Question 4.67: Magnetic frustration. Consider three spin-1/2 particles arra...

(a)

S = S _{1}+ S _{2}+ S _{3} \Rightarrow S^{2}=S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+2 S _{1} \cdot S _{2}+2 S _{1} \cdot S _{3}+2 S _{2} \cdot S _{3}, \text { so } H=\frac{1}{2} J\left(S^{2}-S_{1}^{2}-S_{2}^{2}-S_{3}^{2}\right) .

But S_{i}^{2}=\hbar^{2} s(s+1)=\frac{3}{4} \hbar^{2} (because they all have spin 1/2), so H=\frac{1}{2} J\left(S^{2}-\frac{9}{4} \hbar^{2}\right) .

(b) S^{2}=s(s+1) \hbar^{2} , where s is the total spin. Combine two spin 1/2 states and you get spin 1 or spin 0 ; add spin 1/2 to spin 1 and you get 1+\frac{1}{2}=\frac{3}{2} \text { or } 1-\frac{1}{2}=\frac{1}{2} ; add \frac{1}{2} to 0 and you get \frac{1}{2} . So the total spin is 3/2 (four states) or 1/2 (two states)|and the latter occurs in two different ways. There are 2² = 8 states in all (each of the three could be spin up or spin down), 4 of which carry spin 1/2 (the ground state). The degeneracy of the ground state is 4, and its energy is \frac{1}{2} J\left(\frac{1}{2}\left(\frac{3}{2}\right) \hbar^{2}-\frac{9}{4} \hbar^{2}\right)=-\frac{3}{4} \hbar^{2} J .

(c)  S^{2}=S_{1}^{2}+S_{2}^{2}+S_{3}^{2}+S_{4}^{2}+2 S _{1} \cdot S _{2}+

2 S _{1} \cdot S _{3}+2 S _{1} \cdot S _{4}+2 S _{2} \cdot S _{3}+2 S _{2} \cdot S _{4}+2 S _{3} \cdot S _{4} .

2S_3.S_4-  \cancel{S^2_1} – \cancel{S^2_3} – \cancel{2S_1.S_3}-  \cancel{S^2_2} -\cancel{S^2_4} – \cancel{2S_2.S_4} = 2(S_1.S_2 +S_1 .S_4 +S_2.S_3 + S_3.S_4 ) = 2H/J .

so H = \frac{1}{2} J [S^2 -(S_1+S_3)^2 – (S_2+S_4)^4] . QED.

Write H=\frac{1}{2} \hbar^{2} J\left[s(s+1)-s_{13}\left(s_{13}+1\right)-s_{24}\left(s_{24}+1\right)\right] , where s is the total spin, s_{13} is the spin of the 1,3 combination, and s_{24} that of the 2,4 combination. For the ground state we want s as small as possible, and s_{13}, s_{24} as large as possible: s=0, s_{13}=s_{24}=1 (the triplet combinations). Then E=\frac{1}{2} \hbar^{2} J(0-2-2)= -2 \hbar^{2} J.

[Are we sure we can construct such a state? From 1 and 3 we can make singlet or  triplet combinations; likewise 2 and 4 \frac{1}{2} \otimes \frac{1}{2}=1,0 . If both are in the triplet state we can get a total spin of 2, 1, or 0 : 1 \otimes 1=2,1,0 .

So yes: here is the total spin 0 with each pair having total spin 1. By the way, the other possibilities are 1 \otimes 0=1,0 \otimes 1=1, \text { and } 0 \otimes 0=0 . Thus the total spin could be

s = 2                         (2s + 1) = 5 states,

s = 1                          (three ways) 3 × 3 = 9 states,

s = 0                         (two ways) 2 × 1 = 2 states,

a total of 5 + 9 + 2 = 16 states, which is just right: 2^4 = 16. But the ground state is unique-there’s exactly one combination that puts 1,3 and 2,4 in the triplet configuration and gets a total spin of 0.]