Addition of Spin Angular Momentum

Eric Alcantara, Carlo Paul Morente and Gibson T. Maglasang

If you have two particles of spin $S_1$ and $S_2$ . Let $\bf{S}$ be the combined spin of the particles. You can get values of $\bf{S}$ from $(S_1+S_2)$ down to $(S_1-S_2)$ :

$\bf{S} =(S_1+S_2), (S_1+S_2-1), (S_1+S_2-2),\cdot\cdot\cdot(S_1-S_2)$ (1)

We can get then the combination of states from the total spin state $\bf{S}$ . In particular, a state $|S\quad m\rangle$ with total S and z-component m will be some linear combination of the composite states $|S_1\quad m\rangle$ $|S_2\quad m\rangle$ .

Consider now two spin 3/2 particles at the ground state. To find the possible total spin states, we use (eqn. 1). Thus,

$\bf{S} = (3/2+3/2)= 3 ;$

$\bf{S} = (3/2+3/2-1)= 2;$

$\bf{S}=(3/2+3/2-2)= 1;$

$\bf{S} = (3/2+3/2-3) = 0$ .

(i) For $\bf{S} = 3$ , the corresponding values for $m_s$ are 3, 2, 1, 0, -1, -2, -3.

We have the following states $|\bf{S}$ $m_s\rangle$ ,

$|3$ $3\rangle$ ,

$|3$ $2\rangle$ ,

$|3$ $1\rangle$ ,

$|3$ $0\rangle$ ,

$|3$ $-3\rangle$ ,

$|3$ $-2\rangle$ ,

$|3$ $-1\rangle$ .

(ii) For $\bf{S}$ = 2, $m_s=2, 1, 0, -1, -2$ . The following states are obtained,

$|2$ $2\rangle$ ,

$|2$ $1\rangle$ ,

$|2$ $0\rangle$ ,

$|2$ $-2\rangle$ ,

$|2$ $-1\rangle$ .

(iii) For $\bf{S} = 1$ , $m_s = 1, 0, -1$

$|1$ $1\rangle$ ,

$|1$ $0\rangle$ ,

$|1$ $-1\rangle$ .

(iv) For $\bf{S} = 0$ , $m_s = 0$ . Only one state is available for $\bf{S} = 0$ . We have,

$|0$ $0\rangle$

Thus, we have successfully extracted the 16 possible spin states for two spin 3/2 particles at the ground state. One can also get all the 16 spin states for this particular problem by looking up the Clebsch-Gordan table.

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Addition of Spin Angular Momentum

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