We analyze here the consequence of local rotational-symmetry breaking in the quantum spin (or phase) glass state of the quantum random rotor model. By coupling the spin glass order parameter directly to a vector potential, we are able to compute whether the system is resilient (that is, possesses a phase stiffness) to a uniform rotation in the presence of random anisotropy. We show explicitly that the O(2) vector spin glass has no electromagnetic response indicative of a superconductor at mean-field and beyond, suggesting the absence of phase stiffness. This result confirms our earlier finding [Phys. Rev. Lett. 89, 027001 (2002)] that the phase glass is metallic, due to the main contribution to the conductivity arising from fluctuations of the superconducting order parameter. In addition, our finding that the spin stiffness vanishes in the quantum rotor glass is consistent with the absence of a transverse stiffness in the Heisenberg spin glass found by Feigelman and Tsvelik [Sov. Phys. JETP 50, 1222 (1979)].
|Original language||English (US)|
|Journal||Physical Review B - Condensed Matter and Materials Physics|
|State||Published - Jan 1 2003|
ASJC Scopus subject areas
- Electronic, Optical and Magnetic Materials
- Condensed Matter Physics