efficacy of the entanglement indicators as functions of $\lambda$. For numerical computation we have set $\omega_{1}=0.25$. Consider, as an illustration, plots of the eigenvalues $E(4,k)$ ($k=0,2,4$) as functions of $\lambda$. These plots are exactly the same as those in (a), with $\omega_{1}$ replaced by $\lambda$ on the horizontal axis, since $E(N, k)$ only depends on the parameters $\omega_{1}$ and $\lambda$ in the symmetric combination $\lambda_{1} = (\lambda^{2} 
+ \omega_{1}^{2})^{1/2}$. The avoided crossing of energy levels now occurs at $\lambda = 0$. But this symmetry between $\omega_{1}$ and $\lambda$ does not extend to the unitary transformation $V$, and hence to the eigenstates of $H_{\textsc{bec}}$. (Recall that $V$ involves the parameter $\kappa = \tan^{-1}(\lambda/\omega_{1})$.) When $\lambda=0$ there is no linear interaction between the two modes. $V$ then reduces to the identity operator, and $H_{\textsc{bec}}$ is diagonal in the basis $\lbrace \ket{k,N-k} \rbrace$. We therefore expect the entanglement to vanish at the avoided crossing. This is borne out in in which $\xi_{\textsc{svne}}$ for the state $\ket{\psi(4,k)}$ is plotted for different values of $k$. As before, it suffices to depict the cases $k = 0,1$ and $2$ because of the $k \leftrightarrow N-k$ symmetry. We observe that, in the case $k = 0$, while 