Given the Hamiltonian \begin{equation} H = \frac{p^2}{2m} + V(x) \end{equation} The propagator for a pure harmonic potential of the form \begin{equation} V(x) = \frac{1}{2} m \omega^2 x^2 \end{equation} is given in the wikipedia article about propagators https://en.wikipedia.org/wiki/Propagator.

My Question is: What is the propagator if the potential also contains a linear part? For example: \begin{equation} V(x) = u_1 x + \frac{1}{2} m \omega^2 x^2. \end{equation}


1 Answer 1


You can rewrite that as $$V(x) = \frac{1}{2}m\omega^2x^2 + u_1x = \left(\sqrt\frac{m\omega^2}{2}x+\frac{u_1}{2}\sqrt\frac{2}{m\omega^2}\right)^2 - \frac{u_1}{2m\omega^2}.$$

Then make a coordinate transformation to $\tilde{x} = \sqrt\frac{m\omega^2}{2}x+\frac{u_1}{2}\sqrt\frac{2}{m\omega^2}$ and write the momentum operator in this coordinate frame. All that changes are some prefactors.

EDIT: Sorry, probably better if you write it like that $$V(x) = \frac{m\omega^2}{2}\left(x^2+\frac{2u_1}{m\omega^2}x\right) = \frac{m\omega^2}{2}\left(x+\frac{u_1}{m\omega^2}\right)^2-\frac{u_1^2}{2m\omega^2}$$ and $\tilde{x} = x+\frac{u_1}{m\omega^2}$, such that $$V(\tilde{x}) = \frac{m\omega^2}{2}\tilde{x}^2.$$

  • $\begingroup$ Do you mind adding the resulting propagator to your response? The one which can be applied to the wavefunction in the original coordinate frame? I tried your approach but for me in the limit $\omega \to 0$ I could not reproduce the propagator for the linear potential only. $\endgroup$
    – Luke
    Mar 29 at 12:28
  • $\begingroup$ It's not obvious to me that you should just get to that expression upon taking the limit at the very end. If you take this limit for the propagator of the QHO you also don't arrive at the free particle propagator. The transformation I proposed doesn't make sense when $\omega = 0$. $\endgroup$
    – Samuel
    Mar 29 at 12:39
  • $\begingroup$ but you should, right? $\endgroup$
    – Luke
    Mar 29 at 12:40
  • $\begingroup$ As I wrote in my comment above, doing so for the QHO propagator does not give the free particle propagator, so I don't think so. $\endgroup$
    – Samuel
    Mar 29 at 12:45
  • 1
    $\begingroup$ Oh I'm sorry, yes you are of course right. But still, in the coordinate transformation we divide by $\omega$, so I would say you have to take this limit before the transformation, i.e. you start with $H = p^2/2m + u_1x$. $\endgroup$
    – Samuel
    Mar 29 at 13:04

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