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I want to find the matrix $\hat{H}_d$ which, when exponentiated, leads to a d-dimensional cyclic permutation transformation matrix.


I have solutions for d=2:

$$ \hat{U}_2 =\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right)=-i \exp\left(i\hat{H_2}\right) \to \\ \hat{H}_2 =\left( \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right) $$


d=3:

$$ \hat{U}_3 =\left( \begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\\ \end{matrix} \right)=-i \exp\left(i\hat{H_3}\right) \to \\ \hat{H}_3 =\frac{\pi}{3}\left( \begin{matrix} -\frac{1}{2} & \left(-\frac{1}{\sqrt{3}} + i \right) & \left(\frac{1}{\sqrt{3}} + i \right) \\ \left(\frac{1}{\sqrt{3}} + i \right) & -\frac{1}{2} & \left(-\frac{1}{\sqrt{3}} + i \right)\\ \left(\frac{1}{\sqrt{3}} + i \right) & \left(-\frac{1}{\sqrt{3}} + i \right) & -\frac{1}{2} \end{matrix} \right) $$


d=4:

$$ \hat{U}_4 =\left( \begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ \end{matrix} \right)=-i \exp\left(i\hat{H_4}\right) \to \\ \hat{H}_4 =\frac{\pi}{4}\left( \begin{matrix} i & (1+i) & -i & (-1+i) \\ (-1+i) & i & (1+i) & -i\\ -i & (-1+i) & i & (1+i)\\ (1+i) & -i & (-1+i) & i \\ \end{matrix} \right) $$


Unfortunately, I am missing a general form of $H_d$ for $d>4$, so my question is:

Question: How does $\hat{H}_d$ look when I want that $\hat{U}_d=-i \exp\left(i \hat{H}_d \right)$ is a d-dimensional cyclic permutation transformation?

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    $\begingroup$ Since the matrix is circulant you can diagonalize it. en.wikipedia.org/wiki/Circulant_matrix maybe that helps $\endgroup$ – Markus Sprecher Aug 17 at 6:20
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    $\begingroup$ Because of different branches of the logarithm function, the answer is not unique. For example there is a solution for $\hat H_3$ with zeros on the diagonal. $\endgroup$ – Brendan McKay Aug 17 at 9:27
  • $\begingroup$ Thanks, both of your comments are very useful, and i already see some progress because of them. Thank you! $\endgroup$ – NicoDean Aug 18 at 0:05

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