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

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The discussion revolves around determining the possible total spins for a three-particle system of spin-1/2 particles. The initial guess suggests total spins of 1/2 and 3/2, but confusion arises regarding the derivation of total spin using concepts like triplets and singlets. Participants express uncertainty about the Clebsch-Gordan theorem and its coefficients, which are typically used for composing spins. There is a call for a more intuitive explanation of the spin composition without relying on advanced theorems. Overall, the conversation highlights the complexities of understanding spin combinations in quantum mechanics.
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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,

\tfrac1{\sqrt{2}}(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle,​

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

\mathcal{E}_{\frac{1}{2}}\otimes \mathcal{E}_{\frac{1}{2}}\otimes \mathcal{E}_{\frac{1}{2}}



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

\mathcal{E}_{\frac{1}{2}}\otimes \mathcal{E}_{\frac{1}{2}}\otimes \mathcal{E}_{\frac{1}{2}}

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.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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