# Normalize the even wave functions for the finite square well

1. Feb 27, 2005

### broegger

I'm trying to normalize the even wave functions for the finite square well. The wave function is:

$$\psi(x)= \begin{cases} Fe^{\kappa x} & \text{for } x< a\\ D\cos(lx) & \text{for } -a\leq x \leq a\\ Fe^{-\kappa x} & \text{for } x> a \end{cases}$$

How can I determine D and F? When I set

$$\int_{-\infty}^{\infty}|\psi(x)|^2dx = 1$$,​

I obtain an equation in the two unknown amplitudes D and F. I could apply some boundary conditions to get more equations, but it gets rather complicated and I know there is an easier way...

2. Feb 27, 2005

### James R

We require that the wave function is continuous, so:

$$Fe^{-\kappa a} = D \cos(la)$$

I don't think you can avoid using this, along with the normalisation condition.

3. Feb 27, 2005

### marlon

Also, include the continuity-demand for the FIRST derivative of the wavefunctions at the "walls" of the well.

marlon

4. Feb 27, 2005

### dextercioby

That $$e^{\kappa x}$$ is honestly nonnormalizable...And one more thing,the interval for this function should be:
$$(-\infty,-a)$$

Daniel.

5. Feb 27, 2005

### marlon

Which is indeed the case

marlon

6. Feb 27, 2005

### broegger

Do I need this? The continuity of $$\psi$$, as James R says, gives me two equations in two unknowns.

7. Feb 27, 2005

### dextercioby

How did u get those initial functions...?I mean,why aren't there 3 (a priori) different amplitudes,e.g. F,G and H...?

A little remark.I said earlier that $$e^{\kappa x}$$ was not normalizable.It would have been the case for the $$(a,+\infty)$$ interval.Sure,in your case,because of the negative values that "x" takes,it is VERY NORMALIZABLE...

Daniel.

Last edited: Feb 27, 2005
8. Feb 27, 2005

### broegger

Because of the symmetry of the finite square well (the well is centered at x = 0), I assume that $$\psi$$ must be symmetric (even or odd).

9. Feb 27, 2005

### dextercioby

Yes,well,then that's it...Though the fact that,by imposing continuity along the real axis,you determine completely the wave function and by checking the normalizability,(i think) it will not hold...

There's something fishy.Check Davydov,Cohen-Tannoudji or Flügge...

Daniel.

10. Feb 27, 2005

### polyb

broegger,

When you are normalizing over the full interval, does it look something like this:

$$\int_{-\infty}^{-a}|\psi(x)|^2dx + \int_{-a}^{a}|\psi(x)|^2dx + \int_{a}^{\infty}|\psi(x)|^2dx = 1$$

Between that and the boundary conditions stated by James R, the unknowns should be pretty easy to solve.

Last edited: Feb 27, 2005