mode. When $\gamma = \omega_{\mathrm{a}} 
= \omega_{\mathrm{f}} = 1$ and $g = 0$, the Hamiltonian reduces to $a^{\dagger} a + (b^{\dagger} b)^{2}$. The energy levels $E_{{\textsc{af}}}(4,0)$ and $E_{{\textsc{af}}}(4,1)$ become degenerate at the value $4$. The degeneracy occurs because the operator $\ket{4,0}\bra{3,1} + \ket{3,1}\bra{4,0}$ commutes with $H_{{\textsc{af}}}$ when $\omega_{\mathrm{a}}=\omega_{\mathrm{f}}$ and $g=0$. Mixing of the states $\ket{4,0}$ and $\ket{3,1}$ occurs, and the corresponding energy eigenstates are given by the symmetric linear combination $\ket{\phi_{4,0}} =
(\ket{4,0}+\ket{3,1})/\sqrt{2}$ and the antisymmetric linear combination $\ket{\phi_{4,1}} =  (\ket{4,0}-\ket{3,1})/\sqrt{2}$. As the symmetries of the two states are different, the level crossing does not violate the von Neumann-Wigner no-crossing theorem. At the crossing, each of the states $\ket{\phi_{4,0}}$ and $\ket{\phi_{4,1}}$ remains a manifestly entangled state that is, in fact, a Bell state. This is why the corresponding $\xi_{\textsc{svne}}$ does not vanish at that point, but merely dips to a local minimum with value $1$, characteristic of a Bell state. It is interesting to note that the degeneracy that occurs when $\omega_{\mathrm{f}}=\omega_{\mathrm{a}}$ ensures entanglement even in the absence of any interaction between the two modes.

The level $E_{{\textsc{af}}}(4,2)$, on the other hand, is repelled and has the value $6$ at $g =  0$. The corresponding eigenstate $\ket{\phi_{4,2}}$ becomes the unentangled 