Abstract
We exactly show that the ground state of the Anderson lattice with U=\ensuremath{\infty} is ferromagnetic at quarter filling if the level of localized electrons ${\mathrm{\ensuremath{\varepsilon}}}_{\mathit{f}}$ is deep enough: ${\mathrm{\ensuremath{\varepsilon}}}_{\mathit{f}}$${\mathrm{\ensuremath{\varepsilon}}}_{\mathit{f}\mathit{c}}$, where ${\mathrm{\ensuremath{\varepsilon}}}_{\mathit{f}\mathit{c}}$ is of the order of the bandwidth. Rigorous arguments show that if ${\mathrm{\ensuremath{\varepsilon}}}_{\mathit{f}}$${\mathrm{\ensuremath{\varepsilon}}}_{\mathit{f}\mathit{c}}$, the ground state has the total spin S=(N-1)/2 for ${\mathit{N}}_{\mathit{e}}$=N+1, where N is the number of lattice sites and ${\mathit{N}}_{\mathit{e}}$ is that of electrons. This indicates that a transition to a (incompletely) magnetically ordered ground state will occur for a value of ${\mathrm{\ensuremath{\varepsilon}}}_{\mathit{f}}$ less than ${\mathrm{\ensuremath{\varepsilon}}}_{\mathit{f}\mathit{c}}$. We observe this transition for finite U if -${\mathrm{\ensuremath{\varepsilon}}}_{\mathit{f}}$ is sufficiently large. An extension to more generalized models is discussed. The exact diagonalization technique is applied to a cluster cut out of the ${\mathrm{CuO}}_{2}$ plane. Our analysis shows that the system with one-hole doping has a ferromagnetic phase in the ground state, indicating that a doped hole in the O p orbital is moving around in the ferromagnetic background of Cu spins.
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