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In this (freely available) article, 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?

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    $\begingroup$ From a quick skim, it seems as though this measure is more an indication of "how non-local" an operation is, rather than necessarily its capacity to produce entanglement. All operations that are entangling are non-local, but the converse is not necessarily true (as is the case with SWAP, which is very much non-local, but sends separable states to separable states). $\endgroup$ Commented Jan 30, 2022 at 19:43
  • $\begingroup$ I also had somewhat similar feeling, but the very first sentence of Abstract seems to claim otherwise. "Entanglement properties of bipartite unitary operators are studied via their local invariants, namely the entangling power ..." $\endgroup$
    – User101
    Commented Jan 30, 2022 at 19:53
  • $\begingroup$ Reading beyond that section, the "entangling power" seems to have a definition different from the "entanglement entropy". In fact, it looks like the entanglement power of SWAP, per their definition, is $0$ (the same as local operations). $\endgroup$ Commented Jan 30, 2022 at 19:56

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