In a standard physics course we usually learn that quantization of a system is ambiguous if momentum and position happen to be multiplied in the classical Hamiltonian (i.e. the classical Hamiltonian contains a term like $xp$). This ambiguity of quantization seemed more like of an academic interest to me, until I tried to quantize this simple classical system.

Suppose we want to model the oscillations of an electron bubble in a system of solvated electrons. If an alkali metal atom, for instance sodium, is thrown into liquid ammonia (NH$^3$), the ammonia dissolves the metal atom it into Na$^+$ ion and $e^-$ ion (see a video of dissolution here). The $e^-$ ion is formed by ammonia turning with more positively charged hydrogen towards electron, forming a cavity of the radius $R\approx 0.32$ nm around the electron. The electron occupies the lowest energy level inside the cavity. I would like to model radial quantum oscillation of the bubble within adiabatic approximation, where the energy of the electron is added to the effective potential energy of the bubble (the static bubble model was developed by Cohen and Thompson in 1968; for a free access review paper see here).

From a classical picture, everything is more or less clear. The potential energy of the bubble is assumed to consist of the polarization energy, electron energy and energy of surface tension, and it is a function of the radius of the bubble. The kinetic energy of the bubble is assumed to consist only of the kinetic energy of the liquid around the bubble. The liquid is assumed to be continuous and incompressible, and the bubble is assumed perfectly spherical. Classical Lagrangian looks like this: $$ L(R,\dot{R})=K-\Pi=4\pi \rho R\frac{\dot{R}^2}{2}-\Pi(R).\tag{1} $$ The corresponding classical Hamiltonian looks like this: $$ H=\frac{1}{4\pi\rho R}\frac{p^2}{2}+\Pi(R).\tag{2} $$ Now, I know I cannot quantize the system naively by substituting $p\rightarrow \hat{p}$ and $R\rightarrow \hat{R}$ into the Hamiltonian, because the kinetic energy contains the product $R$ and $p$, which do not commute. Is there any good way of quantizing such system? What do people usually do, if they find that kinetic energy of the classical Hamiltonian depends on the position?


1 Answer 1


The simplest idea is to reorder the Hamiltonian operator in a Hermitian fashion, e.g., $$ \hat{H}~=~\hat{p}\frac{1}{8\pi\rho \hat{R}}\hat{p}+\Pi(\hat{R}).$$ See also e.g. this related Phys.SE post.

  • $\begingroup$ Thank you for the reply. I can see the Hamiltonian is Hermitian, but is there any particular reason to prefer this particular form? From the reference it seems like the proper form should be rather $\hat{R}^{1/4}\hat{p}\frac{1}{R^{3/2}}\hat{p}\hat{R}^{1/4}$ for kinetic energy, if we follow Eq. (7) of your answer. But I am not sure how the logic in that post would be applied here $\endgroup$
    – Pavlo. B.
    May 6, 2021 at 4:47
  • $\begingroup$ Right, quantization is not a unique procedure. $\endgroup$
    – Qmechanic
    May 6, 2021 at 4:51
  • $\begingroup$ Yes, I understand that quantization is not unique in general case. On the other hand, the system I consider is emergent from basic atoms, and it has a measurable oscillatory spectrum. So, one would think there should be "the right Hamiltonian", because all other Hamiltonians would produce a wrong spectrum. From a practical perspective it is really desirable to have a reason to pick one hamiltonian over others, because otherwise I cannot tell if the physical model wrong, or all effects are due to the particular choice of the ordering $\endgroup$
    – Pavlo. B.
    May 6, 2021 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.