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

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Homework Statement


The wave function of a particle satisfies the time-independent schrodinger equation.
If the potential is symmetric and has the form
V(x) = \inf |x|>1.0
V(x) = \frac{\hbar^2V_0}{2m} |x|<0.2
V(x) = 0 Elsewhere
Using the shooting method, I need to find the ground state energy and the normalised group state wave function if V_0 = 50. 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 \phi(-1) = \phi(1) = 0. Other than that I'm clueless.

edit: \phi represents the wavefunction, I just don't know how to write it.
 
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Hi I'm a new member. Where can I find out numerical method for solving Schrodinger 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 Schrodinger 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 Schrodinger 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 boundry 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 boundry 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.
 
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