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I was looking back over an old assignment and I came across a question I wasn't quite sure how to do the problem statement is as follows:

The Hadamard rotation is an element of the group $U(2)$.

(i) Find the global phase with which one needs to multiply the Hadamard gate to obtain an operation that is an element of the group $SU(2)$, and

(ii) propose the evolution operator that implements this operation, that is, propose the suitable Hamiltonian and the duration for which it needs to be turned on.

For (i) I think the answer was either due to this statement

$U(2)$ is the group of $2$ by $2$ unitary matrices or operators. In contrast to the elements $SU(2)$, the determinant of the elements of the group $u ∈ U(2)$ is not ﬁxed to unity. Each element $u∈U(2)$ can be expressed in terms of an element of $SU(2)$ as $u = e^{i\alpha}g$ where $g \in SU(2).$

$$H=\begin{pmatrix}\tfrac{1}{\sqrt{2}} &\tfrac{1}{\sqrt{2}}\\\tfrac{1}{\sqrt{2}} &\tfrac{1}{-\sqrt{2}} \end{pmatrix}$$$$=e^{i \pi/2}\begin{pmatrix}\tfrac{e^{-i\pi/2}}{\sqrt{2}} &\tfrac{e^{-i\pi/2}}{\sqrt{2}}\\\tfrac{e^{-i\pi/2}}{\sqrt{2}} &\tfrac{e^{-i\pi/2}}{-\sqrt{2}} \end{pmatrix}$$$$=e^{i\pi/2}(\cos(\pi/2)-i\sin(\pi/2)(\tfrac{\sigma_x+\sigma_z}{\sqrt{2}}))$$$$=e^{\pi i /2}e^{-\pi i (\tfrac{\sigma_x+\sigma_z}{\sqrt{2}})/2}$$

The right hand side is now being expressed as an element of $SU(2)$. However given the following statement, Considering the determinant of the product of two n-by-n matrices $A and B$, $\det(AB) = \det A \det B$, and the determinant $\det(e^\alpha I )= e^{i2α}$ we obtain the map from the elements of $U(2)$ and $SU(2)$.

$$g=\tfrac{u}{\sqrt{\det(u)}}$$

Perhaps it actually looking for that transformation instead

As for (ii) I know that $$U=e^{\tfrac{-i}{\hbar} \hat{H}t}.$$

and that we can use the Pauli operators to form our Hamiltonian, I know how to do it for the nontrivial term that would appear if we transformed as we did in (i) but I'm not sure how to do it for $H$?