Discretization of Hamiltonian with first derivative In a particular 1D system, the Hamiltonian can be writen as
$$H=\mathrm{i}\left(f(r)\frac{\partial}{\partial r}+\frac{1}{2}f'(r)\right)\; ,$$
wher $\mathrm{i}$ is the imaginary unit, and $f(r)$ is a real function. How could I write this Hamiltonian's matrix using the finite difference method? I thought about replacing the derivative in the first term with
$$\frac{\partial \psi}{\partial r} \rightarrow \frac{\psi_{j+1}-\psi_{j-1}}{2h}\; ,$$
but $f(r)$ depends on the position. Replacing
$$\mathrm{i}f(r)\frac{\partial \psi}{\partial r} \rightarrow \mathrm{i}\frac{f(r_{j+1})\psi_{j+1}-f(r_{j-1})\psi_{j-1}}{2h}$$
wouldn't work either, because the Hamiltonian matrix must be Hermitian. In this case $H_{j,j+1}=\mathrm{i}\frac{f(r_{j+1})}{2h}$ and $H_{j+1,j}=-\mathrm{i}\frac{f(r_{j})}{2h}\neq H_{j,j+1}^*$.
Replacing the first term with
$$\mathrm{i}f(r)\frac{\partial \psi}{\partial r} \rightarrow \mathrm{i}\frac{f((r_{j+1}+r_{j})/2)\psi_{j+1}-f((r_{j}+r_{j-1})/2)\psi_{j-1}}{2h}$$
would give a Hermitian matrix, but is this the correct way to discretize this Hamiltonian?
Edit 1:
I realised that the term
$$\mathrm{i}\frac{1}{2}f'(r)\psi\rightarrow\mathrm{i}\frac{1}{2}f'(r_j)\psi_j$$
would break Hermicity.
Edit 2:
Replacing the first term in the Hamiltonian with the average of the forward and backward derivatives (instead of the central formula):
$$\mathrm{i}f(r)\frac{\partial \psi}{\partial r} \rightarrow \mathrm{i}\frac{f((r_{j+1}+r_{j})/2)\psi_{j+1}-f((r_{j+1}+r_{j})/2)\psi_{j}}{2h}+\mathrm{i}\frac{f((r_{j}+r_{j-1})/2)\psi_{j}-f((r_{j}+r_{j-1})/2)\psi_{j-1}}{2h}=\mathrm{i}\frac{f((r_{j+1}+r_{j})/2)\psi_{j+1}-f((r_{j}+r_{j-1})/2)\psi_{j-1}}{2h}+\mathrm{i}\frac{-f(r_{j}+h/2)+f(r_{j}-h/2)}{2h}$$
Applying finite differences to the second term in the above equation:
$$\mathrm{i}\frac{-f(r_{j}+h/2)+f(r_{j}-h/2)}{2h}=-\mathrm{i}\frac{1}{2}f'(r_{j})$$
which would cancel the $\mathrm{i}\frac{1}{2}f'(r)\psi$ term (see edit 1).
Therefore, the Hamiltonian might be replaced with
$$H\psi=\mathrm{i}\left(f(r)\frac{\partial}{\partial r}+\frac{1}{2}f'(r)\right)\psi \rightarrow \mathrm{i}\frac{f((r_{j+1}+r_{j})/2)\psi_{j+1}-f((r_{j}+r_{j-1})/2)\psi_{j-1}}{2h}$$
 A: Even without the $f(r)$ ordering problem there is no really good way to discretise a first derivative while preserving all its hermiticity properties. This is the notorious "fermion doubling problem" in lattice gauge theories.
By your  ordering problem I mean that your continuum hamiltonian, as written, is not Hermitian.  I suppose that you meant 
$$
H=\frac 12 (i f(r)\partial_r+ i\partial_r f(r)).
$$ 
which is hermitian with respect to the standard $L^2(\mathbb R)$ inner product.
The problem is that forward differences 
$$
(D_+\psi)_n\approx (\psi_{n+1}-\psi_n)/a
$$
or backward differences
$$
(D_-\psi)_n\approx (\psi_{n}-\psi_{n-1})/a
$$
do not lead to symmetric matrices, while center symmetric differences
$$
(D_0\psi)_n\approx (\psi_{n+1}-\psi_{n-1})/2a
$$
map odd sites to even sites and vice versa and so $D_0$ anticommutes with $(-1)^n$, a symmetry absent in the continuum and which leads to high-frequency artifacts. The Nielsen-Ninomiya theorem shows that there  is no satisfactory escape.  
