What Is the Probability of Particle Ionization in a Shifted Finite Square Well?

In summary: So Mathematica is not going to help.In summary, the conversation discusses a particle of mass m in the ground state of a potential well with a width of 2a and depth V0. At t = 0, the bottom of the potential well is shifted down to Vo' from Vo, resulting in a deeper well that can support two bound states with energies -e0 and -e1 > -e0. The question is then asked about the probability of the particle getting ionized, or leaving the potential well, by occupying levels with positive energy. To find this probability, the wave functions for both bound states must be integrated over the entire real line. The amplitude of the odd state is expected to be zero, leaving only
  • #1
photomagnetic
13
0

Homework Statement


Consider a particle of mass m in the ground state of a potential well of width 2 a and depth.
the particle was in the ground state of the potential well with V0 < Vcritical, which means the well is a narrow one.
At t = 0 the bottom of the potential well is shifted down to Vo' from Vo.
The resulting well is deep enough to support two bound states with energies
-e0 and -e1 > -e0.
What is the probability that the particle will get ionized – that is, it will leave the potential well
by occupying levels with positive energy?

Homework Equations

The Attempt at a Solution


I've found the wave function for e0
and the 2nd wave function for e1

Now I have to integrate them (-a to a)
But this is getting outta control. Is there a trick? or I am doing something wrong because there is no way to integrate so many terms. Screw mathematica btw. I need to learn the trick. Thanks!
 
Physics news on Phys.org
  • #2
You should not be integrating from -a to a. This would simply give you the probability of finding the particle inside the well if it is in a bound state. What you want to figure out is the amplitude (and hence probability) of the particle ending up in the bound states.
 
  • #3
I'm pretty sure it's -a to a
"The resulting well is deep enough to support two bound states with energies"

[integral ( first wavefunction) (2nd wave function) dx ]2
 
  • #4
The wave functions will have overlap also in the regions with |x| > a as long as the well is not infinite. Neglecting this contribution is going to introduce errors. This is why you need to integrate over the entire real line.
 
  • #5
I suspect Photo is also chewing on the wrong integral. E0 changes to E0' when V changes...

Particle is in state ##|\psi_{0,V_0}>##, ground state for the V0 well.

V changes, bound states are now the wave functions you claim to have found: One even ##|\psi_{0,V_0'}>##, one odd ##|\psi_{1,V_0'}>##, right?

For a given state ##\phi##, the amplitude of ##\psi_0## is ##<\phi\;|\;\psi_0>##. As Oro says: from ##-\infty## to ##\infty##.

The other state is odd, so I expect that amplitude to be zero.

So there is one (admittedly hideous) integral left to do. Answer is then 1 - (that integral squared)

I would expect that finding the expression, not its numerical value, is the object of the exercise.
 

FAQ: What Is the Probability of Particle Ionization in a Shifted Finite Square Well?

1. What is a finite square well?

A finite square well is a potential energy barrier that is often used in physics to model the behavior of particles in a confined space. It consists of a finite region where the potential energy is constant, surrounded by regions where the potential energy is infinite.

2. How does a finite square well relate to excited states?

In a finite square well, particles can exist in different energy states depending on the depth and width of the well. The excited states refer to the energy levels above the ground state, where the particles have higher energy due to being confined in the well.

3. What determines the number of excited states in a finite square well?

The number of excited states in a finite square well is determined by the size and depth of the well. As the well becomes deeper and narrower, the number of excited states increases.

4. How do excited states affect the behavior of particles in a finite square well?

Excited states play a crucial role in determining the behavior of particles in a finite square well. They can affect the probability of a particle being found in a particular region and can also determine the energy levels at which the particle can exist.

5. Can a particle in a finite square well transition between excited states?

Yes, a particle in a finite square well can transition between excited states by either absorbing or emitting energy. This transition can occur through various processes such as absorption or emission of photons or collisions with other particles.

Back
Top