1. Nov 21, 2004

### 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

Last edited: Nov 21, 2004
2. Nov 21, 2004

### 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 $E_0 / \beta$.

3. Nov 21, 2004

### JamesJames

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

And is that the WKB solution?

Last edited: Nov 21, 2004
4. Nov 22, 2004

### 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

$$e^{\pm i \int^x k(x') dx'}[/itex] I'm being a little sloppy here but you can read more here http://www.du.edu/~jcalvert/phys/wkb.htm (and in lots of other places including your textbook!) 5. Nov 23, 2004 ### JamesJames I thought that this is the form of the SE.....[tex]\frac {d^2 y}{dx^2} + V(x) y = E y$$

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 ?

6. Nov 23, 2004

### Tide

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