# Solving Schrodinger for eingenvalues

• Lee
In summary, the bound energy state energy eigenfunctions for a particle in a 1-D potential well (E<Vo), where; V(x)= (inf, x<0) (0, 0<x<a) (Vo, x>a) can be found in each particular region between 0 and a, but they are discontinuous at x=0. The function for 0<x<a can be found using a exp function.
Lee
I'm looking to solve schrodingers equation for the bound energy state energy eigenfunctions, for a particle in a 1-D potential well (E<Vo), where;

V(x)= (inf, x<0)
(0, 0<x<a)
(Vo, x>a)so we know when x>a our function is going to be a negative exp, and between 0 and a the function is going to be equal to An.sin(kn.z)? from here can we ahead and starting finding solutions to the eqn?

You found the solution in each particular regions; now you going to tie them up together using the boundary conditions and continuity conditions of psi (psi must be a continuous function)

for instance, you need that in the limit of psi as x approaches a- equals the limit as psi approaches a+.

quasar987 said:
You found the solution in each particular regions; now you going to tie them up together using the boundary conditions and continuity conditions of psi (psi must be a continuous function)

for instance, you need that in the limit of psi as x approaches a- equals the limit as psi approaches a+.
I just found this cute Java device for exploring solutions. I probably don't have it totally figured out, but you can model the situation pretty nicely by setting the piecewise constant potentials and choosing the functions in each region. You can't go to infinity, but you can go huge in one region.

From
http://web.phys.ksu.edu/vqm/AVQM Website/AVQMweb.htm

http://web.phys.ksu.edu/vqm/AVQM Website/WFEApplet.html

I had to close my browser window to restart it if I closed it. Expanded it to full screen it is quite useful.

so making sure An.sin(kn.a)=exp(-alpha.a) ?

I think my function for the mid section is wrong, as it will be discontinues if I pick it to be An.sin(kn.a) it shall equal 0 at 0 and a, which is good for the infinite wall to the left (as this is expected) but it that would mean exp(-alpha.a) would have to equal 0.

Can anyone make a better suggestion for the function for 0<x<a? Maybe a exp function?

Perfect. Thanks.

## 1. What is Schrodinger's equation?

Schrodinger's equation is a mathematical equation that describes how a quantum system, such as an electron, evolves over time. It is the fundamental equation of quantum mechanics and is used to calculate the probability of finding a particle in a specific location at a specific time.

## 2. What are eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are solutions to Schrodinger's equation. Eigenvalues represent the possible energy states of a quantum system, while eigenvectors represent the corresponding wavefunctions. The eigenvalues are the coefficients that determine the amplitude of the wavefunction at a particular energy level.

## 3. How do you solve Schrodinger's equation for eigenvalues?

To solve Schrodinger's equation for eigenvalues, you first need to set up the equation for the specific quantum system you are studying. Then, you need to use mathematical techniques, such as separation of variables and boundary conditions, to solve for the eigenvalues and eigenvectors. This process can be complex and often requires advanced mathematical knowledge.

## 4. What is the significance of solving Schrodinger's equation for eigenvalues?

Solving Schrodinger's equation for eigenvalues allows us to understand the quantum mechanical properties of a system, such as the allowed energy states and wavefunctions. This information is crucial for predicting and understanding the behavior of particles at the atomic and subatomic level.

## 5. Are there any limitations to solving Schrodinger's equation for eigenvalues?

Yes, there are limitations to solving Schrodinger's equation for eigenvalues. One of the main limitations is that it can only be applied to systems with a small number of particles. For larger systems, the complexity of the calculations becomes too high. Additionally, Schrodinger's equation does not account for relativistic effects and therefore is not accurate for systems with very high energies.

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