$\mathcal{PT}$-symmetric currents of a Bose-Einstein condensate in a triple well

We study the case of $\mathcal{PT}$-symmetric perturbations of Hermitian Hamiltonians with degenerate eigenvalues using the example of a triple-well system. The degeneracy complicates the question of whether or not a stationary current through such a system can be established, i.e., whether or not the $\mathcal{PT}$-symmetric states are stable. It is shown that this is only the case for perturbations that do not couple to any of the degenerate states. The physical explanation for the inhibition of stable currents is discussed. However, introducing an on-site interaction restores the capability to support stable currents.

Figure 1

Scheme of the three-mode model. The imaginary on-site energies provide a particle influx to well 1 and an outflux from well 3. These two wells are coupled by a constant strength, set to unity, while the coupling $J$ to well 2 is varied.

Figure 2

(a) Real and (b) imaginary parts of the eigenvalue $\mu$ for different values of the coupling strength $J$ as function of the in- and outcoupling parameter $\gamma$. For $J=0$ the ground and second excited state form an exceptional point of second order (EP2) at $\gamma =1$ where both $\mathcal{PT}$-symmetric states vanish and two $\mathcal{PT}$-broken states emerge. The first excited state, located only in well 2, is completely independent from the other eigenstates. For $J>0$ the first excited state forms an EP2 with the second excited state, while the ground state is shifted to smaller energies and retains its $\mathcal{PT}$ symmetry for all values of $\gamma$. For increasing values of $J$, the energy difference between the two excited states decreases and the EP2 is shifted to smaller values of $\gamma$. At $J=1$, two $\mathcal{PT}$-broken states emerge directly at $\gamma =0$ with degenerate eigenvalues $\mu =1$, thus rendering the ground state unstable for $\gamma \ne 0$.

Figure 3

Net current through the system $j_{\mathrm{ext}}$ as function of $\gamma$ for (a) $J=0$, (b) $J=0.1$, (c) $J=0.4$, (d) $J=0.8$. The highlighted areas show regions where $\mathcal{PT}$-broken eigenstates with complex eigenvalues exist, thus rendering all stationary states unstable. The maximum current, $j_{\mathrm{ext}}=1$, is reachable only for $J=0$. For $0<J<1$ the second excited state supports the highest net current slightly before the EP.

Figure 4

Ratio of the current $j_{12}$ through the additional well and $j_{13}$ directly to the loss well as function of $\gamma$ for (a) $J=0$, (b) $J=0.1$, (c) $J=0.4$, and (d) $J=0.8$. While the current $j_{13}$ is always positive, negative ratios represent negative currents $j_{12}$. Again the highlighted areas show regions where $\mathcal{PT}$-broken eigenstates exist. For $J=0$ no particles can be transferred through well 2. After coupling the well to the system, the ground state supports a positive current through the well, while the excited states show negative currents. For $J\to 1$ the ratio approaches $-1$.

Figure 5

Chemical potential $\mu$ of $\mathcal{PT}$-symmetric states against the strength of the parameter $\gamma$ for coupling parameter $J=1$ and nonlinearity parameter (a), (b) $U=1$, (c) $U=2$, and (d) $U=4$. Introducing the nonlinearity (a) gives rise to four new states for $\gamma \ne 0$, which are shown in the enlarged panels (b), (c), and (d). For $U=1$ and $U=2$ three of them are stable (solid lines) and one is unstable (dashed line); for $U=4$ all states except the highest excited one are unstable.

Figure 6

Net current through the system given by the incoupling external current $j_{\mathrm{ext}}$ against $\gamma$ for (a) $U=1$, (b) $U=1.5$, (c) $U=2$, and (d) $U=2.5$. Only the three stable $\mathcal{PT}$-symmetric states introduced in Fig. 5 are shown. For nonlinearities of $U$ up to 2, all three states are able to sustain a similar current, which reaches a maximum of approximately 0.3 at $U\approx 2$. For higher nonlinearities, stable states become unstable until only the highest excited state remains. Its low current is due to the fact that there are only few particles present in the gain and loss well.