Is symmetrization $xp-px$ required for commutation $[H,x]=0$?

I'm not sure this is the right answer, but one problem that I can immediately see is that if you had a Hamiltonian $$hat{H} = k hat{p}hat{x},$$ then it wouldn't be Hermitian, since you could show that $$hat{H}^dagger = k^* hat{x}hat{p},$$ which, since $hat{x}$ and $hat{p}$ don't commute, is not equal to $hat{H}$. Thus, $hat{H} neq hat{H}^dagger$ and so its eigenvalues needn't necessarily be real, which, to my understanding, goes against the postulates of Quantum Mechanics.

Your "symmetric" guy has a slightly weaker problem, since in that case $hat{H} = - hat{H}^dagger$, but this could be fixed by using a purely imaginary $k$, for example. Of course, a nicer "symmetric" Hamiltonian would be $hat{H} = k left(hat{x}hat{p} {color{red}+} hat{p}hat{x}right)$.

Again, I'm not sure if this is the only reason, but it certainly seems like an important one!

As it has been already pointed by others:

• The Hamiltonian $hat{H} = khat{x}hat{p}$ is not Hermitian, i.e. it does not correspond to anything measurable.
• One can obtain Hermitian Hamiltonians by using either symmetrizing or anti-symmetrizing the operator product $hat{x}hat{p}$: $$hat{H}_+ = frac{k}{2}(hat{x}hat{p} + hat{p}hat{x}), hat{H}_- = frac{-ik}{2}(hat{x}hat{p} - hat{p}hat{x})$$ (Of course, in this particular case the second option is trivial, since the commutator is $ihbar$.)

Now my remark:
The symmetrized version of this Hamiltonian is far from exotic - it is the Hamiltonian of a particle in magnetic field: $$hat{H}=frac{1}{2m}left(hat{mathbf{p}} -frac{e}{c}mathbf{A}(mathbf{r})right)^2 = frac{1}{2m}left[hat{mathbf{p}}^2 -frac{e}{c}left(hat{mathbf{p}}mathbf{A}(mathbf{r}) + mathbf{A}(mathbf{r})hat{mathbf{p}}right) + frac{e^2}{c^2}mathbf{A}^2(mathbf{r})right]$$ While in QM books the discussion is usually limited to Landau gauge, where the vector potential and the momentum commute, in general this is not the case.