# Time Dependent Schrodinger Equation

1. Jul 7, 2014

### MostlyHarmless

1. The problem statement, all variables and given/known data
Show that the wave function $\Psi(x,t)=Asin(kx-ωt)$ does not satisfy the time dependent Schrodinger Equation.

2. Relevant equations

$-\frac{\hbar}{2m}\frac{\partial^2\psi(x,t)}{{\partial}x^2}+V(x,t)\psi(x,t)=i\hbar\frac{\partial\psi(x,t)}{{\partial}t}$

3. The attempt at a solution
So first step is I plug in the wave function:

$-\frac{\hbar}{2m}\frac{\partial^2Asin(kx-ωt)}{{\partial}x^2}+V(x,t)Asin(kx-ωt)=i\hbar\frac{{\partial}Asin(kx-ωt)}{{\partial}t}$

From here, I have a differential equation, which, I'm pretty sure I can't solve, but I'm mostly going off the fact that the question says to show that this wave function "does not satisfy.."

But I'm really unsure on how to show this..I don't remember alot from DiffEq, but I can see that I can't do separation of variables here.

2. Jul 7, 2014

### dauto

Quote "I don't remember alot from DiffEq". Time to do a good review, isn't it? To keep it simple assume V = 0 and do the partial derivatives. What do you get?

3. Jul 7, 2014

### MostlyHarmless

I couldn't agree more.

$\frac{\hbar^2}{2m}Ak^2sin(kx-{\omega}t){\partial}x+0=-i{\hbar}A{\omega}cos(kx-{\omega}t){\partial}t$

4. Jul 7, 2014

### vanhees71

Why have you writen $\partial x$ and $\partial t$ at the end of the partial derivatives? That doesn't make sense. Just derive two times wrt. to $x$ on the one side and once by $t$ on the other and check that the given function doesn't fulfill the equation after you've sorted out all the constants.

5. Jul 7, 2014

### MostlyHarmless

Got it, I've forgotten a lot. It's been over year since DiffEq, and a year before that since Cal 3. Thank you.

EDIT:

Actually, one last thing. Why can I let V=0?

EDIT: Nevermind, I think I see what is going on.

This may be overly simplistic, but basically in order for a wave function to satisfy the Time-Dependent Schrodinger equation, do we have to be able to reduce the equation down to:

$\hbar\omega=\frac{\hbar^2k^2}{2m}+V_o$

Last edited: Jul 7, 2014