# Understanding the dynamics of a perturbed quantum harmonic oscillator

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• Rayan
Rayan
TL;DR Summary
Understanding the dynamics of a quantum harmonic oscillator perturbed with a symmetric gaussian potential
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?

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Classically, if you take a pendulum at rest and poke it, does it stay at rest?

Rayan said:
I'm trying to understand how quantum systems behave when they are perturbed, and I'm using the quantum harmonic oscillator as a model.
It seems like you actually might be thinking of it as a classical harmonic oscillator. They're not the same.

Rayan said:
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.
Ok.

Rayan said:
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##
This doesn't make sense if it's a quantum oscillator. The state is a function of position (the wave function), not a position.

Perhaps you mean that the wave function, if it starts in the ground state of the free oscillator (i.e., of ##H_\omega##) at ##t = 0##, becomes the same function again at ##t = \pi## if ##\omega = 1##. But if that's what you mean, you should say so, not least to avoid confusing yourself.

Rayan said:
I thought that it made sense since the sinus-function is zero at the boundaries (0 and ##\pi##)
But the sine function is part of the Hamiltonian, not part of the wave function. So the sine function being zero at some instant means the Hamiltonian is equal to ##H_\omega## at that instant. It does not mean that the wave function must be the ground state wave function of ##H_\omega## at that instant. So your logic here is faulty.

Rayan said:
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!!
Again, your logic here is faulty. See above.

Rayan said:
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?
See above.

Rayan said:
what is the physical explanation of this?
The basic thing you seem to be missing is that, in the presence of the extra driving potential in the Hamiltonian, the state you are calling the "ground state", and more generally the states that are eigenstates of the free harmonic oscillator Hamiltonian, ##H_\omega##, are not eigenstates of the full Hamiltonian ##H##. So as soon as you turn on the driving potential ##V##, the states that were unchanged by time evolution before, now are changed by time evolution. They will start mixing.

The simplest case of this is Rabi oscillations, where there are just two states that are mixed. If the amplitude of your driving potential ##V## is small, that's basically what you are analyzing here--the two states are the ground state and the first excited state of the free harmonic oscillator.

Rayan said:
I start by implementing a symmetric gaussian shaped bump in the middle of the harmonic oscillator
Your "bump" is being added to the Hamiltonian, not the wave function.

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