Solving De Broglie Wave: Direction & F/G

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Homework Help Overview

The discussion revolves around the de Broglie wave function, specifically its direction and the evaluation of certain derivatives related to the wave function. Participants are exploring the implications of the wave function's form and the calculations involving partial derivatives.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to determine the direction of the de Broglie wave and calculates the first and second partial derivatives for functions F and G. Some participants question the clarity and correctness of the expressions provided for F and G, while others suggest clarifying the notation used for partial derivatives.

Discussion Status

The discussion is ongoing, with participants providing feedback on the calculations and expressions presented. Some guidance has been offered regarding the correct notation for partial derivatives, and there is an exploration of the significance of the ratio F/G.

Contextual Notes

Participants are working under the assumption that E and p are fixed, and there is a mention of potential constraints related to the context of the problem, such as the wave function's behavior in a region with zero potential.

UrbanXrisis
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the de Broglie wave has the form: [tex]/psi = e^{i(px-Et)/ \hbar}[/tex]

1. I am asked the direction of this wave. To me, as t increases, the x has to increase as well to keep a constant function on the left hand side so I believe that the direction of this wavelangth is heading towards the positive end.
2. I am to find:

[tex]F(x,t)=-\frac{\delta ^2 \phi}{\delta x^2}[/tex]
and
[tex]G(x,t)=i\frac{\delta \phi}{\delta t}[/tex]

i'm not too good with partial derivatives since I just learned them a week ago, please correct me if I am wrong.

x and t don't depend on each other while E and p are fixed energy and momentum:

first partial derivative:
[tex]F=\frac{ip}{ \hbar} e^{i(px-Et)/ \hbar}[/tex]
second:
[tex]F=-\frac{p^2}{ \hbar ^2} e^{i(px-Et)/ \hbar}[/tex]
[tex]F(x,t)=\frac{p^2}{ \hbar ^2} e^{i(px-Et)/ \hbar}[/tex]

first partial derivative:
[tex]G=-\frac{iE}{\hbar} e^{i(px-Et)/ \hbar}[/tex]
[tex]G(x,t)=\frac{E}{\hbar} e^{i(px-Et)/ \hbar}[/tex]

have I done this correctly?

then I am to evaluate F/G knowing that E=p^2/2m

I get [tex]\frac{F}{G}=\frac{2m}{\hbar}[/tex]?
 
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UrbanXrisis said:
first partial derivative:
[tex]F=\frac{ip}{ \hbar} e^{i(px-Et)/ \hbar}[/tex]
second:
[tex]F=-\frac{p^2}{ \hbar ^2} e^{i(px-Et)/ \hbar}[/tex]
[tex]F(x,t)=\frac{p^2}{ \hbar ^2} e^{i(px-Et)/ \hbar}[/tex]
Here you have three different expressions for F. So which on is it?

I understand what you're doing, but as you've written it down is confusing (and wrong). Your final result for F(x,t) is correct though, which was what you're heading for.
first partial derivative:
[tex]G=-\frac{iE}{\hbar} e^{i(px-Et)/ \hbar}[/tex]
[tex]G(x,t)=\frac{E}{\hbar} e^{i(px-Et)/ \hbar}[/tex]
Same thing here. The first expression is not G, but [itex]\frac{\partial \phi}{\partial t}[/itex], so you should write that.

The rest looks good.

then I am to evaluate F/G knowing that E=p^2/2m

I get [tex]\frac{F}{G}=\frac{2m}{\hbar}[/tex]?[/QUOTE]
 
In the LaTeX, use \ before psi, and use \del or \partial for the partial derivative.

For partial differentiation, one treats the other variable as a contant, and what you have done appears to be correct, except as Galileo pointed out, write the partials of F and G where you are differentiating.

The wave is in the positive direction.

Another way to write the wave equation is [itex]\psi(x,t)\,=\,A\,exp\,[i(kx-\omega t)][/itex], where [itex]k\,=\,2\pi/\lambda[/itex] and [itex]\omega\,=\,2\pi\nu[/itex].
 
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for F and G, I was just showing the first partial derivative and the second partial derivative.

was my calculation of [tex]\frac{F}{G}=\frac{2m}{\hbar}[/tex] correct?

what is the significance of F/G?
 
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In one dimension -

[tex]\frac{d^2\psi}{dx^2}\,+\,\frac{2mE}{\hbar^2}\psi\,=\,0[/tex]

in region with the potential, V(x) = 0.
 
Last edited:

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