Particle in one-dimensional harmonic oscillator

[Muthumanimaran]

Homework Statement

This is a question asked in a entrance examination[/B]
A charged particle is in the ground state of a one-dimensional harmonic oscillator
potential, generated by electrical means. If the power is suddenly switched off, so that the
potential disappears, then, according to quantum mechanics,

Homework Equations

(a) the particle will shoot out of the well and move out towards infinity in one of the two
possible directions
(b) the particle will stop oscillating and as time increases it may be found farther and
farther away from the centre of the well
(c) the particle will keep oscillating about the same mean position but with increasing
amplitude as time increases
(d) the particle will undergo a transition to one of the higher excited states of the
harmonic oscillator

The Attempt at a Solution

My reasoning is When the potential is switched off, the particle is free, so it may be shoot out of the well and move towards infinity in one of the two possible directions. I don't know my reasoning is right, the option I choose is correct or not, please help

 

[NFuller]

I don't know my reasoning is right, the option I choose is correct or not, please help

What is your reasoning here?I would take the ground state wave function and see how it evolves over time when the potential is zero.

 

[Muthumanimaran]

What is your reasoning here?I would take the ground state wave function and see how it evolves over time when the potential is zero.

The ground state wave function is gaussian, when the potential is switched off the particle is free and wavefunction becomes plane waves, so the plane waves must move towards either one of the directions to infinity right?

 

[Muthumanimaran]

The ground state wave function is gaussian, when the potential is switched off the particle is free and wavefunction becomes plane waves, so the plane waves must move towards either one of the directions to infinity right?

One more problem is that the potential is not gradually switched off, it is suddenly switched off.

 

[NFuller]

when the potential is switched off the particle is free and wavefunction becomes plane waves, so the plane waves must move towards either one of the directions to infinity right?

The wave function does not become a plane wave when the potential is switched off. It remains a Gaussian at that time. The Guassian can be written in terms of plane waves however since they form a complete basis for a free particle. You should study how these plane waves behave as a function of time.

 

[hilbert2]

My reasoning is When the potential is switched off, the particle is free, so it may be shoot out of the well and move towards infinity in one of the two possible directions. I don't know my reasoning is right, the option I choose is correct or not, please help

After the binding force is removed, the $|\Psi (x,t)|^2$ behaves like a solution of the heat conduction equation does for a Gaussian initial condition.

 

[Muthumanimaran]

The wave function does not become a plane wave when the potential is switched off. It remains a Gaussian at that time. The Guassian can be written in terms of plane waves however since they form a complete basis for a free particle. You should study how these plane waves behave as a function of time.

I got it. If I write gaussian in the free particle basis the coefficient of the basis (Fourier transform of gaussian) becomes another gaussian (but a broad one), the probability of finding the particle is mod squared of coefficient, so option B is correct not A. The particle will stop oscillating and as time increases it may be found farther and farther away from the centre of the well

 

[NFuller]

After the binding force is removed, the $|\Psi (x,t)|^2$ behaves like a solution of the heat conduction equation does for a Gaussian initial condition.

Right, this is the "clever" way to go about this.

I got it. If I write gaussian in the free particle basis the coefficient of the basis (Fourier transform of gaussian) becomes another gaussian (but a broad one), the probability of finding the particle is mod squared of coefficient, so option B is correct not A. The particle will stop oscillating and as time increases it may be found farther and farther away from the centre of the well

Yes, there are technically two Fourier transforms that need to be done here but the end result will be a Gaussian which broadens with time. Also, I'm not sure what textbook you are using, but this is almost exactly the same as problem 2.22 in Griffith's QM book.