# Decomposition of a 2 qubit gate into standard quantum gates

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 $$\operatorname{det}(\operatorname{CNOT})^3=-1$$)?

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. $$U_{CNOT}\notin \mathcal{SU}(4)$$.