Linear Potential Energy Eigenfunction for Unbound State

  • Thread starter Thread starter Domnu
  • Start date Start date
  • Tags Tags
    Linear Potential
Click For Summary
SUMMARY

The discussion centers on the energy spectrum of a particle in one dimension subjected to a linear potential defined by V = λx. It concludes that the energy spectrum is continuous due to the unbound state of the particle. The proposed energy eigenfunction, ψ(x) = e^{-f(λ)x} sin(x - e^{-h(λ)}), is deemed incorrect. Participants recommend using the WKB approximation and solving for a constant potential in different regions to derive the correct eigenfunction.

PREREQUISITES
  • Understanding of quantum mechanics, specifically potential energy functions.
  • Familiarity with the concept of unbound states in quantum systems.
  • Knowledge of the WKB approximation in quantum mechanics.
  • Ability to solve differential equations related to quantum eigenfunctions.
NEXT STEPS
  • Study the WKB approximation in detail to apply it to quantum systems.
  • Learn about energy eigenfunctions for constant potential scenarios.
  • Explore the implications of continuous versus discrete energy spectra in quantum mechanics.
  • Investigate the mathematical techniques for solving differential equations in quantum mechanics.
USEFUL FOR

Students and professionals in quantum mechanics, physicists analyzing potential energy functions, and researchers exploring unbound states and their implications in quantum systems.

Domnu
Messages
176
Reaction score
0
Problem

A particle in one dimension is subjected to a constant force derivable from

[tex]V = \lambda x[/tex]​

Is the energy spectrum continuous or discrete? Write down an approximate expression for the energy eigenfunction specified by [tex]E[/tex].

Solution attempt
We have an unbound state, so we would have a continuous energy spectrum. Well, I was thinking of something along the lines of [tex]\psi(x)=e^{-f(\lambda) x} \sin (x-e^{-h(\lambda)})}[/tex] for the region [tex]x < E/\lambda[/tex] (of course, for [tex]x > E/\lambda[/tex], we need an exponentially decreasing function); I chose this function based on the following:

1. The function needs to increase exponentially as we go farther left, because the energies are greater than the linear potential by a greater degree.
2. The zeroes need to be bunched together closer as [tex]x\rightarrow -\infty[/tex], because the particle is more energetic here.

Above, [tex]f(\lambda), g(\lambda)[/tex] are positive functions for [tex]\lambda > 0[/tex].

Is the above correct?
 
Physics news on Phys.org
The energy spectrum is continuous, but your guess as to the form of the solution is wrong.

Try writing down the solution for a constant potential (in each region, V<E and V>E), and then plugging in V(x) in place of the constant V.

Also, if you've studied the WKB approximation, this is a good place to apply it. (If not, ignore this comment.)

(I'm going offline for a week, so won't reply further, good luck!)
 
Oh yeah, heh, I forgot about the WKB. Thanks for the help!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
11
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
4
Views
5K
Replies
5
Views
4K