Isospin vs. flavor vs. spin - Difference and similarity?

joechien0218
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Hi All,

I'm wondering what the difference is between flavor and isospin.

If you own a book by Griffiths titled Elementary Particles, on page 183 in the paragraph above EQ 5.129, he mentions isospin and spin together and the former seems to imply the latter.

Does anyone out there know what he's talking about? Or am I simply missing something big here?

Thank you
Joe
 
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joechien0218 said:
Hi All,

I'm wondering what the difference is between flavor and isospin.

If you own a book by Griffiths titled Elementary Particles, on page 183 in the paragraph above EQ 5.129, he mentions isospin and spin together and the former seems to imply the latter.

Does anyone out there know what he's talking about? Or am I simply missing something big here?

Thank you
Joe

Spin is a fundamental property of any elementary particle. Allowed values of spin and its projections follow from isotropy (or rotational symmetry) of space. All massive particles are classified as spin zero (1 possible projection), spin 1/2 (two projections), spin 1 (3 projections), etc.

Isospin is an approximate empirical concept, which is applicable to particles participating in strong interactions. This concept tries to mimic properties of spin. It was recognized long time ago that proton and neutron have (almost) identical properties when it comes to strong interactions. Three pions \pi^+, \pi^0, \pi^- also look identical apart from their charges. So, an idea was proposed that proton/neutron is just the same particle having isospin 1/2. It exists in the form of proton when "isospin projection" is +1/2, and it exists in the form of neutron, when "isospin projection" is -1/2 (or vice versa, I don't remember). Similarly, three pions are just different isospin projections of the same particle having isospin 1. This empirical idea was very useful in description of strongly interacting particles and nuclei.
 
Hi! Thanks for the lightning response!

Actually when I created that thread, I was just trying to understand what Griffiths's saying on page 183, namely how he was able to conclude that

For sigma, (Su+Sd)^2 = 2 h h (EQ 5.129)
For lambda, (Su+Sd)^2 = 0 (EQ 5.130)

If you can help, I'd really appreciate it ^_^ and thanks , have a good day
joe
 
joechien0218 said:
Hi! Thanks for the lightning response!

Actually when I created that thread, I was just trying to understand what Griffiths's saying on page 183, namely how he was able to conclude that

For sigma, (Su+Sd)^2 = 2 h h (EQ 5.129)
For lambda, (Su+Sd)^2 = 0 (EQ 5.130)

If you can help, I'd really appreciate it ^_^ and thanks , have a good day
joe

Sorry, I don't have Griffiths's book. So, I can't help you here.
 
joechien0218 said:
Hi! Thanks for the lightning response!
For sigma, (Su+Sd)^2 = 2 h h (EQ 5.129)
For lambda, (Su+Sd)^2 = 0 (EQ 5.130)
In the Sigma, the two nucleon quarks are in a spin one state.
In the Lambda, they are in a spin zero state.
Griffiths is so user friendly that he doesn't include enough detail to be understandable.
 
Hi Meir,

For lambda, the two nucleon quarks are in spin zero state, but Why?

You've mentioned in the previous post that "one can show that the probability is higher if the nucleons in lambda are in spin zero state". How were you able to conclude that?

thx
joe
 
The PR 172(1968)1807 paper I emailed you answers this.
 
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