Question 7.P.5: Consider a system of four nonidentical spin 1/2 particles. F...

First, we need to couple two spins at a time: \hat{\vec{S} }_{12} = \hat{\vec{S} }_{1}+\hat{\vec{S} }_{2} and \hat{\vec{S} }_{34} = \hat{\vec{S} }_{3}+\hat{\vec{S} }_{4}. Then we couple \hat{\vec{S} }_{12} and \hat{\vec{S} }_{34}: \hat{\vec{S} }=\hat{\vec{S} }_{12} +\hat{\vec{S} }_{34} . From Problem 7.4, page 438, we have s_{12}=0,1 and s_{34}=0,1.

In total there are 16 states |sm 〉 since (2s_{1}+1) (2s_{2}+1) (2s_{3}+1) (2s_{4}+1)=2^{4} =16.

Since s_{12}=0,1 and s_{34}=0,1, the coupling of \hat{\vec{S} }_{12} and \hat{\vec{S} }_{34} yields the following values for the total spin s:

• When s_{12}=0 and s_{34}=0 we have only one possible value, s = 0, and hence only one eigenstate, |sm 〉 = |0,0 〉.

• When s_{12}=1 and s_{34}=0, we have s = 1; there are three eigenstates: |sm 〉 = |1,\pm 1 〉 , and |1,0 〉.

• When s_{12}=0 and s_{34}=1, we have s = 1; there are three eigenstates: |sm 〉 = |1,\pm 1 〉 , and |1,0 〉.

• When s_{12}=1 and s_{34}=1 we have s = 0, 1, 2; we have here nine eigenstates (see Problem 7.3, page 436): |0,0 〉,|1,\pm 1 〉,|1,0 〉,|2,\pm 2 〉,|2,\pm 1 〉, and |2,0 〉.

In conclusion, the possible values of the total spin when coupling four \frac{1}{2} spins are s = 0, 1, 2; the value s = 0 occurs twice, s = 1 three times, and s = 2 only once.