How to Find the Ground State Energy Using the Shooting Method?

Click For Summary

Homework Help Overview

The discussion revolves around finding the ground state energy of a particle using the shooting method applied to the time-independent Schrödinger equation. The potential is defined with specific values for different ranges of x, including infinite potential outside certain bounds.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to start using the shooting method and mentions boundary conditions for the wave function. Some participants discuss the correct boundary conditions and the implications of the potential's symmetry. Others suggest numerical methods and share their experiences with solving differential equations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of boundary conditions and the potential's characteristics. Some guidance on numerical methods has been provided, but there is no explicit consensus on the boundary conditions or the feasibility of an analytical solution.

Contextual Notes

There is a mention of the need for normalizability of the wave function, and some participants question the original poster's boundary conditions. The original poster is also constrained by a lack of familiarity with the shooting method.

Cinimod
Messages
34
Reaction score
0

Homework Statement


The wave function of a particle satisfies the time-independent Schrödinger equation.
If the potential is symmetric and has the form
[tex]V(x) = \inf[/tex] |x|>1.0
[tex]V(x) = \frac{\hbar^2V_0}{2m}[/tex] |x|<0.2
[tex]V(x) = 0[/tex] Elsewhere
Using the shooting method, I need to find the ground state energy and the normalised group state wave function if [itex]V_0 = 50[/itex]. What is the energy of the first excited state?

Homework Equations


The Attempt at a Solution


I have no idea where to even start. I only have the very basic of ideas of how the shooting method works. I am suppose to program this, but the coding shouldn't be a problem. I just don't even know where to begin. I do know from that the boundary conditions are [itex]\phi(-1) = \phi(1) = 0[/itex]. Other than that I'm clueless.

edit: [itex]\phi[/itex] represents the wavefunction, I just don't know how to write it.
 
Physics news on Phys.org
Hi I'm a new member. Where can I find out numerical method for solving Schrödinger equation? Thanks
 
phi(1)=0 is correct. phi(-1)=0 is not. phi(-infinity) needs to be zero in the limit to make the wave function normalizable. The time independent Schrödinger equation is an second order ode for phi(x). So your initial values at x=1 are phi(1)=0 and phi'(x)=c. You pick c and integrate backwards towards -infinity. Find values of c so that phi blows up to +infinity and then to -infinity. Keep splitting the difference until you find one that's relatively stable. You won't find an exact one. You can only guess an estimate. The methods of evolving a second order ode are pretty standard. I usually lean towards a simple predictor-corrector method. Google it.
 
NguyenNguyen said:
Hi I'm a new member. Where can I find out numerical method for solving Schrödinger equation? Thanks

It depends on what the potential are etc. Shrodinger is a differential eq, so search for numerical methods for solving those.

I have never solved the SE nummerical with anything else than matlab's diff eq solver. Which is a Runge-Kutta method.
 
Dick said:
phi(-1)=0 is not

Are you sure? it is a symmetric potential with infinite boundaries
 
mda said:
Are you sure? it is a symmetric potential with infinite boundaries

according to me, your original posted boundary conditions are correct.

why can't you solve this analytically? It is not a hard diff-eq to solve.
 
malawi_glenn said:
according to me, your original posted boundary conditions are correct.

why can't you solve this analytically? It is not a hard diff-eq to solve.

Ooops, you guys are right. Somehow I missed the absolute values in V(x)=inf for |x|>1. But I'm still not sure you can do it analytically, can you? You can still use shooting to approximate the answer by trying to hit phi(-1)=0 numerically.
 

Similar threads

Replies
29
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
3
Views
2K