# Addition of 3 spin 1/2 particles

I'm trying to rationalize what it physically means to add three spin-1/2 particles. I understand that for a system of two spin-1/2 particles that there are four basis vectors in the new space on account of the tensor product. These are the states, $|1,1⟩, |1,0⟩, |1,-1⟩$, and $|0,0⟩$. I'm not sure I'm understanding where the spin-0 state is coming from.

I rationalize it as having three spins running parallel and one state where they are running anti-parallel. Applying similar logic then to the system with three spin-1/2 particles, I first create a system of two particles with the above states, then add the third particle to the system. For spin-1 with spin-1/2, the parallel state would give me a total spin of 3/2, and the anti-parallel state would give total spin, 1/2.

However, why does the spin-0 state when added to a spin-1/2 particle not give a -1/2 total spin state in addition to a spin-1/2 state (0+1/2). Is this because spin-1/2 added to a zero vector in essence gives 1/2 in an arbitrary direction because spin-0 is directionless?

• It appears you are essentially looking for the Clebsh-Gordon coefficients. – ACuriousMind Dec 14 '16 at 1:30
• I came across this question when I read this link. I was just wondering if one could apply similar logic when adding spin-1/2 to a spin-0 state. Edit: But yes, I think maybe that is what I'm after because the article I referenced talks about it. – quanticbolt Dec 14 '16 at 1:37
• Maybe you wish to read 29443. – Cosmas Zachos Dec 14 '16 at 2:34
• "I'm not sure I'm understanding where the spin-0 state is coming from" - Isn't it just that, starting with $|\uparrow\uparrow\rangle$ and apply the lowering operator, one gets (up to a normalization) $|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle$ and then again $|\downarrow\downarrow\rangle$. But there is a state orthogonal to the 'middle' state which is $|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle$ which is the singlet? – Alfred Centauri Dec 14 '16 at 2:44