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Problem 4.87 Magnetic frustration. Consider three spin-1/2 particles arranged on the corners of a triangle and interacting via the Hamiltonian $H = J(S_1 \cdot S_2 + S_2 \cdot S_3 + S_3 \cdot S_1)$, (4.227) where $J$ is a positive constant. This interaction favors opposite alignment of neigh- boring spins (antiferromagnetism, if they are magnetic dipoles), but the triangular arrangement means that this condition cannot be satisfied simultaneously for all three pairs (Figure 4.18). This is known as geometrical \"frustration.\" (a) Show that the Hamiltonian can be written in terms of the square of the total spin, $S^2$, where $S = \sum_i S_i$. (b) Determine the ground state energy, and its degeneracy. (c) Now consider four spin-1/2 particles arranged on the corners of a square, and interacting with their nearest neighbors: $H = J(S_1 \cdot S_2 + S_2 \cdot S_3 + S_3 \cdot S_4 + S_4 \cdot S_1)$. (4.228) In this case there is a unique ground state. Show that the Hamiltonian in this case can be written $H = \frac{J}{2} \left[ S^2 - (S_1 + S_3)^2 - (S_2 + S_4)^2 \right]$. (4.229) What is the ground state energy?

          Problem 4.87 Magnetic frustration. Consider three spin-1/2 particles arranged on the
corners of a triangle and interacting via the Hamiltonian
$H = J(S_1 \cdot S_2 + S_2 \cdot S_3 + S_3 \cdot S_1)$, (4.227)
where $J$ is a positive constant. This interaction favors opposite alignment of neigh-
boring spins (antiferromagnetism, if they are magnetic dipoles), but the triangular
arrangement means that this condition cannot be satisfied simultaneously for all three
pairs (Figure 4.18). This is known as geometrical \"frustration.\"
(a) Show that the Hamiltonian can be written in terms of the square of the total spin,
$S^2$, where $S = \sum_i S_i$.
(b) Determine the ground state energy, and its degeneracy.
(c) Now consider four spin-1/2 particles arranged on the corners of a square, and
interacting with their nearest neighbors:
$H = J(S_1 \cdot S_2 + S_2 \cdot S_3 + S_3 \cdot S_4 + S_4 \cdot S_1)$. (4.228)
In this case there is a unique ground state. Show that the Hamiltonian in this case
can be written
$H = \frac{J}{2} \left[ S^2 - (S_1 + S_3)^2 - (S_2 + S_4)^2 \right]$. (4.229)
What is the ground state energy?

Problem 4.87 Magnetic frustration. Consider three spin-1/2 particles arranged on the
corners of a triangle and interacting via the Hamiltonian
H = J(S1 · S2 + S2 · S3 + S3 · S1), (4.227)
where J is a positive constant. This interaction favors opposite alignment of neigh-
boring spins (antiferromagnetism, if they are magnetic dipoles), but the triangular
arrangement means that this condition cannot be satisfied simultaneously for all three
pairs (Figure 4.18). This is known as geometrical f̈rustration.(̈a) Show that the Hamiltonian can be written in terms of the square of the total spin,
S^2, where S = Si.
(b) Determine the ground state energy, and its degeneracy.
(c) Now consider four spin-1/2 particles arranged on the corners of a square, and
interacting with their nearest neighbors:
H = J(S1 · S2 + S2 · S3 + S3 · S4 + S4 · S1). (4.228)
In this case there is a unique ground state. Show that the Hamiltonian in this case
can be written
H = (J)/(2)[ S^2 - (S1 + S3)^2 - (S2 + S4)^2 ]. (4.229)
What is the ground state energy?

Added by Marc G.

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