What Determines the Range of Bound States in a Spherical Finite Well?

Click For Summary
SUMMARY

The discussion focuses on the determination of bound states in a spherical finite potential well defined by the potential V = {-V_{0}: 0 < r < a} and V = {0: r ≥ a}. The key equation derived is k_{2} = k_{1} cot(k_{1} a), which is crucial for understanding the relationship between the parameters. The participants explore the implications of the condition R = π/2, concluding that no bound states exist under this scenario, particularly when considering the restriction -V_{0} < E < 0. The analysis reveals that at R = π/2, the only solutions occur at α = ±π/2, leading to k_{2} = 0.

PREREQUISITES
  • Understanding of quantum mechanics, specifically potential wells
  • Familiarity with spherical coordinates in quantum systems
  • Knowledge of the Schrödinger equation and bound state solutions
  • Basic grasp of trigonometric functions and their applications in physics
NEXT STEPS
  • Study the implications of the Schrödinger equation in spherical coordinates
  • Research the concept of bound states in quantum mechanics
  • Learn about the mathematical properties of cotangent functions in physical applications
  • Explore the behavior of potential wells and their impact on energy levels
USEFUL FOR

This discussion is beneficial for physics students, quantum mechanics researchers, and educators focusing on potential wells and bound state analysis in quantum systems.

Void123
Messages
138
Reaction score
0

Homework Statement



I was reviewing some homework problems and looking at the solutions. There is one problem with a tiny step I just cannot rationalize and I am hoping someone can point me in the right direction.

I have a spherical finite well:

[tex]V = {- V_{0}: 0 < r < a}[/tex],

[tex]= {0: r \geq a}[/tex]

[tex]- k_{2} = k_{1} cot (k_{1} a)[/tex] (1)

Refining the notation,

[tex]\alpha = a \sqrt{(2m(E + V_{0})}/hbar = k_{1} a[/tex]

[tex]R = a \sqrt{(2m(V_{0})}/hbar[/tex] and [tex]k_{2} = \sqrt{(2m(V_{0})}/hbar[/tex]

So (1) may be rewritten as [tex]\sqrt{R^{2} - \alpha^{2}} = - \alpha cot (\alpha)[/tex]

Homework Equations



From part 1.

The Attempt at a Solution



I don't understand how at [tex]R = \pi/2[/tex] there are no bound states.

Also, I am given this restriction: [tex]-V_{0} < E < 0[/tex]

How is this justified and how is the precise range of bound states determined?
 
Physics news on Phys.org
When [itex]R=\pi/2[/itex], the only solutions are at [itex]\alpha=\pm\pi/2[/itex]. In these cases, you get k2=0.
 

Similar threads

Replies
14
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
5K
Replies
7
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K