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I'm trying to understand the relationship between the factorability of a 2 qubit gate and that gate's ability to cause entanglement.

I've begun by considering the controlled Pauli Z gate. After playing around with the computational basis inputs and the Hadamard gate, it seems like the CZ gate cannot cause entanglement. But I'm not sure if this idea holds in general (i.e. is there some input that would become entangled if run through a CZ gate).

My intuition is that since the CZ matrix is diagonal it is not able to cause entanglement. Meanwhile the CX and CY are not diagonal and therefore can cause entanglement. But I am hoping to find a more formal definition/distinction or counterexample.

If anyone has any thoughts or references in this area, they would be greatly appreciated.

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Yes, CZ gate can entangle its inputs. For example

$$ CZ|+\rangle|+\rangle = \frac{|0\rangle|+\rangle+|1\rangle|-\rangle}{\sqrt2} $$

which is entangled because the reduced density matrix of either qubit is maximally mixed.

By the spectral theorem all normal matrices are diagonalizable. In particular, every unitary matrix is diagonalizable. Thus, the existence of a diagonal matrix for a given gate has no bearing on whether the gate can entangle its inputs.

That said, there do exist representations of unitary operators which are related to their ability to create entanglement, see e.g. operator Schmidt decomposition and KAK decomposition.

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    $\begingroup$ Thank you! One follow-up: is there anything on the relationship between the ability to write a gate as the tensor product of two matrices and that gate's ability to cause entanglement? Clearly, if I can write a gate as the tensor product of two gates, it cannot cause entanglement (since I could apply those gates separately). Was just wondering about the other direction (i.e. if a gate cannot be factored it certainly causes entanglement) $\endgroup$
    – Flipper
    Nov 3, 2021 at 21:03
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    $\begingroup$ Yes, a gate can create entanglement if and only if it cannot be written as a tensor product. In fact, this is a special case of operator Schmidt decomposition. $\endgroup$ Nov 3, 2021 at 21:13

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