In this paper (arXiv), just below equation 14, it says
The SWAP operator has the maximum operator entanglement entropy...
The operator entanglement entropy, denoted by $E(U)$ for an operator $U$, is such that $E(u_A \otimes u_B) = 0$, which makes sense, since local operations do not create entanglement. However, to the best of my understanding, SWAP just re-arranges a product state $|\psi\rangle \otimes |\phi\rangle$ to $|\phi\rangle \otimes |\psi\rangle$, which is still a separable state. Then why should it have maximum operation entanglement entropy?
