Are the uds particle wavefunctions correct in explaining the particle octet?

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Discussion Overview

The discussion revolves around the wavefunctions of uds particles and their role in explaining the particle octet. Participants explore the theoretical framework behind the quark compositions, spin states, and the implications of symmetry in particle physics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant expresses uncertainty about their wavefunction calculations for uds particles and questions the validity of their symmetrical approach.
  • Another participant states that the only totally symmetric state for a three-quark system occurs when all spins are aligned, leading to a total spin of 3/2.
  • There is a discussion about the coupling of quark spins and how this affects the total spin states, with references to different configurations yielding either 1/2 or 3/2 total spins.
  • Some participants mention that quarks exist in an asymmetric color state, which leads to effective Bose statistics, termed "quark statistics." This is contrasted with the requirement for symmetric states in certain configurations.
  • Two specific uds spin functions are proposed for the Lambda and Sigma zero particles, with a focus on their mathematical representation.
  • Concerns are raised about the symmetry of the proposed states and how to determine which state vectors correspond to actual particles, highlighting a lack of clarity in flavor symmetry as presented in the literature.
  • There is a suggestion that particles with the same quark composition can be distinguished by their spin states, emphasizing the importance of spin in defining particle identity.

Areas of Agreement / Disagreement

Participants express differing views on the symmetry of the wavefunctions and the implications for particle identity. There is no clear consensus on the correctness of the proposed wavefunctions or the interpretation of spin states.

Contextual Notes

Participants note limitations in their understanding of flavor symmetry and the criteria for identifying particle states based on their wavefunctions and spin configurations.

Who May Find This Useful

This discussion may be of interest to those studying particle physics, particularly in the context of quark models, spin statistics, and the classification of baryons in the particle octet.

lark
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I'm interested in comments on thishttp://camoo.freeshell.org/25.6.gif" .
I tried to upload it to the forum but apparently that doesn't work well with .gifs.
It's an exercise in a book asking you to explain the particle octet, why there aren't uuu, ddd and sss spin 1/2 particles, there are 1 each of uud, uss, etc. etc., and 2 types of uds particles. I'm wondering whether I got the wavefunctions for the uds particles right. I was just guessing, came up with something nice and symmetrical that looks right, but I don't know if there's a way to tell if it is right.
Yes, I know that according to the color theory, the quarks aren't actually following boson statistics. This exercise was before that was explained in the book, all we were told was that the quarks are acting like bosons.
Laura
 
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I did not follow all your reasoning, I would just simply state that the only totally symmetric state with respect to two particle interchange in 3quark system is if you have all three quark spins in the same direction, which gives you S = 3/2

Are you using "Particle Physics" by Martin & Shaw ?

The wave function of uds will just be the total antisymmetric wave function when you couple 3 spin 1/2 particles. I.e you can first couple u & d to S = 1 or 0.
Then you couple the s quark to S = 1 to get S_tot = 3/2 or 1/2.
Or you couple the s quark to S = 0 to get S_tot = 1/2.
i.e You'll get three such states for uds
 
Quarks are in a completely asymmetric color state, so what they follow, if color is not explicit, is effective Bose statistics. This could be called "quark statistics".
For 6 of the octet, two quarks are the same, and must be in the symmetric spin one state. Coupling this spin one to spin 1/2 gives either 1/2 or 3/2.
The decuplet arises because the 3 uuu type states and the uds can be included with the other 6. Since the spin 1/2 state is not completely symmetric, it can't be coupled with states like uuu.
The 8 octet states include 2 uds states. That is because the ud pairing can be in either spin one or spin zero.
The two uds spin functions are [tex][aba-baa]/\sqrt{2}[/tex] for the Lambda, and
[tex][2aab-aba-baa]/\sqrt{6}[/tex] for the Sigma zero.
 
pam said:
For 6 of the octet, two quarks are the same, and must be in the symmetric spin one state. Coupling this spin one to spin 1/2 gives either 1/2 or 3/2.
The 8 octet states include 2 uds states. That is because the ud pairing can be in either spin one or spin zero.
The two uds spin functions are [tex][aba-baa]/\sqrt{2}[/tex] for the Lambda, and
[tex][2aab-aba-baa]/\sqrt{6}[/tex] for the Sigma zero.
It sounds like I got the basic idea of the spin states for the proton, neutron, etc. right, but my guess about the 2 possible uds particles was maybe too symmetric to be right. The two states I thought might be the particles are the complex conjugate of each other, at least if you're ignoring anything besides spin.
I mean, you have this vector space of dimension 2 containing 3 possible spin 1/2 states for uds, and I wasn't sure how you decide which vectors are the "particles"? You make a Hamiltonian for the vector space and you find 2 eigenvectors, and they have different eigenvalues, and you go "Aha, those are the particles?"
I think deciding which state vectors correspond to particles is something we don't know how to do yet in the book, there's been no explanation of flavor symmetry and that would be involved.
I'm reading the Penrose book "The Road to Reality" supplemented by outside looking around when an exercise puzzles me.
Laura
 
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The Sigma zero spin state should be the same as the Sigma plus.
 
pam said:
The Sigma zero spin state should be the same as the Sigma plus.
I see. So the particles that have the same letter have the same spin state, if symmetrized right? Then the [tex]\Lambda^0[/tex] is orthogonal to the [tex]\Sigma^0[/tex]. Seems like the spins are considered more important to what the particle "is" than the quark composition.
Laura
 
They play an equal role. If two particles have the same quark composition, then spin state distinguishes them.
 

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