Abstract

We compute the zero-temperature conductivity in the two-dimensional quantum $\mathrm{O}(N)$ model using a nonperturbative functional renormalization-group approach. At the quantum critical point we find a universal conductivity $\sigma^*/\sigma_Q$ (with $\sigma_Q=q^2/h$ the quantum of conductance and $q$ the charge) in reasonable quantitative agreement with quantum Monte Carlo simulations and conformal bootstrap results. In the ordered phase the conductivity tensor is defined, when $N\geq 3$, by two independent elements, $\sigma_{\mathrm{A}}(\omega)$ and $\sigma_{\mathrm{B}}(\omega)$, respectively associated to $\mathrm{O}(N)$ rotations which do and do not change the direction of the order parameter. Whereas $\sigma_{\mathrm{A}}(\omega\to 0)$ corresponds to the response of a superfluid (or perfect inductance), the numerical solution of the flow equations shows that $\lim_{\omega\to 0}\sigma_{\mathrm{B}}(\omega)/\sigma_Q=\sigma_{\mathrm{B}}^*/\sigma_Q$ is a superuniversal (i.e. $N$-independent) constant. These numerical results, as well as the known exact value $\sigma_{\mathrm{B}}^*/\sigma_Q=\pi/8$ in the large-$N$ limit, allow us to conjecture that $\sigma_{\mathrm{B}}^*/\sigma_Q=\pi/8$ holds for all values of $N$, a result that can be understood as a consequence of gauge invariance and asymptotic freedom of the Goldstone bosons in the low-energy limit.

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