Register to reply

Wave funtions

by UrbanXrisis
Tags: funtions, wave
Share this thread:
Mar5-06, 03:35 PM
P: 1,200
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.
Phys.Org News Partner Science news on
'Office life' of bacteria may be their weak spot
Lunar explorers will walk at higher speeds than thought
Philips introduces BlueTouch, PulseRelief control for pain relief
Physics Monkey
Mar5-06, 05:05 PM
Sci Advisor
HW Helper
Physics Monkey's Avatar
P: 1,334
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
Mar5-06, 05:44 PM
P: 1,200
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]

Mar6-06, 04:32 AM
Sci Advisor
HW Helper
P: 11,952
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]


Register to reply

Related Discussions
Series with Trigonometric funtions Calculus & Beyond Homework 1
Vector Funtions Calculus & Beyond Homework 3
How to add more funtions on Ti-84 silver Computing & Technology 2
Eigen Funtions Calculus 2
Trig Funtions Problems Introductory Physics Homework 4