- #1
bumclouds
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Hey guys,
I have an assignment, which is to Solve Schrodinger's equations, for a certain potential distribution, which can be divided up into three regions.
A solution for one of the regions is of the form: Ae[tex]^{kx}[/tex]
If you substitute this into Schrodinger's equation (time independant, one dimension) and solve for k, you get this:
Schrodingers:
[tex]\frac{-h}{2m} \frac{d^{2}}{dx^{2}} \Psi (x) + Vo \Psi (x) = E \Psi (x)[/tex]
Solve for k:
k = [tex]\frac{\sqrt{2mE}}{h}[/tex]
I know this part is right because I've seen it written on the board a couple of times, and it's also what I get on paper.
But then there's the next bit, which I don't get. Apparently it's 'normalising k' which I just don't get..
k = [tex]\frac{\sqrt{2mE}}{h}[/tex]
[tex]k\overline{^}\overline{}[/tex] = [tex]\frac{\sqrt{2mE}}{h}.\frac{L}{2}[/tex]
[E] = [tex]\frac{\pi^{2}h^{2}}{2ML^{2}}[/tex]
k_hat = [tex]\frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{E}[/tex]
k_hat = [tex]\frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{\frac{E[E]}{[E]}}[/tex]
k_hat = [tex]\frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{[E]} \sqrt{Ehat} [/tex]
k_hat = [tex]\frac{\pi}{2} \sqrt{Ehat}[/tex]
None.
If I rearrange k = [tex]\frac{\sqrt{2mE}}{h}[/tex] and make E the subject,
I get ..
E = [tex]\frac{h^{2}khat^{2}}{2m}[/tex]
and maybe this is where I go wrong.. because I assume E = [E] ?
Subtituting [E] into the second last step in section 2 above yields:
[tex]khat = \frac{L}{2} khat \sqrt{Ehat}[/tex]
and that doesn't equal the last step >_<
EDIT: ahh.. if I use their definition of [E], [E] = [tex]\frac{\pi^{2}h^{2}}{2ML^{2}}[/tex]
I arrive at the right answer..
so how did they come up with [E] = [tex]\frac{\pi^{2}h^{2}}{2ML^{2}}[/tex]??
Homework Statement
I have an assignment, which is to Solve Schrodinger's equations, for a certain potential distribution, which can be divided up into three regions.
A solution for one of the regions is of the form: Ae[tex]^{kx}[/tex]
If you substitute this into Schrodinger's equation (time independant, one dimension) and solve for k, you get this:
Schrodingers:
[tex]\frac{-h}{2m} \frac{d^{2}}{dx^{2}} \Psi (x) + Vo \Psi (x) = E \Psi (x)[/tex]
Solve for k:
k = [tex]\frac{\sqrt{2mE}}{h}[/tex]
I know this part is right because I've seen it written on the board a couple of times, and it's also what I get on paper.
But then there's the next bit, which I don't get. Apparently it's 'normalising k' which I just don't get..
k = [tex]\frac{\sqrt{2mE}}{h}[/tex]
[tex]k\overline{^}\overline{}[/tex] = [tex]\frac{\sqrt{2mE}}{h}.\frac{L}{2}[/tex]
[E] = [tex]\frac{\pi^{2}h^{2}}{2ML^{2}}[/tex]
k_hat = [tex]\frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{E}[/tex]
k_hat = [tex]\frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{\frac{E[E]}{[E]}}[/tex]
k_hat = [tex]\frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{[E]} \sqrt{Ehat} [/tex]
k_hat = [tex]\frac{\pi}{2} \sqrt{Ehat}[/tex]
Homework Equations
None.
The Attempt at a Solution
If I rearrange k = [tex]\frac{\sqrt{2mE}}{h}[/tex] and make E the subject,
I get ..
E = [tex]\frac{h^{2}khat^{2}}{2m}[/tex]
and maybe this is where I go wrong.. because I assume E = [E] ?
Subtituting [E] into the second last step in section 2 above yields:
[tex]khat = \frac{L}{2} khat \sqrt{Ehat}[/tex]
and that doesn't equal the last step >_<
EDIT: ahh.. if I use their definition of [E], [E] = [tex]\frac{\pi^{2}h^{2}}{2ML^{2}}[/tex]
I arrive at the right answer..
so how did they come up with [E] = [tex]\frac{\pi^{2}h^{2}}{2ML^{2}}[/tex]??
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