Finding Exact and WKB Wavefunction for V(x) in E=10 Situation

[JamesJames]

Consider the potential
V(x) = \beta x for x \geq\ 0
V(x) = 0 for x < 0.

Find the exact and WKB wavefunction for the situation where a particle has
E = 10 in units where \beta = \hbar = m = 1.

Any suggestions guys?
James

 

[Tide]

For x > 0 you are looking for approximate solutions of an equation having this form:

\frac {d^2 y}{dx^2} + k^2\left(1-\frac {\beta x}{E_0}\right) y = 0

which will be of the form

e^{\pm i \frac {2}{3} k\left(1-\frac {\beta x}{E_0}\right)^{\frac {3}{2}}

which is valid for x not close to E_0 / \beta.

 

[JamesJames]

Ok, what are you calling k because here is what I would think the Shrodinger equation is
<br /> \frac {d^2 y}{dx^2} + \beta x y = E_0 y<br />

And is that the WKB solution?

 

[Tide]

Yes, I abbreviated k and just gave the form of the equation. I think your signs are reversed on the \beta and E_0 terms.

Your equation is actually a variant of the Airy differential equation but the point of your problem is to use the WKB approximation which essentially says that the solution of equations like

\frac {d^2 y}{dx^2} + k^2(x) y = 0

are approximately of the form

e^{\pm i \int^x k(x&#039;) dx&#039;}[/itex]<br /> <br /> I&#039;m being a little sloppy here but you can read more here http://www.du.edu/~jcalvert/phys/wkb.htm (and in lots of other places including your textbook!)

 

[JamesJames]

I thought that this is the form of the SE...\frac {d^2 y}{dx^2} + V(x) y = E y

So subbing in \beta x y for the potential, I think the signs are rigt.
I just don' t see how to get
\frac {d^2 y}{dx^2} + k^2\left(1-\frac {\beta x}{E_0}\right) y = 0

Also what is the k that goes in your integral? Shouldn 't it be
\beta x y - E ?

 

[Tide]

No, you have the signs reversed on the E and V terms in the Schrodinger equation. I used k^2 = \frac {2m}{\hbar^2}(E - V)