# Time-Dependent Schrodinger Equation

1. Oct 13, 2007

### Peter5897

1. The problem statement, all variables and given/known data

Show that $$\Psi(x,t) = Ae^{i(kx-\omega t)}$$ is a solution to the time-dependent Schrodinger equation for a free particle [ $$U(x) = U_0 = constant$$ ] but that $$\Psi(x,t) = Acos(kx-\omega t)$$ and $$\Psi(x,t) = Asin(kx-\omega t)$$ are not.

2. Relevant equations

$$- \frac{h^2}{4\pi m}\frac{\partial^2\Psi(x,t)}{\partial x^2}+U(x)\Psi(x,t)=i\frac{h}{2\pi}\frac{\partial\Psi(x,t)}{\partial t}$$

3. The attempt at a solution

I tried to solve the first part of the problem by figuring out if both sides are equal when plugging in $$\Psi(x,t) = Ae^{i(kx-\omega t)}$$ for $$\Psi(x,t)$$ but I'm not even sure if that's how I'm supposed to go about solving this problem. I can take the partial deriviatives OK but I get to a point where I have a bunch of unknowns and I can't tell if both sides are equal.

It's quite possible that I'm going about solving this problem completely wrong and, if that's the case, I'd really like to know.

2. Oct 13, 2007

### nrqed

It will be a solution at the condition that there is a certain relation between U_0, k and omega. So k and omega could not be any old number. They must be related by a relation that you will find by plugging the solution in Schrodinger's equation

3. Oct 13, 2007

### Peter5897

Thanks for the reply, I'm still unsure of what relation I'm looking for. Here are my steps for solving the partial derivatives.

$$- \frac{h^2}{4\pi m}\frac{\partial^2\Psi(x,t)}{\partial x^2}+U(x)\Psi(x,t)=i\frac{h}{2\pi}\frac{\partial\Psi(x,t)}{\partial t}$$

$$- \frac{h^2}{4\pi m}\frac{\partial^2Ae^{i(kx-\omega t)}}{\partial x^2}+U(x)Ae^{i(kx-\omega t)}=i\frac{h}{2\pi}\frac{\partialAe^{i(kx-\omega t)}}{\partial t}$$

$$- \frac{h^2}{4\pi m}k^2Ae^{i(kx-\omega t)}+U(x)Ae^{i(kx-\omega t)}=i\frac{h}{2\pi}(- \omega)Ae^{i(kx- \omega t)}$$

$$- \frac{h^2}{4\pi m}k^2+U(x)=i\frac{h}{2\pi}(- \omega)$$

$$U_0=\frac{h}{2\pi}(\frac{1}{2m}k^2-i\omega)$$

Is there some relation that I'm missing? Well... I guess obviously there is. Either that or I solved the partial derivatives wrong.

**edit**

After plugging in the Acos() equation I noticed that I'm going to have two cos functions on one side and a sin on the other. I assume that when I solve for Asin() I'll have two sin functions on one side and a cos on the other so I obviously won't be able to get down to my U_0 = blah relation I found earlier.

I knew this going in but I'm still not sure what the relation is telling me.

Last edited: Oct 13, 2007
4. Oct 14, 2007

### HungryChemist

Well, you're missing i (imaginary) when you take partial derivative! I noticed this because you're potential function is imaginary.

Anyway, once you fix your partial derivative and find that your potential is indeed real function, you will immediately see that it is exactly equal to zero(here, you have to invoke de brogile's relation).

In other words, the solution given in problem is indeed free particle (free of potential field)

Last edited: Oct 14, 2007
5. Oct 14, 2007

### Peter5897

Wow, I'm a little embarrassed that I forgot to distribute that constant but at least I wasn't totally lost with what I was doing. I haven't redone my work yet since I'm sick and I want to get some sleep but I'm sure it will all work out fine now.

Thanks!

6. Mar 5, 2009

### HOZEFATINWALA

Hi there i want to prepare the fortran program of time dependent of schrodinger equation, can u help?

7. Mar 5, 2009

### turin

This is not correct. A free particle does not require vanishing potential. I free particle only requires a total energy above the largest value of the potential energy. A plane wave solution (as is the case in this problem), which is a type of free particle solution, does not require zero potential, either. It requires a constant and uniform potential, as specified by the problem.