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In here it is claimed that any entangling two qubit gate of the form

$$ U =e^{-iH}, $$

where $H = h_x \sigma_x\otimes \sigma_x + h_y \sigma_y\otimes \sigma_y + h_z \sigma_z\otimes \sigma_z$ can be decomposed was the product of unitaries

$$ U = \operatorname{CNOT} (u_2\otimes v_2)\operatorname{CNOT} (u_3\otimes v_3)\operatorname{CNOT}(w\otimes w^{-1}) $$

with $u_2=\frac{i}{\sqrt{2}}(\sigma_x + \sigma_z)e^{-i(h_x + \pi/2)\sigma_z}$, $v_2=e^{-ih_z \sigma_z}$, $u_3=-\frac{i}{\sqrt{2}}(\sigma_x + \sigma_z)$, $v_3=e^{i h_y\sigma_z}$, $w=\frac{\mathbb{I}- i\sigma_x}{\sqrt{2}}$ where $\sigma_i$ are the Pauli matrices and $\mathbb{I}$ is the $2\times 2$ identity. The $\operatorname{CNOT}$ gate takes as a control qubit 1 and target qubit 2.

Now, if $U\in SU(4)$, as far as I can see the $u_j\otimes v_j$ products are $SU(4)$ matrices. How can it be that the decomposition is correct, if the determinant on the left side is $1$ and on the right side is $-1$ (since $\operatorname{det}(\operatorname{CNOT})^3=-1$)?

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The authors are neglecting the unobservable global phase, see $[11]$ at the very end of references and mentioned above equation $(15)$

$[11]$ In Eq. $(15)$ we have neglected a physically irrelevant, global phase $e^{i\pi/4}$ originating in that $\det(U_{CNOT}) =−1$ in Eq. $(1)$, i.e. $U_{CNOT}\notin \mathcal{SU}(4)$.

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