A unitary matrix \(U\) satisfies the following relation
\begin{equation*}
U^{\dagger} = U^{-1}
\end{equation*}
A hermitian matrix \(V\) satisfies the following relation
\begin{equation*}
V^{\dagger} = V
\end{equation*}
The simplest case involves 2×2 matrices. The most general 2×2 hermitian matrix is given by
\begin{equation*}
V = \frac{1}{2} \begin{bmatrix}
w + z & x - yi \\
x + yi & w - z
\end{bmatrix} = \frac{w}{2} I_{2} + \frac{x}{2} X + \frac{y}{2} Y + \frac{z}{2} Z
\end{equation*}
Recall that
\begin{equation*}
[X, I_{2}] = [Y, I_{2}] = [Z, I_{2}] = 0
\end{equation*}
A unitary matrix can be written in exponential form:
\begin{equation*}
U = \exp{(iV)} = \exp{\left( \frac{iw}{2} \right)} \exp{\left( \frac{i}{2} \left[x X + y Y + z Z \right] \right)}
\end{equation*}
Using the familiar result for exponentiation of Pauli matrices gives
\begin{equation*}
U = \exp{\left( \frac{iw}{2} \right)} \left[ I_{2} \cos{\left( \frac{\sqrt{x^{2} + y^{2} + z^{2}}}{2} \right)} + i \left( \frac{x X + y Y + z Z}{\sqrt{x^{2} + y^{2} + z^{2}}} \right) \sin{\left( \frac{\sqrt{x^{2} + y^{2} + z^{2}}}{2} \right)} \right]
\end{equation*}
As an example, consider the 2-dimensional Hadamard transform:
\begin{equation*}
H_{2} = \frac{1}{\sqrt{2}} \begin{bmatrix}
1 & 1 \\ 1 & -1
\end{bmatrix} = \frac{1}{\sqrt{2}}\left( X + Z \right)
\end{equation*}
Thus, we need
\begin{align*}
w &= -\pi, & y &= 0, & \sqrt{x^{2} + z^{2}} &= \pi
\end{align*}
Then
\begin{equation*}
U = \frac{xX + zZ}{\pi}
\end{equation*}
Finally, choose
\begin{equation*}
x = z = \frac{\pi}{\sqrt{2}}
\end{equation*}
Thus,
\begin{align*}
H_{2} &= \exp{\left(i V \right)}, & V &= -\frac{\pi}{2} I_{2} + \frac{\pi}{2} H_{2}
\end{align*}
Recall that \(H_{2}\) is equivalent to the 2-dimensional Fourier transform, so the above result gives you the Hamiltonian of the 2-dimensional Fourier transform. In general, the \(N\)-dimensional discrete Fourier transform \(F_{N}\) satisfies the following relations:
\begin{align*}
(F_{N})^{2} &= S_{N}, & (S_{N})^{2} &= (F_{N})^{4} = I_{N}
\end{align*}
Let
\begin{equation*}
F_{N} = \exp{(i G_{N})}
\end{equation*}
Then
\begin{equation*}
(F_{N})^{2} = \exp{(2i G_{N})} = S_{N}
\end{equation*}
and also
\begin{equation*}
(F_{N})^{4} = \exp{(4i G_{N})} = I_{N}
\end{equation*}
Note that this does not necessarily implies that \(G_{N} = 0\). Indeed, looking back at our result for 2×2 matrices, setting
\begin{align*}
w &= 4 \pi m, & \sqrt{x^{2} + y^{2} + z^{2}} &= 4 \pi n, & m, n &\in \mathbb{Z}
\end{align*}
you get
\begin{align*}
\exp{(i V(m, n))} &= I_{2}, & V(m, n) &= 2 \pi m I_{2} + 2 \pi n \left(xX + yY + zZ \right), & x^{2} + y^{2} + z^{2} &= 1
\end{align*}
Thus, the exponential of a 2×2 hermitian matrix of the above form will be the 2×2 identity matrix.
For two qubits you can have the 4-dimensional Fourier transform. The Hamiltonian for this matrix would presumably require understanding 4×4 hermitian matrices in more detail. However, since the 4-dimensional Hadamard transform is given by a Kronecker product,
\begin{equation*}
H_{4} = H_{2} \otimes H_{2}
\end{equation*}
you can find the Hamiltonian for this transform with what we have seen so far. There is an important identity that I will not prove:
\begin{equation*}
\exp{(A)} \otimes \exp{(B)} = \exp{(A \oplus B)}
\end{equation*}
Here \(\oplus\) is the Kronecker sum. If \(A\) is \(M \times M\) and \(B\) is \(N \times N\), then the Kronecker sum is given by
\begin{equation*}
A \oplus B = (A \otimes I_{N}) + (I_{M} \otimes B)
\end{equation*}
Thus
\begin{equation*}
H_{4} = \exp{(iG_{2})} \otimes \exp{(iG_{2})} = \exp{(i G_{2} \oplus iG_{2})} = \exp{(i G_{4})}
\end{equation*}
That is
\begin{align*}
G_{4} &= G_{2} \oplus G_{2}, & G_{2} = -\frac{\pi}{2} I_{2} + \frac{\pi}{2} H_{2} = \frac{\pi}{2\sqrt{2}} \begin{bmatrix}
1 - \sqrt{2} & 1 \\
1 & -1 - \sqrt{2}
\end{bmatrix}
\end{align*}
More explicitly,
\begin{equation*}
G_{4} = \frac{\pi}{2 \sqrt{2}} \begin{bmatrix}
2 - \sqrt{8} & 1 & 1 & 0 \\
1 & -\sqrt{8} & 0 & 1 \\
1 & 0 & -\sqrt{8} & 1 \\
0 & 1 & 1 & -2 - \sqrt{8}
\end{bmatrix} = -\pi I_{4} + \frac{\pi}{2} \left( H_{2} \oplus H_{2} \right)
\end{equation*}
This is the Hamiltonian for the 4-dimensional Hadamard transform.
