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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?

Changu Kang
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  • 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? – DaftWullie Jul 14 '21 at 11:12
  • @DaftWullie I actually wanted to be general at best, but I also wonder if there is two-qubit system specific method. – Changu Kang Jul 14 '21 at 11:14
  • 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 $U=(U_A\otimes U_B).SWAP$?) – DaftWullie Jul 14 '21 at 11:14
  • what's your definition of $E$ here? More specifically, if it is the entanglement entropy, why does it take two inputs? – glS Jul 15 '21 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.

Ben
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