In this paper, Vedral shows that the entanglement in second quantized states can depend on the choice of modes. As a simple example, he points out that the ground state of two non-interacting harmonic oscillators is not entangled. But if one does a mode transformation to couple the oscillators, the ground state becomes entangled. How to operationally make sense of this feature of entanglement in second quantization? Surely the amount of entanglement in a state should not depend on a mathematical transformation.
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This may not be the full answer, but after scanning the paper, here are a couple thoughts that might be of some value.
If one rewrites Eq. (8) (the original Hamiltonian) the following is obtained:
$$ H=\hbar\omega \left(b_{1}^{\dagger}b_{1}+b_{2}^{\dagger}b_{2}\right)+\hbar\lambda\left(b_{2}^{\dagger}b_{2}-b_{1}^{\dagger}b_{1} \right). $$
This highlights the role of $\lambda.$ One can see that the total energy of the system also depends on the difference of the operators scaled by $\lambda$. Thus, it appears the $\lambda$ is some type of coupling constant.
Furthermore, if $\lambda=0$, then the "interaction" term in Eq. (13) vanishes.
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$\begingroup$ So that mode transformation corresponds to turning on and off an interaction? But then, we can think of an analogous transformation for a pair of classical oscillators, where we change coordinates x1 --> y_1 + y_2 and x2 --> y_1 - y_2. Such a transformation can couple a pair of uncoupled oscillators described by x_1 and x_2 coordinates. There the transformation is a purely mathematical operation. We didn't change anything about the system. I don't know if the same could be said about the quantum oscillators. $\endgroup$ Commented May 26 at 19:50