What Are the Possible Total Spins for a Three-Particle Spin-1/2 System?

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Discussion Overview

The discussion revolves around determining the possible total spins for a three-particle system composed of spin-1/2 particles. Participants explore theoretical aspects of spin composition, including the application of the Clebsch-Gordan theorem and the implications of total spin values.

Discussion Character

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

Main Points Raised

  • One participant proposes that the possible total spins for the system are 1/2 and 3/2, but expresses uncertainty about this conclusion.
  • Another participant suggests using the Clebsch-Gordan theorem to analyze the spin composition, indicating that this approach would yield eight states instead of four.
  • Some participants express confusion about the Clebsch-Gordan coefficients and the theorem itself, indicating that they have not covered this material in their studies.
  • There is a question regarding whether the total spin of a spin-1/2 particle is exactly 1/2 or if it could be larger, relating this to the uncertainty principle.
  • A later reply mentions that there are three irreducible representations associated with the spin composition, referencing the Clebsch-Gordan theorem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the total spins for the three-particle system, with differing views on the use of the Clebsch-Gordan theorem and the interpretation of total spin values.

Contextual Notes

Some participants express a lack of familiarity with the Clebsch-Gordan theorem and its application, which may limit their ability to engage fully with the discussion. There is also uncertainty regarding the implications of total spin values in relation to the uncertainty principle.

broegger
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Hi,

I have to find out the possible total spins for a three-particle system composed of spin-1/2-particles. My guess is that there are two possible spins; 1/2 (one up, the others down or vice versa) and 3/2 (all up or all down), but I'm not sure.

In my book they show how to find the total spin of a system composed of two spin-1/2-particles, but I don't understand the derivation. He talks about triplets and singlets (what is that!?) and apparently the state,

[tex]\tfrac1{\sqrt{2}}(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle[/tex],​

represents a system of total spin 1. How come? I don't get it.

Also, another question: Is the total spin of a spin-1/2 particle s = 1/2 or is it slightly bigger (like for orbital angular momentum, where the total is always bigger than the z-component). I would think that it is, since if it is 1/2 you would know the direction of the spin vector completely (Sx = 0, Sy = 0, Sz = +/-1/2), which would violate the uncertainty principle.
 
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What did u make of Clebsch-Gordan theorem and the C-G coefficients...?

There's one way to do it.Use the 2 1/2 spins case & compose it with a spin 1/2 case.Instead of 4,u'll have 8 states...

Daniel.
 
Huh? We aren't suppose to use the Clebsch-Gordon coefficients (we skipped that part).
 
Composing spins (and angular momenta in general) is done starting with the theorem of Clebsch-Gordan...Read it and compute

[tex]\mathcal{E}_{\frac{1}{2}}\otimes \mathcal{E}_{\frac{1}{2}}\otimes \mathcal{E}_{\frac{1}{2}}[/tex]



Daniel.
 
Can anyone give a more intuitive explanation? Am I right in my initial guess?

And what about my last question?

dextercioby said:
Composing spins (and angular momenta in general) is done starting with the theorem of Clebsch-Gordan...Read it and compute

[tex]\mathcal{E}_{\frac{1}{2}}\otimes \mathcal{E}_{\frac{1}{2}}\otimes \mathcal{E}_{\frac{1}{2}}[/tex]

I'm not familiar with that notation or the Clebsch-Gordan theorem. We're not supposed to use that (trust me).
 
There are 3 irreducible representations (3 irreducible spaces) spanned by the vectors given by the C-G theorem...

Daniel.
 

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