wave funtions


by UrbanXrisis
Tags: funtions, wave
UrbanXrisis
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#1
Mar5-06, 03:35 PM
P: 1,214
I am to show that neither of the two wave functions [tex]\psi_1 (x,t) = M_1 e^{kx-\omega t}[/tex] and [tex]\psi_2 (x,t) = M_2 e^{i(kx-\omega t)}[/tex] solve the de Broglie form of Schr. Eqn:

[tex]-\frac{\hbar ^2}{2m} \frac{\partial ^2 \psi}{\partial x^2}=i \hbar \frac{\partial \psi}{\partial t}[/tex]

for the first wave, i got:

[tex]-\frac{\hbar ^2}{2m} M_1 k^2 e^{kx-wt}=-i \omega \hbar M_1 e^{kx-\omega t}[/tex]

for the second wave, i got:
[tex]\frac{\hbar ^2}{2m} M_2 k^2 e^{i(kx-\omega t)}= \omega \hbar M_2 e^{i(kx-\omega t)}[/tex]

i was just wondering if I did these differentiation correct.
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Physics Monkey
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#2
Mar5-06, 05:05 PM
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Yes, you did the differentiations correctly. I am confused by your task to show that neither function satisfies the Schrodinger equation when in fact both do as you have just shown
UrbanXrisis
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#3
Mar5-06, 05:44 PM
P: 1,214
well, all I have to do is to show that they are not equal. Because if i simplify both of those equations, do not get the de Broglie relation of: [tex]\hbar \omega = \frac{\hbar ^2 k^2}{2m}[/tex]

dextercioby
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#4
Mar6-06, 04:32 AM
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wave funtions


What do you mean...? You do get the deBroglie relation

[tex]p=\hbar k [/tex]

and so [tex] E=\frac{p^{2}}{2m} [/tex]

Daniel.


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