
#1
Mar506, 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^{kxwt}=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. 



#2
Mar506, 05:05 PM

Sci Advisor
HW Helper
P: 1,322

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




#3
Mar506, 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]




#4
Mar606, 04:32 AM

Sci Advisor
HW Helper
P: 11,866

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. 


Register to reply 
Related Discussions  
[SOLVED] Series with Trigonometric funtions  Calculus & Beyond Homework  1  
Vector Funtions  Calculus & Beyond Homework  3  
How to add more funtions on Ti84 silver  Computing & Technology  2  
Eigen Funtions  Calculus  2  
Trig Funtions Problems  Introductory Physics Homework  4 