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Given a separable state, $|\psi\rangle$ = $|a\rangle\otimes|b\rangle$, operating on this state with a local operator of the form, $A\otimes B$ will not lead to an entangled state. Is the converse true? i.e., given that I know that action of an operator on a separable state is a separable state, can I conclude that the operator must've been a local operator?

or will the action of a non-local operator always entangle a separable state?

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  • $\begingroup$ The "or" in the last sentence does not seem to make sense logically. $\endgroup$
    – Norbert Schuch
    Commented Nov 11, 2022 at 16:03

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No.

The swap operator will never entangle any product state, yet it is not a local operator.

(Note, however, that it will create entanglement if it acts on part of a larger system.)

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