Consider an \(N \times N\) matrix \(A\) that squares to the identity:
\begin{equation*}
A^{2} = I_{N}
\end{equation*}
Now consider the exponential of such a matrix:
\begin{equation*}
\exp{\left( \frac{i}{2} \theta A \right)} = \sum_{n = 0}^{\infty} \frac{1}{\Gamma(n+1)} \left( \frac{i}{2} \theta A \right)^{n}
\end{equation*}
Using the familiar trick of splitting the infinite sum into even and odd powers gives
\begin{equation*}
\exp{\left( \frac{i}{2} \theta A \right)} = I_{N} \cos{\left( \frac{\theta}{2} \right)} + i A \sin{\left( \frac{\theta}{2} \right)}
\end{equation*}
Setting \(\theta = \pi (4n + 1)\) with \(n \in \mathbb{Z}\) gives
\begin{equation*}
\exp{\left( \frac{i}{2} \pi (4 n + 1) A \right)} = i A
\end{equation*}
Writing
\begin{align*}
-i I_{N} &= \exp{\left( \frac{i}{2} \pi (4 m - 1) I_{N} \right)}, & m &\in \mathbb{Z}
\end{align*}
lets you write \(A\) as an exponential:
\begin{align*}
A &= \exp{\left[ i B(m, n) \right]}, & B(m, n) &= \frac{\pi}{2} (4 m - 1) I_{N} + \frac{\pi}{2} (4 n + 1) A
\end{align*}
This procedure works for any matrix that squares to the identity, not just Pauli matrices!
Recall that the quantum Fourier transform matrix \(F_{N}\) satisfies the following properties
\begin{align*}
(F_{N})^{2} &= S_{N}, & (S_{N})^{2} &= (F_{N})^{4} = I_{N}
\end{align*}
Here \(S_{N}\) is a certain permutation matrix
\begin{equation*}
S_{N} = \begin{bmatrix}
1 & 0 & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & 0 & 0 & \cdots & 0 & 1 \\
0 & 0 & 0 & 0 & \cdots & 1 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 1 & \cdots & 0 & 0 \\
0 & 0 & 1 & 0 & \cdots & 0 & 0 \\
0 & 1 & 0 & 0 & \cdots & 0 & 0 \\
\end{bmatrix}
\end{equation*}
If you write the Fourier transform matrix in an exponential form,
\begin{equation*}
F_{N} = \exp{(i E_{N})}
\end{equation*}
then it follows that
\begin{align*}
S_{N} &= \exp{(2 i E_{N})}, & I_{N} &= \exp{(4 i E_{N})}
\end{align*}
Since the permutation matrix \(S_{N}\) squares to the identity, it follows that
\begin{align*}
S_{N} &= \exp{[i R_{N}(m, n)]}, & R_{N}(m, n) &= \frac{\pi}{2}(4 m - 1) I_{N} + \frac{\pi}{2}(4 n + 1) S_{N}
\end{align*}
Thus, the "Hamiltonian" for the quantum Fourier transform is given by
\begin{equation*}
E_{N}(m, n) = \frac{\pi}{4}(4 m - 1) I_{N} + \frac{\pi}{4}(4 n + 1) S_{N}
\end{equation*}
Let us check this for the first few cases.
For \(N = 3\), you get:
\begin{equation*}
E_{3}(m, n) = \frac{\pi}{4}(4 m - 1) \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{bmatrix} + \frac{\pi}{4}(4 n + 1) \begin{bmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{bmatrix} = \frac{\pi}{4} \begin{bmatrix}
4(m+n) & 0 & 0 \\
0 & 4m-1 & 4n+1 \\
0 & 4n+1 & 4m-1
\end{bmatrix}
\end{equation*}
In particular,
\begin{equation*}
E_{3}(0, 0) = \frac{\pi}{4} \begin{bmatrix}
0 & 0 & 0 \\
0 & -1 & 1 \\
0 & 1 & -1
\end{bmatrix}
\end{equation*}
Well, checking with Wolfram Alpha or NumPy/SciPy, this Hamiltonian does not produce the 3-dimensional Fourier transform:
\begin{equation*}
\exp{(i E_{3})} = \frac{1}{2} \begin{bmatrix}
2 & 0 & 0 \\
0 & 1-i & 1+i \\
0 & 1+i & 1-i
\end{bmatrix}
\end{equation*}
Indeed, this is just a square root of \(S_{3}\).
Let us see if this also does not work for even values of \(N\). For \(N = 4\), you get:
\begin{equation*}
E_{4}(m, n) = \frac{\pi}{4}(4 m - 1) \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix} + \frac{\pi}{4}(4 n + 1) \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0
\end{bmatrix}
\end{equation*}
That is,
\begin{equation*}
E_{4}(m, n) = \frac{\pi}{4} \begin{bmatrix}
4(m+n) & 0 & 0 & 0 \\
0 & 4m-1 & 0 & 4n+1 \\
0 & 0 & 4(m+n) & 0 \\
0 & 4n+1 & 0 & 4m-1
\end{bmatrix}
\end{equation*}
In particular,
\begin{equation*}
E_{4}(0, 0) = \frac{\pi}{4} \begin{bmatrix}
0 & 0 & 0 & 0 \\
0 & -1 & 0 & 1 \\
0 & 0 & 0 & 0 \\
0 & 1 & 0 & -1
\end{bmatrix}
\end{equation*}
Again, checking with Wolfram Alpha or NumPy/SciPy, this Hamiltonian does not produce the 4-dimensional Fourier transform:
\begin{equation*}
\exp{(i E_{4})} = \frac{1}{2} \begin{bmatrix}
2 & 0 & 0 & 0 \\
0 & 1-i & 0 & 1+i \\
0 & 0 & 2 & 0 \\
0 & 1+i & 0 & 1-i
\end{bmatrix}
\end{equation*}
Indeed, this is just a square root of \(S_{4}\).
I guess you are not going to get the Hamiltonian for the quantum Fourier transform this way.
