a c-number term, ${H_{\mathrm{KPO}}(t_0) |{\pm \alpha_0} \rangle \propto \alpha_0^2 - p_0/K = 0}$, where ${a |{\pm \alpha_0} \rangle = \pm \alpha_0 |{\pm \alpha_0} \rangle}$ [@Leonhardt]. Thus at $t_0$, the amplitude starts decreasing as $\displaystyle {\pm \alpha (t) = \pm \alpha_0 e^{-\kappa_{\tiny{\mbox{ex}}} (t-t_0)/2}}$ because of the external coupling. If we set the pump amplitude as $p(t) = p_0 e^{-\kappa_{\tiny{\mbox{ex}}} (t-t_0)}$, $H_{\mathrm{KPO}}(t) |{\pm \alpha (t)} \rangle \propto \alpha (t)^2 - p(t)/K = 0$, that is, the Kerr term and the pumping term are cancelled out at any time $t \ge t_0$. Thus, the KPO behaves like a linear cavity during the release of the cat state, and therefore the cat state prepared inside the KPO is faithfully released as a traveling cat state.

Numerical simulation
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To examine the above method quantitatively, we numerically solve the Schrödinger equation with the Hamiltonian in Eq. (\[eq-H\]) and evaluate the fidelity between the output pulse and an ideal cat state. An approach to the numerical simulation is based on the discretization of frequency $\omega$ (or wave number $\omega /v_p$) [@Duan2003a; @Goto2005a]. In the present case, however, the photon number in the one-dimensional system is larger than one, and consequently such discretization results in a complicated 