# Proving solutions to the schrodinger equation

• Bacat
In summary, the conversation discusses verifying that solutions to the Schrodinger equation for a free particle are not solutions to the given equations, (a) \Psi(x,t) = A*Cos(kx-\omega t) and (b) \Psi(x,t) = A*Sin(kx-\omega t). The conversation includes calculations and a graph to show that (a) is not a solution, and suggests a simpler method using the equation \Psi(x,t) = A*ei(kx-\omega t) to prove that neither (a) nor (b) are solutions.
Bacat

## Homework Statement

Verify that the following are not solutions to the Schrodinger equation for a free particle:

(a) $$\Psi(x,t) = A*Cos(kx-\omega t)$$

(b) $$\Psi(x,t) = A*Sin(kx-\omega t)$$

## Homework Equations

Schrodinger equation: $$\frac{-hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = \frac{i*hbar*\partial \Psi}{\partial t}$$

## The Attempt at a Solution

For part a:

Let $$y=kx-\omega t$$

Calculating the derivatives gives...

$$\frac{\partial^2 \Psi}{\partial x^2}=-Ak^2*Cos y$$
$$\frac{\partial \Psi}{\partial t} = A*\omega*Sin y$$

Substituting and rearranging, we see that:

$$Sin y = \frac{hbar * k^2}{2mi\omega}Cos y$$

Let $$f=\frac{hbar * k^2}{2mi\omega}$$

Then $$Sin y = f Cos y$$

The equality holds when $$y=\pm ArcCos(\pm\frac{1}{\sqrt{1+f^2}})$$

If we substitute back in...

$$kx-\omega t = \pm ArcCos(\pm\frac{1}{\sqrt{1+\frac{hbar^2k^4}{4m^2\omega^2}}})$$

If I assign some values to omega, m, hbar, and k, I can graph x as a function of t for one of the cases. It shows up as a straight line with a positive slope.

This does not seem to prove that (a) is not a solution to Schrodinger's equation. I think I may be making this more complicated than it is. How should I approach this problem differently?

Bacat said:
[
Substituting and rearranging, we see that:

$$Sin y = \frac{hbar * k^2}{2mi\omega}Cos y$$

Let $$f=\frac{hbar * k^2}{2mi\omega}$$

Then $$Sin y = f Cos y$$

This right here should tell you that $\Psi(x,t) = A*Cos(kx-\omega t)$ is not a solution to the schrodinger equation. If it were, you would expect it to satisfy the schodinger equation everywhere (i.e. for all values of y)--- which it clearly doesn't.

a faster way would be to express cos(kx-$$\omega$$ t) in terms of ei(kx-$$\omega$$ t)
and substitute in the time dependant equation both will never satisfy it(coz the wavefunc. is of the form Aei(kx-$$\omega$$ t )
any ideas on this one mate..?

## 1. What is the Schrodinger equation?

The Schrodinger equation is a mathematical equation that describes the behavior of quantum particles, such as electrons, in a given system. It is a fundamental equation in quantum mechanics and is used to calculate the probability of finding a particle in a certain location.

## 2. Why is it important to prove solutions to the Schrodinger equation?

Proving solutions to the Schrodinger equation is important because it allows us to understand and predict the behavior of quantum systems. By solving the equation, we can determine the energy levels and wave functions of particles in a given system, which is crucial in fields such as nanotechnology and quantum computing.

## 3. How is the Schrodinger equation solved?

The Schrodinger equation is solved using mathematical techniques such as separation of variables and perturbation theory. These methods allow us to break down the equation into simpler parts and find solutions for each part, which can then be combined to find a complete solution.

## 4. Are there any limitations to the Schrodinger equation?

Yes, there are limitations to the Schrodinger equation. It is a non-relativistic equation, meaning it does not take into account the effects of special relativity. It also only applies to non-relativistic particles with spin 1/2, such as electrons. For more complex systems, the equation becomes increasingly difficult to solve.

## 5. How do scientists validate the solutions to the Schrodinger equation?

Scientists validate the solutions to the Schrodinger equation by comparing them to experimental data. If the predicted energy levels and wave functions match with the observed values, then the solutions are considered valid. Additionally, the solutions should also be consistent with other known principles and laws of physics.

Replies
9
Views
848
Replies
2
Views
805
Replies
1
Views
183
Replies
4
Views
923
Replies
5
Views
969
Replies
30
Views
1K
Replies
1
Views
623
Replies
1
Views
2K