Frequency regulators for the nonperturbative renormalization group: A general study and the model A as a benchmark

We derive the necessary conditions for implementing a regulator that depends on both momentum and frequency in the nonperturbative renormalization-group flow equations of out-of-equilibrium statistical systems. We consider model A as a benchmark and compute its dynamical critical exponent $z$. This allows us to show that frequency regulators compatible with causality and the fluctuation-dissipation theorem can be devised. We show that when the principle of minimal sensitivity (PMS) is employed to optimize the critical exponents $\eta , \nu$, and $z$, the use of frequency regulators becomes necessary to make the PMS a self-consistent criterion.

Figure 1

Values of the critical exponents $\eta$ (a), $\nu$ (b), and the dynamical exponent $z=2+{\eta }_X-\eta$ (c) in $d=3$ for the frequency-independent regulator $R_{1,k}(q,\omega )=R_{1,k}\left(q\right),R_{2,k}(q,\omega )=0$ and different values of the regulator parameter $\alpha$ in Eq. (37). The PMS value is reached for $\alpha \simeq 2$ for the two static exponents $\eta$ and $\nu$ and for $\alpha \simeq 0.6$ for the dynamical exponent $z$.

Figure 2

Typical shape of the time-dependent part ${\rho }_1\left(t\right)$ of the three regulators studied: the first regulator is defined in Eq. (43), the second in (46), and the third in (47).

Figure 3

Values of the critical exponent $z$ in $d=3$ for the flow regulated in frequencies for different values of the regulator parameter $\alpha$. For each value of $\alpha$, the value of $\beta$ has been chosen such that $z$ is extremal. The three curves are obtained from top to bottom by the regulators defined in Eqs. (43), (46), and (47). The PMS value is reached for $\alpha \simeq 1.5$ for the three regulators.

Figure 4

Exponent $z$ for $N=1$ in $d=3$ as a function of the parameter $\beta$ of the regulator (43). This curve is obtained for $\alpha =0.6$, which corresponds to the stationary point of $z$ at $\beta =0$.