Hamiltonian of Qiskit QFT is not hermitian

I am trying to generate the Hamiltonian of a quantum Fourier transform by taking the log of the corresponding unitary using qiskit and scipy.

I don't find a hermitian matrix. Why?

import numpy as np
from qiskit.circuit.library import QFT
from scipy.linalg import logm, norm
from qiskit.quantum_info import Operator

circuit = QFT(num_qubits=4, do_swaps=True)
op = Operator(circuit)
U = op.data
H = 1j*logm(U)
print(norm(U@U.T.conj()-np.identity(2**4))) #check if U is unitary
print(norm(H-H.T.conj())) #check if H is hermitian


Note that I find U to be unitary and that there is no issue when do_swaps=False.

This is due to the fact that if $$U$$ is unitary with negative eigenvalues (which is the case here), its logarithm is not uniquely defined. Note that the fact that $$U$$ is unitary only ensures that there is some hermitian matrix $$H$$ such that: $$U=\exp(\mathrm{i}H).$$ It does not, however, ensure that every matrix $$H$$ verifying the previous equation is Hermitian.
There is, however, in this case, a method to obtain such an $$H$$. According to this answer, the eigenvalues of a $$QFT$$ matrix are $$\pm1$$ and $$\pm\mathrm{i}$$. Let $$Q_n$$ be the $$QFT$$ matrix on $$n$$ qubits. We know that $$Q_n$$ can be written as: $$Q_n = VDV^\dagger$$ with $$D$$ being a diagonal matrix whose entries belong to the set $$\{1;-1;\mathrm{i},-\mathrm{i}\}$$. We know that the matrix: $$L=V\log(D)V^\dagger$$ is a logarithm of $$Q_n$$. Computing the logarithm of a diagonal matrix is easily done by taking the $$\log$$ of its entries. As such, we will map $$1$$ to $$0$$, $$-1$$ to $$\mathrm{i}\pi$$ (or $$-\mathrm{i}\pi$$, it does not matter), $$\mathrm{i}$$ to $$\mathrm{i}\frac\pi2$$ and $$-\mathrm{i}$$ to $$-\mathrm{i}\frac\pi2$$. Finally, we compute $$H$$ with: $$H=-\mathrm{i}L.$$ Since we know that $$L$$ is a logarithm of $$Q_n$$, it is easy to see that $$H$$ satisfies the desired equation $$Q_n=\exp(\mathrm{i}H)$$. We thus simply have to prove that it is Hermitian. We have: $$H^\dagger=\mathrm{i}L^\dagger=\mathrm{i}P\log(D)^\dagger P^\dagger=\mathrm{i}P\overline{\log(D)}P^\dagger.$$ Thus, $$H=H^\dagger$$ holds if and only if $$\mathrm{i}\overline{\log(D)}=-\mathrm{i}\log(D)$$, which is true if and only if every element of $$\log(D)$$, which is a diagonal matrix, can be written as $$\alpha\mathrm{i},\alpha\in\mathbb{R}$$. Since this is the case, we have found one matrix $$H$$ which is Hermitian and which statisfies $$Q_n=\exp(\mathrm{i}H)$$.