I want to express the square root of NOT as a time-dependent unitary matrix such that each $n$ units of time, the square root of NOT is produced.

More precisely, I want to find a $U(t_0,t_1)$ such that $U(t_0,t_1) = \sqrt{\text{NOT}}$, if $t_1-t_0=n$ for some $n$.

One possible solution is to express $\sqrt{\text{NOT}}$ as a product of rotation matrices, and then, parametrize the angles in a clever way to depend on the time. But I do not know how to express $\sqrt{\text{NOT}}$ as a product of rotation matrices.

Any help?

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    $\begingroup$ Use $U(t)=\exp[-it\sigma_x]$. $\endgroup$ – Norbert Schuch Jul 4 '18 at 18:03

$$ \sqrt{NOT} = e^{(\frac{i \pi}{4} I_2 - \frac{i \pi}{4} \sigma_x)}\\ U(t) = e^{\frac{t-t_0}{t_1 - t_0} (\frac{i \pi}{4} I_2 - \frac{i \pi}{4} \sigma_x)} $$

  • $\begingroup$ I think the $U(t)$ you proposed is not unitary (so not valid). Am I right? $\endgroup$ – Alejandro Díaz-Caro Jul 5 '18 at 18:33
  • $\begingroup$ If you write that $U(t)=e^{i (t-t_0) H}$, $H$ would be $\frac{\pi}{4(t_1-t_0)} (I_2 - \sigma_x)$. Check that $H$ is Hermitian. $\endgroup$ – AHusain Jul 5 '18 at 18:43
  • $\begingroup$ Sorry for all the questions: Doesn't have to be skew-Hermitian to ensure unitarity? That is, if $A^\dagger=-A$ then $e^A$ is unitary. Here the matrix $\frac{t-t_0}{t_1-t_0}\frac{i\pi}4(I_2-\sigma_x)$ seems to be Hermitian, which just ensure that $U(t)$ is Hermitian, but not unitary. $\endgroup$ – Alejandro Díaz-Caro Jul 5 '18 at 22:28
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    $\begingroup$ $(i H)^\dagger = -i H^\dagger = - i H$ $\endgroup$ – AHusain Jul 5 '18 at 22:33
  • $\begingroup$ Great. Thanks. Then I am doing something wrong. Take $\delta_t\frac{i\pi}4(I_2-\sigma_x)$ and diagonalize it, getting $SDS^{-1}$, then $U(t)=Se^DS^{-1}$, and so you can transform the expression of $U(t)$ into $\frac 12\begin{bmatrix} 1+i^{\delta_t} & 1-i^{\delta_t} \\ 1-i^{\delta_t} & 1+i^{\delta_t}\end{bmatrix}$, but that matrix is not unitary. So I do not understand where I am making a mistake. $\endgroup$ – Alejandro Díaz-Caro Jul 5 '18 at 22:36

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