For the time-dependent Hamiltonian

$$H = \frac{\hat{P}^2}{2m} + \frac{1}{2} m\omega^2\hat{X}^2 + m\omega^2vt\hat{X} +v\hat{P}$$

I would like to calculate the ground state, more precise, the stationary point up to a moving origin. The Hamiltonian looks like a quantum mechanical oscillator but has 2 additional terms which makes it impossible to solve it right away. Since $\hat{H}$ is an even function, I was hoping for some symmetry properties that makes it easy to calculate it but so far I couldn't find one.

Does somebody has an idea of how to solve the Schrödinger equation for that Hamiltonian or does somebody know a property that makes it easy to determine the ground state?

  • 3
    $\begingroup$ If $t$ here is time, then you are dealing with a non-stationary problem, so the question about teh ground state does not make much sense. Apart from that, one usually completes the square and introduces new momentum and position operators. $\endgroup$
    – Roger V.
    Nov 10, 2020 at 13:30

1 Answer 1


For a time-dependent hamiltonian, you can define the ground state as the instantaneous eigenstate (which will therefore depend on time), but this is only useful in certain specific contexts.

For your specific hamiltonian, you can try completing the square to get $$ \hat H = \frac1{2m}(\hat P+mv)^2 + \frac12 m\omega^2(\hat X+vt)^2 - \frac12mv^2(1+\omega^2t^2) , $$ and this will generally help.

But the specific steps to take then depend on exactly what type of solution you want. If you want fully stationary solutions, it's basically guaranteed that you won't find them. If you want solutions which are e.g. stationary up to a moving orign, then you can look for those -- but until you specify what kind of solution is useful in your context, there isn't really much more to say about the problem.

  • $\begingroup$ Isn't that just the Galilean transformation of the harmonic oscillator? $\endgroup$
    – gamma
    Nov 10, 2020 at 16:57
  • $\begingroup$ @gamma Indeed it is. $\endgroup$ Nov 10, 2020 at 18:24

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.