Abstract
Polarized Raman spectra of isotope substituted $\mathrm{Sr}{\mathrm{Ti}}^{18}{\mathrm{O}}_{3}$ (STO18) crystals were studied in the ferroelectric phase below ${T}_{c}=24.5\phantom{\rule{0.3em}{0ex}}\mathrm{K}$. Samples were confirmed to be a single domain in the tetragonal phase $({T}_{c}<T<{T}_{0}=108\phantom{\rule{0.3em}{0ex}}\mathrm{K})$. Raman spectra in various phonon propagation directions ${\stackrel{P\vec}{K}}_{p}$ were found to satisfy the selection rules for orthorhombic ${C}_{2v}$ symmetry with the spontaneous polarization $\stackrel{P\vec}{P}$ parallel to the tetragonal axes, either ${[110]}_{c}$ or ${[1\overline{1}0]}_{c}$. Below ${T}_{c}$, STO18 is inhomogeneous and consists of a number of small ferroelectric domains as well as remnants of the tetragonal structure. Soft TO modes originated from the lowest ${\ensuremath{\Gamma}}_{15}$ mode in cubic phase were observed at 20 and $12.5\phantom{\rule{0.3em}{0ex}}{\mathrm{cm}}^{\ensuremath{-}1}$. On heating, these modes soften but do not freeze at ${T}_{c}$. Near ${T}_{c}$, a quasielastic scattering component appears and its intensity critically increases. Although for temperatures far from ${T}_{c}$, the transition is promoted by the softening of the TO modes, the appearance of $\stackrel{P\vec}{P}$ at ${T}_{c}$ cannot be attributed to the freezing of Slater mode since $\stackrel{P\vec}{P}$ is not parallel to the cubic axis ${[100]}_{c}$. Origin of the relaxational mode near ${T}_{c}$ and the direction of $\stackrel{P\vec}{P}$ are discussed in terms of the orientational ordering of locally polarized clusters which already exist above ${T}_{c}$. These results indicate that the phase transition in STO18 is a peculiar transition and has both order-disorder and displacive nature.
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