Fermion bag approach for the massive Thirring model at finite density

We consider the $2+1$-dimensional massive Thirring model with one flavor at finite density. Two numerical methods, the fermion bag approach and complex Langevin dynamics, are used to calculate the chiral condensate and fermion density of this model. The numerical results obtained by the fermion bag approach are compared with those obtained by complex Langevin dynamics. They are also compared with those obtained under a phase-quenched approximation. We show that in some range of fermion coupling strength and chemical potential, the sign problem in the fermion bag approach is mild, while it becomes severe for the complex Langevin dynamics.

Figure 1

Frequency of negative $C$ where the configurations $k=(k_{x,\alpha })$ are chosen randomly.

Figure 2

Comparison of the chiral condensate and fermion density obtained by complex Langevin dynamics and by the fermion bag approach. In the fermion bag approach, the sampling starts after $1\times 10^6$ Monte Carlo steps and ends after $1\times 10^7$ steps. Two sequential samples are separated by $10\times 3\times N^3$ Monte Carlo steps. In complex Langevin dynamics, $\mathrm{\Delta }\mathrm{\Theta }=0.001$, the equilibrium Langevin time $t_{\mathrm{eq}}=20$ and the sampling end time $t_{\mathrm{end}}=40$. Two sequential samples are separated by 10 complex Langevin steps.

Figure 3

Parameters are the same as those in Fig. 2, except $U$.

Figure 4

Parameters are the same as those in Fig. 2, except $U$, for FB and CL, with phase-quenched approximations (pq).

Figure 5

The real part of phase $⟨e^{i\phi }{⟩}_{\mathrm{pq}}$ obtained by the fermion bag approach for $U=0.25$, 1.0, 2.0, 5.0 and by complex Langevin dynamics for $U=0.25$ [$U=0.25(\mathrm{CL})$].

Figure 6

Comparison among the fermion bag approach, complex Langevin dynamics, and the exact solution for the one-dimensional Thirring model (see Ref. [56]).

Figure 7

The chiral condensate in the heavy quark limit for coupling strength $U=0.25$ (left) and $U=1/12=0.08333$ (right). Parameters are the same as those in Fig. 2, except $U$ and $m$.