# De Broglie waves

1. Feb 14, 2006

### UrbanXrisis

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

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:

$$F(x,t)=-\frac{\delta ^2 \phi}{\delta x^2}$$
and
$$G(x,t)=i\frac{\delta \phi}{\delta t}$$

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 dont depend on each other while E and p are fixed energy and momentum:

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

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

have I done this correctly?

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

I get $$\frac{F}{G}=\frac{2m}{\hbar}$$?

Last edited: Feb 14, 2006
2. Feb 14, 2006

### Galileo

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:
Same thing here. The first expression is not G, but $\frac{\partial \phi}{\partial t}$, so you should write that.

The rest looks good.

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

I get $$\frac{F}{G}=\frac{2m}{\hbar}$$?[/QUOTE]

3. Feb 14, 2006

### Astronuc

Staff Emeritus
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 $\psi(x,t)\,=\,A\,exp\,[i(kx-\omega t)]$, where $k\,=\,2\pi/\lambda$ and $\omega\,=\,2\pi\nu$.

Last edited: Feb 14, 2006
4. Feb 14, 2006

### UrbanXrisis

for F and G, I was just showing the first partial derivative and the second partial derivative.

was my calculation of $$\frac{F}{G}=\frac{2m}{\hbar}$$ correct?

what is the significance of F/G?

Last edited: Feb 14, 2006
5. Feb 14, 2006

### Astronuc

Staff Emeritus
In one dimension -

$$\frac{d^2\psi}{dx^2}\,+\,\frac{2mE}{\hbar^2}\psi\,=\,0$$

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

Last edited: Feb 14, 2006