on the fact that any quantum state inside a *linear* cavity results in a traveling pulse in the same quantum state through the leak to the output mode, where the pulse shape is exponential corresponding to the exponential decay [@Eichler2011a]. This property of linear cavities removes the concern with the entanglement. The issue with the decay can also be solved by generating a cat state faster than the decay, which is possible if $K$ is much larger than $\kappa_{\mathrm{ex}}$. The remaining problem is that the KPO has a large Kerr effect, namely, it is not a linear cavity.

Our solution is to switch off the parametric pumping as ${p(t) \propto \exp (-\kappa_{\mathrm{ex}} t)}$ after the cat-state preparation. Then the Kerr term and the pumping term are cancelled out each other, and hence the KPO can be regarded as a linear cavity. This is confirmed as follows. Suppose that at time $t_0$, the KPO is prepared in a cat state ${|\alpha_0 \rangle + |{-\alpha_0} \rangle}$, where $\displaystyle {\alpha_0 = \sqrt{p_0/K}}$ and ${p_0 = p(t_0)}$. Since $H_{\mathrm{KPO}}(t)$ in Eq. (\[eq-HKPO\]) is rewritten as $\displaystyle {H_{\mathrm{KPO}}(t) = -\frac{K}{2} 
\! \left( a^{\dagger 2} - \frac{p(t)}{K} \right)
\! \left( a^{2} - \frac{p(t)}{K} \right)}$ by dropping 