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Define a quantum gate(or unitary) $U$ as entangling if there is a product state that $U$ produce entangling state when applied on.

I've referred previous answer that notion of 'entangling power' can be criterion for entanglement testing for any unitary, which is defined as

$$ K_E(U) = max_{|\phi\rangle, |\psi\rangle } E(U|\phi\rangle, |\psi\rangle). $$ (here E denotes von Neumann entropy)

However, I think it is unsatisfactory to be used, since it seems that it cannot be computed with method other than brute force method in general (enumerating on every product states).

Is there another criterion of entangling gate or maybe no general (fancier then brute-force) method exist?

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  • $\begingroup$ Are you talking specifically about a bipartite unitary, i.e. a unitary acting on exactly two systems and (possibly) creating entanglement between the two? Or are you wanting it to be more general? $\endgroup$
    – DaftWullie
    Jul 14 at 11:12
  • $\begingroup$ @DaftWullie I actually wanted to be general at best, but I also wonder if there is two-qubit system specific method. $\endgroup$
    – 강찬구
    Jul 14 at 11:14
  • $\begingroup$ My initial reaction is that you would simply try to identify if $U=U_A\otimes U_B$. If not, there are only a limited number of non-entangling gates left. (Is it only $U=(U_A\otimes U_B).SWAP$?) $\endgroup$
    – DaftWullie
    Jul 14 at 11:14
  • $\begingroup$ what's your definition of $E$ here? More specifically, if it is the entanglement entropy, why does it take two inputs? $\endgroup$
    – glS
    Jul 15 at 11:13
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The only invertible linear maps (and particular, unitaries) $U$ acting on $\mathbb{C}^A \otimes \mathbb{C}^B$ that do not entangle any product state are those of the form $V \otimes W$, and the SWAP gate if $A=B$. See, e.g., Section 3 of this paper.

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