Proving solutions to the schrodinger equation

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Bacat
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Homework Statement



Verify that the following are not solutions to the Schrödinger 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



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

This right here should tell you that [itex]\Psi(x,t) = A*Cos(kx-\omega t)[/itex] 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.