Particle in one-dimensional harmonic oscillator

In summary, the particle in the ground state of a one-dimensional harmonic oscillator potential, when the potential is suddenly switched off, will stop oscillating and its wave function will behave like a solution of the heat conduction equation for a Gaussian initial condition, spreading out over time. Therefore, option (b) is the correct choice.
  • #1
Muthumanimaran
81
2

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
 
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  • #2
Muthumanimaran said:
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.
 
  • #3
NFuller said:
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?
 
  • #4
Muthumanimaran said:
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.
 
  • #5
Muthumanimaran said:
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.
 
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  • #6
Muthumanimaran said:
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.
 
  • #7
NFuller said:
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
 
  • #8
hilbert2 said:
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.
Muthumanimaran said:
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.
 

Related to Particle in one-dimensional harmonic oscillator

1. What is a one-dimensional harmonic oscillator?

A one-dimensional harmonic oscillator is a physical system that exhibits simple harmonic motion, where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction of the displacement. It is often described as a particle moving back and forth along a straight line.

2. How is a particle in a one-dimensional harmonic oscillator described mathematically?

The motion of a particle in a one-dimensional harmonic oscillator can be described using Hooke's Law, which states that the force (F) is equal to the negative of the spring constant (k) multiplied by the displacement (x): F = -kx. This can be further simplified using Newton's Second Law of Motion (F = ma) to give the equation of motion: m(d^2x/dt^2) = -kx, where m is the mass of the particle and t is time.

3. What is the potential energy function for a one-dimensional harmonic oscillator?

The potential energy function for a one-dimensional harmonic oscillator is given by U(x) = 1/2kx^2, where k is the spring constant and x is the displacement from equilibrium. This potential energy function is a parabola, with the minimum point at x = 0, representing the equilibrium position of the particle.

4. What is the significance of energy levels in a one-dimensional harmonic oscillator?

In a one-dimensional harmonic oscillator, the allowed energy levels are quantized, meaning that the energy of the particle can only take on certain discrete values. This is because the energy of the particle is directly related to its frequency of oscillation, and the frequency must be a multiple of the fundamental frequency to maintain simple harmonic motion. The energy levels are evenly spaced, with the lowest energy level (ground state) corresponding to the particle at rest at the equilibrium position.

5. How do the properties of a one-dimensional harmonic oscillator change with increasing energy levels?

As the energy level of a one-dimensional harmonic oscillator increases, the amplitude of the particle's oscillation also increases. This means that the particle will move further away from the equilibrium position and its motion will become more "violent". Additionally, the frequency of oscillation will also increase, meaning that the particle will complete more cycles of motion in the same amount of time. However, the period of oscillation (time taken for one complete cycle) will remain constant regardless of the energy level.

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