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 $SU(2)$.


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 $H$?


1 Answer 1


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


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

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.


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