# Normalizing Wave Functions Over Multiple Regions

1. Nov 13, 2013

### SHISHKABOB

1. The problem statement, all variables and given/known data

I need to normalize the following wave function in order to determine the value of the coefficients. This is from the basic finite square well potential.

$\Psi(x) = Ae^{k_{1}x},for \ x < -a/2$
$\Psi(x) = Csin(k_{2}x),for \ -a/2 \leq x \leq a/2$
$\Psi(x) = De^{-k_{1}x}, for \ x > a/2$

2. Relevant equations

$\int\left|\Psi(x)\right|^{2} dx = 1$

3. The attempt at a solution

Do I do an integral for each region, with the limits of integration being the boundaries of each region, and that integral normalized to 1 for each of those regions? Or do I add up those integrals with the same limits of integration and then set that equal to 1?

Last edited: Nov 13, 2013
2. Nov 13, 2013

### qbert

You use the fact that the wavefunction is continuous at each connecting point,
(-a/2), a/2, to write C and D in terms of A.

Then you integrate the piecewise-function squared over the whole interval to tell
you what A should be.

[also this might be a solution to the finite square well but it's not the most general solution, which
allows cos(kx) in the middle as well. <-- totally meant this to be punny
]

3. Nov 14, 2013

### Staff: Mentor

This. More explicitly:

$$\int_{-\infty}^{+\infty} {|\Psi|^2 dx} = 1 \\ \int_{-\infty}^{-a/2} {|\Psi_1|^2 dx} + \int_{-a/2}^{+a/2} {|\Psi_2|^2 dx} + \int_{+a/2}^{+\infty} {|\Psi_3|^2 dx} = 1$$

This is the same thing as if you were to integrate a single function over the entire range from -∞ to +∞, by splitting up that range into three pieces and doing those as separate integrals for whatever reason.