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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 ?

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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.

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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!

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    $\begingroup$ 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. $\endgroup$ – Ruslan May 25 at 12:00
  • $\begingroup$ 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 ? $\endgroup$ – Qbuoy May 25 at 12:03
  • $\begingroup$ 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? $\endgroup$ – Philip May 25 at 13:02
  • $\begingroup$ @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 . $\endgroup$ – Qbuoy May 25 at 13:44

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