Abstract
For a 180\ifmmode^\circ\else\textdegree\fi{} domain wall in a Heisenberg ferromagnet the Hamiltonian containing exchange, anisotropy, and boundary interaction terms is treated quantum mechanically. This leads to results which differ qualitatively from those of classical studies in which the spins are considered as vectors with fixed length $S$. A complete system of eigenstates, for which the transverse magnetization vanishes, exists. Selecting the eigenstates using further symmetry of the Hamiltonian $H$ leads in some cases to $〈{\stackrel{\ensuremath{\rightarrow}}{\mathrm{S}}}_{n}〉=0$ in the middle layer of the wall, where ${\stackrel{\ensuremath{\rightarrow}}{\mathrm{S}}}_{n}$ is the spin operator at site $n$. For a one-dimensional chain, with spin $S=\frac{1}{2}$ at each site, the properties of the ground state are investigated. For the infinite chain the lowest eigenvalues ${E}_{M}(\ensuremath{\alpha})$ ($\ensuremath{\alpha}$ is the coupling constant of the boundary interaction of $H$) in each subspace with total magnetization ${S}^{z}=\ensuremath{\Sigma}{n}^{}{S}_{n}^{z}=M$ are calculated exactly. The same is carried out numerically for 2, 3, 4, and 5 sites. For these cases ${E}_{M}(\ensuremath{\alpha})$ have the property: ${E}_{M}(\ensuremath{\alpha})l{E}_{M+1}(\ensuremath{\alpha})$ for all $M\ensuremath{\ge}0$ and all $\ensuremath{\alpha}$. There is considerable evidence that this is true for any number of sites. From this follows that the ground state of the wall is not degenerate for an even number of sites and therefore has vanishing tranverse magnetization, while the $z$ component of the magnetization has opposite sign at sites reflected at the middle layer of the wall.
Published Version
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