I'm trying to understand how quantum systems behave when they are perturbed, and I'm using the quantum harmonic oscillator as a model. I start by implementing a symmetric gaussian shaped bump in the middle of the harmonic oscillator, and then i propagate the wave functions in time.
the hamiltonian takes the form ($ \omega$ stands for hamiltonian for harmonic oscillator):
$$ \hat{H} = \hat{H}_{\omega} + V(x,t) = \hat{H}_{\omega} + 10e^{-5x^2} \cdot \sin(\Omega t ), \qquad \qquad 0 \leq t \leq \frac{ \pi }{\Omega} $$
I set the angular frequency of the harmonic oscillator to 1, so $\omega = 1$.
Now when I set the angular frequency of the sinus function to the same, i.e $\omega = \Omega =1$, I observe that the ground state ( when propagated between 0 and $\pi$) goes back to its initial position at $t=\pi$.
I thought that it made sense since the sinus-function is zero at the boundaries (0 and $\pi$), but here is the interesting thing! I then set the angular frequency of the sinus-function to $\Omega = 2\omega =2$ and instead propagate the ground state between 0 and $\pi /2$, which should also give us zero at the boundaries!!
But what I observe now is that the ground state does not go back to its initial position even after the perturbation is over, instead, its in superposition of the first two states, how is that possible? And what is the physical explanation of this?
