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

Question

For the time-dependent Hamiltonian
$$H = frac{hat{P}^2}{2m} + frac{1}{2} momega^2hat{X}^2 + momega^2vthat{X} +vhat{P}$$
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?

Emilio Pisanty · Accepted Answer

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
$$
hat H
=
frac1{2m}(hat P+mv)^2 + frac12 momega^2(hat X+vt)^2 - frac12mv^2(1+omega^2t^2) ,
$$
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.

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

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