# Spin problem

1. Apr 24, 2005

### broegger

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.

Last edited: Apr 24, 2005
2. Apr 24, 2005

### dextercioby

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.

3. Apr 24, 2005

### broegger

Huh? We aren't suppose to use the Clebsch-Gordon coefficients (we skipped that part).

4. Apr 24, 2005

### dextercioby

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.

5. Apr 24, 2005

### broegger

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

And what about my last question?

I'm not familiar with that notation or the Clebsch-Gordan theorem. We're not supposed to use that (trust me).

6. Apr 24, 2005

### dextercioby

There are 3 irreducible representations (3 irreducible spaces) spanned by the vectors given by the C-G theorem...

Daniel.