Abstract

We have studied what happens to dense excess electrons bound to doubly charged donors in magnetic semiconductors by using the Kohn-Sham density-functional method for excess electrons and the molecular-field approximation for localized spins (f spins). Detailed calculations have been performed at T=0 K for the antiferromagnetic semiconductor with strong interaction between excess electrons and f spins, with a special interest in Eu-rich EuTe. Two models are considered: one is dense He-like donors coupled with f spins, and the other is dense bound molecular magnetic polarons with degenerate orbitals. The calculation for the former case, with the parameters relevant to EuTe, reveals that the electronic state is constructed from the tightly and (rather) loosely bound excess electrons, and the first-order transition from a singlet to a triplet electron-spin configuration occurs at the excess electron concentration of ${\mathit{n}}_{0}$\ensuremath{\sim}${10}^{19}$/${\mathrm{cm}}^{3}$. After the transition, the loosely bound electrons in the triplet state nearly spread over the crystal, merging with the perturbed conduction band. The singlet-triplet transition, thus, seems to correlate to a nonmetal-metal transition (NMMT). On the other hand, in the latter model, the calculation shows the strongly localized state at ${\mathit{n}}_{0}$\ensuremath{\sim}${10}^{19}$/${\mathrm{cm}}^{3}$, and the NMMT may occur at ${\mathit{n}}_{0}$\ensuremath{\sim}${10}^{20}$--${10}^{21}$/${\mathrm{cm}}^{3}$, where the ferromagnetic ordering of f spins is established throughout the crystal. The existence of magnetic polaron effects is evident for both models. A comparison with experiment shows that the former model seems quite suitable for Eu-rich EuTe with ${\mathit{n}}_{0}$\ensuremath{\lesssim}${10}^{19}$/${\mathrm{cm}}^{3}$ when the values of the parameters used for the calculation are appropriate for EuTe. We conclude that the localization of the extended excess electrons at low temperatures in the samples with ${\mathit{n}}_{0}$\ensuremath{\sim}${10}^{19}$/${\mathrm{cm}}^{3}$ is of a purely magnetic origin.

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