Two ions 1 and 2, at fixed separation, with spin angular momentum operators S1 and S2, have the interaction Hamiltonian H = −J S1·S2, where J > 0. The values of S1² and S2² are fixed at S1(S1 + 1) and S2(S2 + 1), respectively. Which of the following is the energy of the ground state of the system?
A. 0
B. –JS1S2
C. –J[S1(S1 + 1) – S2(S2 + 1)]
D. –(J/2)[(S1 + S2)(S1 + S2 + 1) – S1(S1+1) – S2(S2+1)]
E. –(J/2)[(S1(S1 + 1) + S2(S2 + 1))/(S1 + S2)(S1 + S2 + 1)]
(GR9677 #77)
Solution:
S1² ψ = S1(S1 + 1) ψ
S2² ψ = S2(S2 + 1) ψ
Si² ψ = Si(Si + 1) ψ
H = −J S1·S2
Using general arithmetic equation: ab = ½ [(a + b)² − a² − b²]
H = −(J/2)[(S1 + S2)² − S1² − S2²]Since Si² ψ = Si(Si + 1) ψ
For (S1 + S2)² → replace Si with S1 + S2
→ (S1 + S2)² ψ = (S1 + S2)(S1 + S2 + 1) ψ
H = −(J/2)[(S1 + S2)(S1 + S2 + 1) − S1(S1 + 1) − S2(S2 + 1)]
Answer: D
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