# Showing that the ground state of the Heisenberg ferromagnet is an eigenstate of the Hamiltonian

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.

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