# How to determine the ground state of quantum harmonic oscillator like Hamiltonian?

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?

• 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. Nov 10, 2020 at 13:30

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.