# Finding a global phase that transform the Hadamard gate to an element of $SU(2)$ and propose an evoultion operator which implents the operation

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 $$H$$?

You did more than you needed to in part i. You effectively did part ii already.

Because

$$g = \frac{u}{\sqrt{\det{u}}}$$

is in $$SU(2)$$, the phase you need to multiply by is $$\frac{1}{\sqrt{\det{u}}}$$.

$$\det H = -1/2-1/2 = -1$$

So to get in $$SU(2)$$, you need to multiply $$\sqrt{-1}=\pm i$$. That is seen as the $$e^{\pi i /2}$$ you see as the prefactor in the way you did it. You can use either $$e^{\pm \pi i/2}$$.

For ii, you can drop the prefactor and only consider

$$H \propto e^{-\pi i \frac{\sigma_x+\sigma_z}{2\sqrt{2}}}$$

You can match that exponent with $$\frac{-it}{\hbar} H$$ in order to suggest $$H$$ and $$t$$.

$$-\pi i \frac{\sigma_x+\sigma_z}{2\sqrt{2}} = \frac{-it}{\hbar} H\\ \pi \frac{\sigma_x+\sigma_z}{2\sqrt{2}} = \frac{t}{\hbar} H\\ t H = \frac{\pi}{\sqrt{2}} (\frac{\hbar}{2} \sigma_x+\frac{\hbar}{2} \sigma_z)$$

So you can use $$t=\frac{\pi}{\sqrt{2}}$$ and $$H=(\frac{\hbar}{2} \sigma_x+\frac{\hbar}{2} \sigma_z)$$.

The reason for putting the Hamiltonian with $$\frac{\hbar}{2} \sigma$$ terms is because this is the spin operator

$$\vec{S} = \frac{\hbar}{2} \vec{\sigma}$$

Of course you could replace $$t \to \lambda t$$ and $$H \to \frac{1}{\lambda} H$$ for any $$\lambda$$, but if you do it with very small $$\lambda$$ like $$\hbar$$, then you will get unreasonably short time-scales for applying the evolution. You're not going to have precision control to only apply a signal for $$10^{-30}$$ seconds.