The quantum Fourier transform is a unitary matrix \(F\) that satisfies the following properties:
\begin{align*}
F^{2} &= S, & F^{3} &= F^{\dagger} = F S, & F^{4} &= S^{2} = I
\end{align*}
Here \(S\) is a particular (hermitian) permutation matrix. Let
\begin{align*}
F &= \exp{(i G)}, & G^{\dagger} = G
\end{align*}
Here \(G\) would be the hermitian Hamiltonian for the quantum Fourier transform. We are going to assume that \(G\) has a particular form:
\begin{equation*}
G = w I + x (F + F^{\dagger}) + iy (F - F^{\dagger}) + z S
\end{equation*}
Here \(w\), \(x\), \(y\), and \(z\) are real numbers. You can re-write \(G\) as
\begin{align*}
G &= w I + u F + v F^{\dagger} + z S, & u &\equiv x + i y, & v &\equiv x - i y
\end{align*}
Note that
\begin{align*}
[F, S] &= 0, & [F^{\dagger}, S] &= 0, & [F, F^{\dagger}] &= 0
\end{align*}
Thus,
\begin{equation*}
\exp{(iG)} = \exp{(iw)} \exp{(iuF)} \exp{(ivF^{\dagger})} \exp{(izS)}
\end{equation*}
Each of these factors can be expanded using trigonometric and quartic functions:
\begin{align*}
\exp{(i u F)} &= I p_{0}(u) + i F p_{1}(u) - S p_{2}(u) - i F^{\dagger} p_{3}(u) \\
\exp{(i v F^{\dagger})} &= I p_{0}(v) + i F^{\dagger} p_{1}(v) - S p_{2}(v) - i F p_{3}(v) \\
\exp{(i z S)} &= I \cos{(z)} + i S \sin{(z)}
\end{align*}
Let
\begin{equation*}
\exp{(iuF)} \exp{(ivF^{\dagger})} \exp{(izS)} = I a + F b + F^{\dagger}c + S d
\end{equation*}
In order for \(G\) to be the hermitian Hamiltonian, we must have
\begin{align*}
a &= 0 & b &\neq 0 & c &= 0 & d &= 0
\end{align*}
The \(I\) coefficient will be
\begin{multline*}
a = \cos{(z)} \left[p_{0}(u) p_{0}(v) - p_{1}(u) p_{1}(v) + p_{2}(u) p_{2}(v) - p_{3}(u) p_{3}(v) \right] \\
- i \sin{(z)} \left[p_{0}(u) p_{2}(v) - p_{1}(u) p_{3}(v) + p_{2}(u) p_{0}(v) - p_{3}(u) p_{1}(v) \right]
\end{multline*}
This coefficient must vanish. The \(F\) coefficient will be
\begin{multline*}
b = -i\cos{(z)} \left[p_{0}(u) p_{3}(v) - p_{1}(u) p_{0}(v) + p_{2}(u) p_{1}(v) - p_{3}(u) p_{2}(v) \right] \\
- \sin{(z)} \left[p_{0}(u) p_{1}(v) - p_{1}(u) p_{2}(v) + p_{2}(u) p_{3}(v) - p_{3}(u) p_{0}(v) \right]
\end{multline*}
This coefficient cannot vanish. The \(F^{\dagger}\) coefficient will be
\begin{multline*}
c = i \cos{(z)} \left[p_{0}(u) p_{1}(v) - p_{1}(u) p_{2}(v) + p_{2}(u) p_{3}(v) - p_{3}(u) p_{4}(v) \right] \\
+ \sin{(z)} \left[p_{0}(u) p_{3}(v) - p_{1}(u) p_{0}(v) + p_{2}(u) p_{1}(v) - p_{3}(u) p_{2}(v) \right]
\end{multline*}
This coefficient must vanish. The \(S\) coefficient will be
\begin{multline*}
d = -\cos{(z)} \left[p_{0}(u) p_{2}(v) - p_{1}(u) p_{3}(v) + p_{2}(u) p_{0}(v) - p_{3}(u) p_{1}(v) \right] \\
+ i \sin{(z)} \left[p_{0}(u) p_{0}(v) - p_{1}(u) p_{1}(v) + p_{2}(u) p_{2}(v) - p_{3}(u) p_{3}(v) \right]
\end{multline*}
This coefficient must vanish. These coefficients can be written in a simpler form:
\begin{align*}
a &= \alpha(u, v) \cos{(z)} - i \beta(u, v) \sin{(z)} & b &= -i\gamma(u, v) \cos{(z)} - \delta(u, v) \sin{(z)} \\
c &= i \delta(u, v) \cos{(z)} + \gamma(u, v) \sin{(z)} & d &= -\beta(u, v) \cos{(z)} + i \alpha(u, v) \sin{(z)}
\end{align*}
with
\begin{align*}
\alpha(u, v) &\equiv p_{0}(u) p_{0}(v) - p_{1}(u) p_{1}(v) + p_{2}(u) p_{2}(v) - p_{3}(u) p_{3}(v) \\
\beta(u, v) &\equiv p_{0}(u) p_{2}(v) - p_{1}(u) p_{3}(v) + p_{2}(u) p_{0}(v) - p_{3}(u) p_{1}(v) \\
\gamma(u, v) &\equiv p_{0}(u) p_{3}(v) - p_{1}(u) p_{0}(v) + p_{2}(u) p_{1}(v) - p_{3}(u) p_{2}(v) \\
\delta(u, v) &\equiv p_{0}(u) p_{1}(v) - p_{1}(u) p_{2}(v) + p_{2}(u) p_{3}(v) - p_{3}(u) p_{0}(v)
\end{align*}
Actually, these are much simpler:
\begin{align*}
2\alpha(u, v) &= p_{0}(u - v) + p_{2}(u - v) + p_{0}(u + v) - p_{2}(u + v) \\
2\beta(u, v) &= p_{0}(u - v) + p_{2}(u - v) - p_{0}(u + v) + p_{2}(u + v) \\
2\gamma(u, v) &= -p_{3}(u - v) - p_{1}(u - v) + p_{3}(u + v) - p_{1}(u + v) \\
2\delta(u, v) &= -p_{3}(u - v) - p_{1}(u - v) - p_{3}(u + v) + p_{1}(u + v)
\end{align*}
Indeed, using the definitions of the \(p\)-functions gives
\begin{align*}
2\alpha(u, v) &= \cosh{(u-v)} + \cos{(u+v)} = \cos{(2y)} + \cos{(2x)} \\
2\beta(u, v) &= \cosh{(u-v)} - \cos{(u+v)} = \cos{(2y)} - \cos{(2x)} \\
2\gamma(u, v) &= -\sinh{(u - v)} - \sin{(u + v)} = -i\sin{(2y)} - \sin{(2x)} \\
2\delta(u, v) &= -\sinh{(u - v)} + \sin{(u + v)} = -i\sin{(2y)} + \sin{(2x)}
\end{align*}
Thus, \(\alpha\) and \(\beta\) are real and \(\gamma\) and \(\delta\) are complex. Since \(\alpha\) and \(\beta\) are real, and \(a\) and \(d\) must vanish, we must have
\begin{equation*}
\alpha = \beta = 0
\end{equation*}
This can be achieved when
\begin{align*}
x_{m} &= \frac{(2m + 1) \pi}{4} & y_{n} &= \frac{(2n + 1) \pi}{4}
\end{align*}
Note that
\begin{align*}
\sin{(2x_{m})} &= (-1)^{m} & \sin{(2y_{n})} &= (-1)^{n}
\end{align*}
and thus
\begin{align*}
\gamma_{mn} &= -\frac{(-1)^{m} + i (-1)^{n}}{2} & \delta_{mn} &= \frac{(-1)^{m} - i (-1)^{n}}{2}
\end{align*}
Hence
\begin{align*}
b_{mn} &= \frac{1}{2} \cos{(z)} \left[i(-1)^{m} - (-1)^{n} \right] + \frac{1}{2} \sin{(z)} \left[i(-1)^{n} - (-1)^{m} \right] \\
c_{mn} &= \frac{1}{2} \cos{(z)} \left[i(-1)^{m} + (-1)^{n} \right] - \frac{1}{2} \sin{(z)} \left[i(-1)^{n} + (-1)^{m} \right]
\end{align*}
There are four distinct possibilities:
\begin{align*}
b_{00} &= (i - 1) \left[ \frac{\cos{(z)} + \sin{(z)}}{2} \right] & c_{00} &= (i + 1) \left[ \frac{\cos{(z)} - \sin{(z)}}{2} \right] \\
b_{01} &= (i + 1) \left[ \frac{\cos{(z)} - \sin{(z)}}{2} \right] & c_{01} &= (i - 1) \left[ \frac{\cos{(z)} + \sin{(z)}}{2} \right] \\
b_{10} &= (1 + i) \left[ \frac{\sin{(z)} - \cos{(z)}}{2} \right] & c_{10} &= (1 - i) \left[ \frac{\cos{(z)} + \sin{(z)}}{2} \right] \\
b_{11} &= (1 - i) \left[ \frac{\cos{(z)} + \sin{(z)}}{2} \right] & c_{11} &= (1 + i) \left[ \frac{\sin{(z)} - \cos{(z)}}{2} \right]
\end{align*}
Note that \(c_{00} = 0\) or \(c_{11} = 0\) whenever
\begin{equation*}
z_{k} = \frac{(4k + 1) \pi}{4}
\end{equation*}
Then
\begin{align*}
b_{00k} &= (-1)^{k} \left( \frac{-1 + i}{\sqrt{2}} \right) = \exp{\left(\frac{3 \pi i}{4} + k \pi i \right)} \\
b_{11k} &= (-1)^{k} \left( \frac{1 - i}{\sqrt{2}} \right) = \exp{\left(-\frac{\pi i}{4} + k \pi i \right)}
\end{align*}
Note that
\begin{align*}
b_{000} &= b_{111} = \exp{\left(\frac{3 \pi i}{4} \right)} & b_{001} &= b_{110} = \exp{\left(-\frac{\pi i}{4} \right)}
\end{align*}
In a similar way, \(c_{01} = 0\) or \(c_{10} = 0\) whenever
\begin{equation*}
z_{l} = \frac{(4l + 3) \pi}{4}
\end{equation*}
Then
\begin{align*}
b_{01l} &= (-1)^{l} \left( \frac{-1 - i}{\sqrt{2}} \right) = \exp{\left(-\frac{3 \pi i}{4} + l \pi i \right)} \\
b_{10l} &= (-1)^{l} \left( \frac{1 + i}{\sqrt{2}} \right) = \exp{\left(\frac{\pi i}{4} + l \pi i \right)}
\end{align*}
Note that
\begin{align*}
b_{010} &= b_{101} = \exp{\left(\frac{5 \pi i}{4} \right)} & b_{011} &= b_{100} = \exp{\left(\frac{\pi i}{4} \right)}
\end{align*}
Thus, there are four possible outcomes for the hermitian Hamiltonian:
\begin{align*}
G_{000} &= -\frac{3 \pi}{4}I + \frac{\pi}{4}(1 + i) F + \frac{\pi}{4}(1 - i) F^{\dagger} + \frac{\pi}{4} S \\
G_{001} &= \frac{\pi}{4}I + \frac{\pi}{4}(1 + i) F + \frac{\pi}{4}(1 - i) F^{\dagger} - \frac{3 \pi}{4} S \\
G_{010} &= \frac{3 \pi}{4}I + \frac{\pi}{4}(1 - i) F + \frac{\pi}{4}(1 + i) F^{\dagger} + \frac{3 \pi}{4} S \\
G_{011} &= -\frac{\pi}{4}I + \frac{\pi}{4}(1 - i) F + \frac{\pi}{4}(1 + i) F^{\dagger} - \frac{\pi}{4} S
\end{align*}
I have checked that these four hermitian Hamiltonians work for the 3-dimensional and 4-dimensional discrete Fourier transform. Indeed, my result works for any unitary matrix whose quartic power is the identity.
