Re-examining the self-contained quantum refrigerator in the strong-coupling regime

We revisit the self-contained quantum refrigerator in the strong-internal-coupling regime by employing the quantum optical master equation. It is shown that strong internal coupling reduces the cooling ability of the refrigerator. In contrast to the weak-coupling case, strong internal coupling could lead to quite different and even converse thermodynamic behaviors.

Figure 1

Heat currents ${\dot{Q}}_{\mu }[J/s]$ versus $T_H\left[K\right]$ in the weak-coupling regime. Solid line, ${\dot{Q}}_C$; dashed line, ${\dot{Q}}_H$; and dash-dotted line, ${\dot{Q}}_R$. Here $g=0.001{\omega }_H$. In particular, we set $\gamma =0.001{\omega }_H$ throughout the paper.

Figure 2

Efficiency $\eta$ of the refrigerator versus $T_H\left[K\right]$ in the weak-coupling regime. The efficiency changes slightly and it can be considered to be almost invariant within a good approximation, which is also supported by Fig. 5.

Figure 3

Heat currents ${\dot{Q}}_C[J/s]$ versus $T_H\left[K\right]$ for different coupling constants. From top to bottom, $g=0.001{\omega }_H$, $g=0.1{\omega }_H$, $g=0.2{\omega }_H$, $g=0.25{\omega }_H$, $g=0.3{\omega }_H$, and $g=0.35{\omega }_H$. The straight line represents zero heat current.

Figure 4

Heat currents versus $T_H[J/s]$ in the strong-coupling regime. Here $g=0.3{\omega }_H$ and $\gamma =0.001{\omega }_H$. The dotted line represents the zero heat current and the other lines denote the same heat currents as in Fig. 1. ${\dot{Q}}_C$ is first pushed in the positive direction and then suppressed back to the negative direction.

Figure 5

Efficiency $\eta$ versus $T_H\left[K\right]$ with different coupling constants. The lower straight line corresponds to zero efficiency. From top to bottom, the lines correspond to $g=0.001{\omega }_H$, $g=0.1{\omega }_H$, $g=0.15{\omega }_H$, $g=0.2{\omega }_H$, $g=0.25{\omega }_H$, and $g=0.3{\omega }_H$. In particular, $\eta =\mathrm{const}.$ within acceptable approximations for $g=0.001{\omega }_H$, which is consistent with Fig. 2.

Figure 6

Heat currents ${\dot{Q}}_C[J/s]$ versus $T_H\left[K\right]$ for different coupling constants. Here we let $T_C=10$ K and $T_R=40$ K in order to satisfy $\frac{{\omega }_C}{T_C}=\frac{{\omega }_R}{T_R}=\frac{1}{10}$. The upper straight line represents zero heat current. From top to bottom, the lines correspond to $g=0.001{\omega }_H$, $g=0.1{\omega }_H$, $g=0.2{\omega }_H$, $g=0.3{\omega }_H$, $g=0.4{\omega }_H$, and $g=0.5{\omega }_H$, respectively.