Suppression of superconductivity by the localization of hole carriers in Ti-dopedLa1.85Sr0.15CuO4

Abstract

The effects of Ti doping in the single-doped samples of ${\mathrm{La}}_{1.85}{\mathrm{Sr}}_{0.15}{\mathrm{Cu}}_{1\ensuremath{-}x}{\mathrm{Ti}}_{x}{\mathrm{O}}_{4}\phantom{\rule{0.2em}{0ex}}(0\ensuremath{\leqslant}x\ensuremath{\leqslant}0.06)$ and the double-doped samples of ${\mathrm{La}}_{1.85\ensuremath{-}2x}{\mathrm{Sr}}_{0.15+2x}{\mathrm{Cu}}_{1\ensuremath{-}x}{\mathrm{Ti}}_{x}{\mathrm{O}}_{4}\phantom{\rule{0.2em}{0ex}}(0\ensuremath{\leqslant}x\ensuremath{\leqslant}0.4)$ have been studied. The critical Ti doping level (above which the pure phase is not formed) of double-doped samples are much higher than that of the single-doped samples. For single-doped samples, the high valence ${\mathrm{Ti}}^{4+}$ would introduce extra electrons into the ${\mathrm{CuO}}_{2}$ plane, which leads to severe valence mismatch and prevents the formation of a pure phase sample at $xg0.06$ and the superconductivity is depressed at $xg0.04$. While for double-doped samples, the extra electrons introduced by the doping of Ti are compensated by increasing holes due to the increase of the Sr content, which lead to the valence balance in the system. Thus, the pure-phase samples are formed up to $x=0.4$ and the superconductivity can survive up to $x=0.1$. The resistivity for double-doped samples is three orders of magnitude smaller than that for single-doped samples at the same Ti doping content. The reason that Ti dopants suppress superconductivity is attributed to the localization of hole carriers.