Quark Model (Supermultiplet and States questions)

  • Context: Graduate 
  • Thread starter Thread starter Elwin.Martin
  • Start date Start date
  • Tags Tags
    Model Quark States
Click For Summary

Discussion Overview

The discussion centers around the concept of supermultiplets in the context of baryons, specifically exploring the classification of baryon states based on quark symmetries and the implications of these classifications. Participants delve into the definitions and properties of baryon states, including their symmetry under quark interchange and the formation of specific combinations of quark flavors.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Elwin questions the definition of a supermultiplet and the meaning of quark label interchange, expressing confusion about how the states arise from the 27 possible configurations.
  • Some participants clarify that a supermultiplet relates to particles under SU(3) and that the ten states mentioned form a baryon decuplet, constructed from symmetric combinations of states with similar charge and strangeness.
  • There is a discussion about the nature of the state "uud+udu+duu," with some participants asserting that it represents the Δ+ resonance rather than a proton.
  • Questions arise regarding the symmetry properties of the decuplet, particularly concerning color and flavor symmetries, with some participants noting that hadrons are colorless and thus must be color singlets.
  • Participants discuss the implications of symmetry in fermion states, including how the overall symmetry must be antisymmetric and how this relates to the spin and flavor states of baryons.

Areas of Agreement / Disagreement

Participants exhibit a mix of agreement and uncertainty regarding the definitions and implications of supermultiplets and baryon states. While some clarifications are provided, there remains a lack of consensus on certain aspects, particularly concerning the symmetry properties and the classification of states.

Contextual Notes

Limitations include potential misunderstandings about the nature of baryon states and the specific definitions of symmetry in this context. The discussion does not resolve the complexities surrounding the octet and decuplet classifications.

Elwin.Martin
Messages
202
Reaction score
0
From Ryder:
"We can now see how a supermultiplet of ten baryons may arise. Baryons are made of three identical fermions, so the possible states may be classified according to their symmetry under interchange of quark labels. Altogether there are 27 states. One of these is totally anti-symmetric: uds+dsu+sud-usd-sdu-dus,
and ten symmetric states:
uuu,
ddd,
sss,
uud+udu+duu,
uus+usu+suu,
udd+ddu+dud,
uss+ssu+sus,
dds+dsd+sdd,
dss+ssd+sds,
uds+dsu+sud+usd+sdu+dus"

I am a little bit lost here, first question:
what is a supermultiplet...? Also, what does he mean by interchange of quark labels?

second question:
How are these states supposed to clearly follow from the existence of 27 states...? I'm not really sure how I'd come up with these states.

For "uud+udu+duu"
(assuming I'm not missing something really obvious) my first thought was "2 up quarks, 1 down...proton"? I feel like I'm missing something obvious here, besides the idea that the terms don't commute (which I'm guessing is part of this). Wait, it said these were non-normalized...normalized would we have 1/sqrt 3(uud+udu+duu) and then the whole thing would just be the different states of a proton? I feel like I'm seriously over simplifying at this point.

Thanks for any and all help,

Elwin
 
Physics news on Phys.org
Elwin, In Ryder's terminology, a multiplet is a group of particles related to each other under isospin, and a supermultiplet is a group of particles related to each other under SU(3). (Usually, I think, the term 'supermultiplet' is reserved for SU(4) and higher, but Oh well.)

By quark labels he means the flavors u, d, s. There are 27 possible states (33 = 27.)

Each of the 27 states has a certain charge and strangeness. The ten states he's listing form the baryon decuplet, and to form the ten, he takes the states having similar charge and strangeness and forms the symmetric combination. So starting with uus we get uus + usu + suu.

Since the overall symmetry of a fermion state (space, spin, flavor, color) must be antisymmetric, the decuplet is symmetric under spin, i.e. spin 3/2. So uud+udu+duu is not a proton, it's a resonance usually called Δ+, or the "3,3" resonance.
 
Bill_K said:
Elwin, In Ryder's terminology, a multiplet is a group of particles related to each other under isospin, and a supermultiplet is a group of particles related to each other under SU(3). (Usually, I think, the term 'supermultiplet' is reserved for SU(4) and higher, but Oh well.)

By quark labels he means the flavors u, d, s. There are 27 possible states (33 = 27.)

Each of the 27 states has a certain charge and strangeness. The ten states he's listing form the baryon decuplet, and to form the ten, he takes the states having similar charge and strangeness and forms the symmetric combination. So starting with uus we get uus + usu + suu.

Since the overall symmetry of a fermion state (space, spin, flavor, color) must be antisymmetric, the decuplet is symmetric under spin, i.e. spin 3/2. So uud+udu+duu is not a proton, it's a resonance usually called Δ+, or the "3,3" resonance.

How do we know that the decuplet won't be antisymmetric under something like color...? Is it because hadrons are colorless? I know it's antisymmetric under space (Pauli exclusion principle). What does it mean to be symmetric under flavor? I know what the flavors are, is it just symmetric because of the additive nature of the states?

Thanks for your help :) Sorry if I'm asking slightly stupid questions
 
Elwin, Not at all. Good questions. A hadron is colorless, which means it's a color singlet, i.e. the color part is totally antisymmetric.
The space part is (almost always) L=0, so totally symmetric.
That leaves the spin part and flavor part, which must both have the same symmetry "S". So that (antisymmetric x symmetric x S x S) works out to be antisymmetric overall, for a fermion state.

As I said, the decuplet has a symmetric flavor state, so the spin state must be the same - symmetric also.

Ryder focused on the decuplet because the octet which contains the proton is a bit more complicated to describe. It has "mixed" symmetry (neither totally symmetric nor totally antisymmetric) As likewise the coupling of the three spins to total spin 1/2 is mixed.
 

Similar threads

Graduate Resoance states and quark annihilation
  • · Replies 1 ·
Replies
1
Views
3K