The Hamiltonian of a Heisenberg ferromagnet in terms of $S^+, S^-, S^z$ is given by:

$$H = -\frac{1}{2}|J| \sum_{i,\vec{\delta}} \left[\frac{1}{2}(S_i^+S^-_{i+\vec{\delta}} + S_i^-S^+_{i+\vec{\delta}}) + S_i^zS^z_{i+\vec{\delta}}\right]$$

where we assume a 3D cubic lattice, $\vec{\delta}$ is a vector connecting nearest-neighbor sites and $J$ is an exchange constant which is negative since we are considering the ferromagnetic case. To avoid double counting, an additional factor of $\frac{1}{2}$ is included.

We want to show that the ground state, i.e. the state with all spins pointing along $z$ ($S_j^z = S)$ is an eigenstate of the Hamiltonian. For simplicity, we can write the ground state as $|0>$.

First, we note that $S_j^+$ kills the ground state since it already has maximum $S_j^z$ which therefore cannot be increased further. However, what about $S_j^-$? In order for $|0>$ to be the ground state, this should also kill the ground state, but why is that true? Because the $S_j^-$ state should just lower the z-component of the spin, which is possible, since in the ground state the z-component is maximized.


1 Answer 1


The $S_j^-$'s don't kill the ground state, but the $S_j^+$'s that they're sitting next to do.


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