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

Given a Quantum Hamiltonian: $$\hat{H}=ax^2+bp^2$$ It does not commute with either $$x$$ or $$p$$. Suppose we have a Hamiltonian :$$H = k \hat{p}\hat{x}$$ why do we need it to be: $$H = k (\hat{p}\hat{x} - \hat{x}\hat{p})$$ And why can't we leave it in the former form? What can be the possible reason we don't have Hamiltonians of the format :$$H = k \hat{p}\hat{x}$$ and what can be the eigensolutions?

Edit1: Operators commuting with $$H$$ does not evolve with time by the Heisenberg Equation of motion :$$i \hbar\frac{dA_H}{dt}=[A_H,H]$$ so I wished to construct Hermitian Hamiltonians such as :$$H = k (\hat{p}\hat{x} - \hat{x}\hat{p})$$ which commute with one of the conjugating variables.

To be precise, does symmetrizing like this surely result in commuting Hamiltonians ? What if we have $$\hat{H}(\sigma_x$$,$$\sigma_y$$,$$\sigma_z)$$ instead of {$$x,p$$} ? What would be the commutable symmetric Hamiltonian and how to find its eigensolutions ?

As it has been already pointed by others:

• The Hamiltonian $$\hat{H} = k\hat{x}\hat{p}$$ 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 $$i\hbar$$.)

Now my remark:
The symmetrized version of this Hamiltonian is far from exotic - it is the Hamiltonian of a particle in magnetic field: $$\hat{H}=\frac{1}{2m}\left(\hat{\mathbf{p}} -\frac{e}{c}\mathbf{A}(\mathbf{r})\right)^2 = \frac{1}{2m}\left[\hat{\mathbf{p}}^2 -\frac{2e}{c}\left(\hat{\mathbf{p}}\mathbf{A}(\mathbf{r}) + \mathbf{A}(\mathbf{r})\hat{\mathbf{p}}\right) + \frac{e^2}{c^2}\mathbf{A}^2(\mathbf{r})\right]$$ 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.

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!

• There's one more problem with the anti-Hermitian Hamiltonian after you converted it to Hermitian: it reduces to $2kpx$ in the classical limit, unlike the supposed $kpx$ from the original proposed non-Hermitian quantum Hamiltonian. – Ruslan May 25 at 12:00
• I understand the Hermitian condition is the primary problem, but my question was does symmetrization necessarily mean it will commute with the Hamiltonian ? Why don't these kind of valid hermitian symmetric Hamiltonians $H=k(xp+px)$ don't exist and what can be possible solutions to those Hamiltonians ? – Qbuoy May 25 at 12:03
• I agree that I haven't completely answered your question, but that was partly because I didn't quite understand it. It is true that $H = k (xp - px)$ does commute with $x$, but for example, $H = k (xp + px)$ doesn't. I can't think of any specific reason why we'd want $H$ to commute with $x$ or $p$, indeed all the standard Hamiltonians we learn in an introductory QM course don't. Is there a specific reference that you could provide in your question to make it a little clearer? – Philip May 25 at 13:02
• @Philip I wished to construct Hermitian hamiltonians such as $H=k(px−xp)$ which commute with one of the conjugating variables. I edited the question further to be more precise and I wanted to generally know Other than just to make it hermitian , will symmetrization/anti-symmetrization help in commutation to be zero surely ? I think I'm generally asking a recipe to make hamiltonians conjugable with variables ,and if symmetrization helps or what helps . – Qbuoy May 25 at 13:44