Proving solutions to the schrodinger equation

[Bacat]

Homework Statement

Verify that the following are not solutions to the Schrödinger 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

Schrödinger 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 Schrödinger's equation. I think I may be making this more complicated than it is. How should I approach this problem differently?

 

[gabbagabbahey]

[
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 Schrödinger equation. If it were, you would expect it to satisfy the Schrödinger equation everywhere (i.e. for all values of y)--- which it clearly doesn't.

 

[Bacat]

Ah, good point. So it IS a lot simpler than I was making it. Haha.

Any ideas on this one:

https://www.physicsforums.com/showthread.php?t=292957

?

 

[vaibhav1803]

a faster way would be to express cos(kx-\omega t) in terms of e^{i(kx-\omega t)}
and substitute in the time dependent equation both will never satisfy it(coz the wavefunc. is of the form Ae^{i(kx-\omega t} )
any ideas on this one mate..?
https://www.physicsforums.com/showthread.php?p=2218109#post2218109