipulate liquid metals (w/ Video)
Vanadium 50
#2
Aug26-13, 09:33 AM
Emeritus
Sci Advisor
PF Gold
Vanadium 50's Avatar
P: 16,461
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.
Bill_K
#3
Aug26-13, 10:34 AM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
Quote Quote by SteDolan View Post
I'm a little confused about what is actually meant by the JP = 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 I3 = 3/2 but why couldn't we have this but maintain spin 1/2?
Let me give it a try. The overall wavefunction of the three fermions must be totally antisymmetric.
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.

SteDolan
#4
Aug26-13, 11:26 AM
P: 14
Quark Model Spectroscopic Notation

Thanks! That makes sense, I didn't think about the wavefunction symmetry at all.
lpetrich
#5
Sep10-13, 01:06 PM
P: 533
What's what in JP:
J = total spin
P = parity

As to the light-baryon octet, it's a rather long story.

Quarks have color, flavor, spin, and orbit states.

Since they have spin 1/2, they follow Fermi-Dirac 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 longer-lived baryons; the quarks are all in ground-state 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 spin-flavor states. This is the light-baryon 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 angular-momentum 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 spin-1/2 state.

If all three quarks have the same flavor, then no spin-1/2 state is possible.

We thus get these states: uud, udd, uus, 2(uds), dds, uss, dss

These 8 states are the light-baryon 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.
SteDolan
#6
Sep10-13, 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?
Bill_K
#7
Sep10-13, 02:28 PM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
Quote Quote by SteDolan View Post
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?
It's because, as I said above, physical states must be colorless. Colorless means they are invariant under rotations in the SU(3) color symmetry.

A quark by itself has any one of three colors, in fact a quark state transforms under SU(3) as a 3-vector Va 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 εabc. Thus the baryon wavefunction is like εabcUaVbWc, which is totally antisymmetric under interchange of the quarks.
lpetrich
#8
Sep10-13, 03:13 PM
P: 533
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 coupling-constant strength g reveals that it has this behavior when it is relatively small:

g2 ~ b/log(p/p0) ~ b/log(r0/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 one-loop 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 ~ p0 ~ a few hundred MeV and r ~ r0 ~ 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.


Register to reply

Related Discussions
Spectroscopic notation - again Introductory Physics Homework 0
Spectroscopic notation Introductory Physics Homework 1
Question about spectroscopic notation Chemistry 3
Atomic Spectroscopic Notation Atomic, Solid State, Comp. Physics 0
Question on spectroscopic notation Advanced Physics Homework 1