Given the Hamiltonian $$\hat{H} = \frac{1}{2} \hat{p}^2 + \hat{p}\hat{q}^4.\tag{1}$$ I would like to know how do I find the Heisenberg equation for the operators $\hat{p}$ and $\hat{q}$.

I know that the Heisenberg equation of motion is given by $$\frac{d}{dt}A(t) = \frac{i}{\hbar}\left[H,A(t)\right] + \left( \frac{\partial A}{\partial t}\right)_H,\tag{2} $$ where $A$ is an observable and $H$ is the Hamitlonian. So for $\hat{p}(t)$,

\begin{align*} \frac{d}{dt} \hat{p}(t) &= \frac{i}{\hbar}\left[ H , \hat{p}(t)\right] + \left( \frac{\partial \hat{p}(t)}{\partial t}\right)_H \\ &=\frac{i}{\hbar}\left[\left( \frac{1}{2}\hat{p}^2 + \hat{p}\hat{q}^4 \right)\hat{p} - \hat{p}\left(\frac{1}{2}\hat{p}^2 + \hat{p}\hat{q}^4 \right)\right] + \frac{\partial \hat{p} }{\partial t}\tag{3} \end{align*} but I'm unsure of how to simplify this or whether there is an easier way to do this without expanding the commutator. What is the easiest way to determine the Heisenberg equations of motion?

  • 2
    $\begingroup$ Have you tried expanding the thing inside square brackets? $\endgroup$
    – John Donne
    Dec 4, 2017 at 20:36
  • 3
    $\begingroup$ You are misinterpreting the partial derivative symbols: The partial derivatives vanish as the full derivative is purely convective. On your last line, evaluate the square bracket, and trash the partial derivative. $\endgroup$ Dec 4, 2017 at 20:44

2 Answers 2


Another issue is that OP's Hamiltonian (1) is not Hermitian. This renders the Heisenberg EOM (2) inconsistent.


Don't bother to do any calculation. The problem is ill-posed because your H cannot be a Hamiltonian. There's a direct proof stemming from the fact that, because of the Groenewold-van Hove no-go theorem, there's no clear quantization (rendering an essentially self-adjoint operator) of a polynomial of degree higher than 2 on the phase space of a simple system. See the related problem and answer here: Operator Ordering Ambiguities.

Another option would be to check that the fifth order term is essentially self-adjoint in $L^2 (\mathbb R)$. (thus, completely overlooking that there's no way to quantize a classical theory to get the quantum H posted in the OP). It isn't. It is not even formally symmetric, due to the fact that q and p do not commute on the commutator domain.


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