- #1
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]?
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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