Does Heisenberg-chain superdiffusion survive symmetric integrability-breaking perturbations
In plain words
In the solvable Heisenberg spin chain, magnetization spreads faster than ordinary diffusion. If the chain is perturbed so that it is no longer solvable but keeps its rotation symmetry, it is unclear whether the fast spreading persists forever or eventually becomes ordinary diffusion.
Precise statement
Take $H = H_{\mathrm{XXX}} + g V$ with $V$ local, integrability breaking and $\mathrm{SU}(2)$ invariant (e.g. $V = \operatorname{sum}_i S_i . S_{i+2}$), at infinite temperature and zero magnetization. De Nardis, Gopalakrishnan, Vasseur, Ware (PRL 127, 057201, 2021) argued that large spin-carrying quasiparticles survive, giving spin conductivity $\sigma(\omega) \sim \omega^{-1/3}$ at low frequency instead of a finite DC value, while perturbations breaking $\mathrm{SU}(2)$ restore diffusion. Determine the low-frequency form of $\sigma(\omega)$ in the quantum chain, i.e. whether the spin diffusion constant is finite. Answer: finite $D(g)$ or divergent, with the exponent.
What would settle it
Real-time computation of the spin current correlator in the perturbed quantum chain at times long enough to distinguish $\omega^{-1/3}$ from a finite DC limit, for at least two values of $g$.
Status in the literature
Unverified note
Kinetic theory predicts persistence (2021), while in the nonintegrable classical Heisenberg chain superdiffusion lasts a time diverging as a power of 1/T but finite (McRoberts and Moessner, PRL 133, 256301, 2024); long-time quantum results are lacking.