I have a one-dimensional Heisenberg chain with a Magnetic field with $N$ sites with $J>0$ \begin{equation} \mathcal{H} = -J \sum_{i = 1}^{N-1} \vec{S_i}\cdot \vec{S_{i+1}}- \sum_{i = 1}^N \vec{H}\cdot S_i \end{equation} where the spins $\vec{S_i} = (S_i^x, S_i^y, S_i^z)$ live in 3 dimensions. What is the ground state of such a system. I know that without the external magnetic field one of the ground states is just all the spins pointing upwards such that $S_i^z = S$ for all $i$. But how does the ground state change in the presence of an external field?
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The spins will all point along the field, i.e., $\displaystyle\langle \vec S_i\rangle = S\frac{\vec H}{|\vec H|}$.
(That this is indeed the ground state is easily verified: The corresponding state -- a product state of all spins pointing in that direction -- already minimizes the energy of each of the terms in the Hamiltonian individually.)
answered Oct 1, 2023 at 10:17