I am trying to calculate the Equations of Motion in the Heisenberg picture for $\hat{x}$ and $\hat{p}$ in a perturbed Hamiltonian, $$ \tag{1} \hat{H} = \hat{H}_0 + \hat{H}' $$

Assume the Heisenberg Equations of Motion are given by,

$$ \tag{2} \frac{\partial \hat{A}(t)}{\partial t} = -i\left[ \hat{A}(t),\hat{H} \right] $$ with $\hbar=1$ to simplify.


Can I use the operators $\hat{x}(t)$, $\hat{p}(t)$ for the unperturbed hamiltonian, $\hat{H}_0$, in Equation $(2)$ to get the perturbed equations of motion?

  • Is this the correct approach?
  • Do I need a correction in $\hat{x}(t)$, $\hat{p}(t)$ for the perturbation?


In my case, $\hat{H}_0$ is an oscillator such that,

$$ \hat{H}_0 = \frac{1}{2m}\hat{p}^2 + \frac{k}{2}\hat{x}^2 $$ and the perturbation, $\hat{H}'$, is quartic in $\hat{x}$ like $\lambda\frac{1}{4}\hat{x}^4$ with a coupling constant of $\lambda$

  • $\begingroup$ ... equations of motion for what, exactly? $\endgroup$ Nov 15, 2015 at 16:44
  • $\begingroup$ any Heisenberg operator $\hat{A}(t)$ but specifically $\hat{x}(t)$ and $\hat{p}(t)$. I've just added the time dependences of the operators as I just realised it might not have been immediately clear $\endgroup$ Nov 15, 2015 at 16:45

2 Answers 2


It seems like your answer sidestepped the whole question.

When you do the same for $\hat p$ you'll find that its derivative depends on $\hat x$, and on $\lambda$.

But these coupled equations can then be solved as a second order equation for the terms individually, which should be what you are looking for.

  • $\begingroup$ yeah agreed, thanks for your time :). The question was whether I needed special treatment to $\hat{x}$ because of the perturbation. You're right though, the solution is incomplete unless you solve for $\hat{p}$ and relate the two. I've edited the end of my answer slightly as a result. $\endgroup$ Nov 16, 2015 at 7:38

Basically, you do need to treat it as a perturbation and no correction is necessary...

Calculating the commutator, \begin{equation} \left[\hat{x}, \hat{H}\right] = \left[ \hat{x}, \frac{\hat{p}^2}{2m} - \frac{k}{2}\hat{x}^2 + \frac{\lambda}{4}\hat{x}^4 \right] \end{equation} but as $\left[\hat{x},\hat{x}\right]=0$, \begin{equation} \left[\hat{x}, \hat{H}\right] = \left[ \hat{x}, \frac{\hat{p}^2}{2m} \right] = \frac{1}{2m}\left[\hat{x},\hat{p}\hat{p}\right] = \frac{1}{2m} \left[\hat{x},\hat{p}\right]\hat{p} + \frac{1}{2m} \hat{p}\left[\hat{x},\hat{p}\right] \end{equation} rmembering I'm setting $\hbar=1$ so we get, \begin{equation} \tag{1} \left[\hat{x}, \hat{H}\right] = \frac{i}{2m} \hat{p} + \frac{i}{2m} \hat{p} = \frac{i}{m} \hat{p} \end{equation} Using Equation $(1)$ with the Heisenberg equations of motion, \begin{equation} \frac{\partial}{\partial t} \hat{x} = -i\left[\hat{x}, \hat{H}\right] = \frac{\hat{p}}{m} \end{equation}

The same can be done for $\hat{p}$. The two can then be related to get two pairs of $1^\text{st}$ and $2^\text{nd}$ order coupled ODEs


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