- #1

Bacat

- 151

- 1

## Homework Statement

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

(a) [tex]\Psi(x,t) = A*Cos(kx-\omega t)[/tex]

(b) [tex]\Psi(x,t) = A*Sin(kx-\omega t)[/tex]

## Homework Equations

Schrodinger equation: [tex]\frac{-hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = \frac{i*hbar*\partial \Psi}{\partial t}[/tex]

## The Attempt at a Solution

For part a:

Let [tex]y=kx-\omega t[/tex]

Calculating the derivatives gives...

[tex]\frac{\partial^2 \Psi}{\partial x^2}=-Ak^2*Cos y[/tex]

[tex]\frac{\partial \Psi}{\partial t} = A*\omega*Sin y[/tex]

Substituting and rearranging, we see that:

[tex]Sin y = \frac{hbar * k^2}{2mi\omega}Cos y[/tex]

Let [tex]f=\frac{hbar * k^2}{2mi\omega}[/tex]

Then [tex]Sin y = f Cos y[/tex]

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

If we substitute back in...

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

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?