How Many Bound States Exist in a Half Finite Square Well?

[Bobbo Snap]

Homework Statement

A particle of mass m is in the potential
$$V(x) = \left\{ \begin{array}{rl} \infty & \text{if } x < 0\\ -32 \hbar / ma^2 & \text{if } 0 \leq x \leq a \\ 0 & \text{if } x > a. \end{array} \right.$$

(a) How many bound states are there?

(b) In the highest energy bound state, what is the probability that the particle would be found outside the well?

Homework Equations

Schrodinger's Time Independent Eqn. $-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} - V(x)\psi = E \psi$

The Attempt at a Solution

Starting with $0<x<a$, I rewrote the S.T.I. eqn to read $\frac{d^2 \psi}{dx^2} = -k^2\psi$
where $k=\sqrt{2m(E-V)/\hbar^2}$.
The solutions being $\psi=A\cos{kx} + B\sin{kx}$, I dropped the cosine term because $\psi(0) = 0$ leaving $$\psi(x) = B\sin{kx} \text{ (eqn 1)}$$
When $x>a, V=0$ and the S.T.I. eqn. gives $\frac{d^2\psi}{dx^2} = l^2\psi$ where $l = \sqrt{-2mE/\hbar^2}$ The solutions are exponential and I dropped the positive one because it blows up as x gets big. So I have
$$\psi = Ce^{-lx} \text{ (eqn 2)}$$
I set eqns. 1 and 2 equal by continuity at a. I set their derivatives equal by continuity of the derivative at a. So I have
$$B\sin{ka}=Ce^{-la} \text{ (eqn 3)} \qquad \text{ and } \qquad kB\cos{ka} = -lCe^{-la} \text{ (eqn 4) }$$
Now I divide (4) by (3) and get
$$k\cot{ka} = -l$$

From here I'm not sure where to go. The book (I'm using Griffith's) has a similar example for the fully finite square well. In it, he defines $z=ka, z_0 = (ka\text{ with }E=0)$ and then does some wizardry where he plots solutions and finds their intersection points. I have tried to do this with my slightly different problem and failed. But do I even need to plot intersection points to answer part (a)? I'm kind of lost. And then how would I proceed with part (b), just calculate the probability density?

*I can't edit my typo in the thread title? Embarrassing.

 

[vela]

Looks good so far. Try writing that last equation in terms of ka. You should be able to show that
$$ka \cot ka = -la = -\sqrt{64-(ka)^2}.$$ You can't solve that equation analytically for ka, but you can see whether it has solutions by plotting the two sides of the equation and seeing where the curves intersect. If you need a numerical answer, you'll need to solve it numerically.

 

[Bobbo Snap]

Thanks vela, I was able to work that out. It lead to three bound states. Does anyone have any advice on how to calculate the probability in part (b)?

 

[vela]

What expression gives the probability for finding the particle between x and x+dx? Integrate that over the appropriate limits.

 

[Bobbo Snap]

I integrate the probability density from a to infinity. In this case,
$$C^2 \int_a^{\infty}e^{-2lx}dx$$
I need to normalize to find C right? I'm not sure what to do about B?
$$B^2 \int_0^a sin^2{(kx)}dx + C^2 \int_a^{\infty}e^{-2lx}dx = 1$$
Do I have this right?

 

[vela]

Use your equation 3 to write B in terms of C. Then use the normalization condition to solve for C.

 

[Bobbo Snap]

Ok, I see. Thank you for your help.