The Duan Criterion, when written as a function of creation and annihilation operators $b$ and $b^\dagger$ depends only on $\langle b_1^\dagger b_1\rangle$, $\langle b_2^\dagger b_2\rangle$ and $\langle b_1 b_2\rangle$. I am ignoring displacement as it plays no role in entanglement.
More specifically, the Duan criterion, for variables $x_1 = (b_1 + b_1^\dagger)/\sqrt{2}$, $p_1 = -i(b_1 - b_1^\dagger)/\sqrt{2}$, $x_2= (e^{i \phi}b_2 + e^{-i \phi} b_2^\dagger)/\sqrt{2}$ and $p_2 = -i(e^{i\phi}b_2 + e^{-i\phi}b_2^\dagger)/\sqrt{2}$, can be cast into the form $$ a^2 \langle b_1^\dagger b_1\rangle + \sigma (e^{i\phi}\langle b_1 b_2\rangle +e^{-i \phi}\langle b_1^\dagger b_2^\dagger\rangle) + a^{-2}\langle b_2^\dagger b_2\rangle \leq 0, $$ where $\sigma = \mathrm{sign}({a})$ and $a$ is a non-zero real number. This can be minimized with a convenient choice of $a$ to be $$ a^2 \langle b_1^\dagger b_1\rangle + a^{-2}\langle b_2^\dagger b_2\rangle - 2 \vert\langle b_1 b_2\rangle\vert \leq 0, $$ For Gaussian states this is stronger than an entanglement witness. A state is entangled if and only if this criterion is met (see Duan's paper).
For Gaussian states, in order to get non-zero $\langle b_1 b_2\rangle$ then it is necessary to have two mode squeezing. Does this necessarily imply that two-mode squeezing is required to entangle them?
It seems to me that all the other terms are necessarily positive, the best case scenario being when they are zero.
That being said, an interaction such as $b_1 b_2^\dagger + \mathrm{h.c.}$ can create states such as $|01\rangle + |10\rangle$ from a separable (though admitedly non-Gaussian) states $|11\rangle$, which by themselves would be entangled, yet how would these contribute to the Duan criterion? Or do these states appear is such way that in Gaussian states they do not generate entanglement? (or as always a possibility, was any of my statements wrong?)
