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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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