How Is k Normalized in Schrodinger's Equation?

[bumclouds]

Hey guys,

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^{kx}



If you substitute this into Schrodinger's equation (time independant, one dimension) and solve for k, you get this:

Schrodingers:
\frac{-h}{2m} \frac{d^{2}}{dx^{2}} \Psi (x) + Vo \Psi (x) = E \Psi (x)

Solve for k:
k = \frac{\sqrt{2mE}}{h}

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 = \frac{\sqrt{2mE}}{h}

k\overline{^}\overline{} = \frac{\sqrt{2mE}}{h}.\frac{L}{2}

[E] = \frac{\pi^{2}h^{2}}{2ML^{2}}

k_hat = \frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{E}

k_hat = \frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{\frac{E[E]}{[E]}}

k_hat = \frac{L}{2} . \frac{\sqrt{2m}}{h} . \sqrt{[E]} \sqrt{Ehat}

k_hat = \frac{\pi}{2} \sqrt{Ehat}

Homework Equations

None.

The Attempt at a Solution

If I rearrange k = \frac{\sqrt{2mE}}{h} and make E the subject,

I get ..

E = \frac{h^{2}khat^{2}}{2m}

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:

khat = \frac{L}{2} khat \sqrt{Ehat}

and that doesn't equal the last step >_<

EDIT: ahh.. if I use their definition of [E], [E] = \frac{\pi^{2}h^{2}}{2ML^{2}}
I arrive at the right answer..


so how did they come up with [E] = \frac{\pi^{2}h^{2}}{2ML^{2}}??

 

[Cyosis]

What is k_hat and what are you trying to achieve, an expression for E?

 

[bumclouds]

Sorry! I've edited my above post.. Now it makes more sense :)

What I'm really after is how did they (my teacher) come up with [E] and what is it?


k_hat is apparently a normalised k