# Wavefunction normalization

1. Dec 31, 2007

### raintrek

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

I'm trying to determine a normalization value, A, for the following wavefunction:

$$\Psi = Ax{^2}exp(-\alpha x)}, x>0$$
$$\Psi = 0, x<0$$

In the past, I've had an i term in my exponential, so when applying the Normalization Condition:

$$\int|\Psi(x)|^2 dx = \int\Psi{^*}(x) \Psi(x) dx$$

the exponentials always multiply to equal one, leaving me with an easy route to getting the normalization factor.

However in this case, I'm left with the following integral:

$$\int|\Psi(x)|^2 dx = \int A{^2}x{^4}exp(-2\alpha x)} dx$$

...which seems horrible!

Can anyone advise what I'm doing wrong here? I'm sure there's a simpler way...

2. Dec 31, 2007

### George Jones

Staff Emeritus
This looks OK.

Have you ever seen tabular integration by parts?

3. Dec 31, 2007

### Erythro73

No, you're doing in the right way. Just do your integral by parts now.

You can use the gamma function to do it faster.

Last edited: Dec 31, 2007
4. Dec 31, 2007

### raintrek

Never...

I tried it in an online integrator and got a horrendous answer so assumed I'd gone drastically wrong somewhere...

5. Dec 31, 2007

### George Jones

Staff Emeritus
I tried (maybe not very successfully) to explain it http://groups.google.ca/group/sci.math/msg/ebd6104dfcc6263c?dmode=source".

Most of the terms will be zero; remember, as $x \rightarrow \infty$, an exponential dominates any power of $x$.

Last edited by a moderator: Apr 23, 2017
6. Dec 31, 2007

### George Jones

Staff Emeritus
Last edited by a moderator: Apr 23, 2017
7. Dec 31, 2007

### raintrek

OK, I think I've followed that George Jones;

$$\int|\Psi(x)|^2 dx = \int A{^2}x{^4}exp(-2\alpha x)} dx$$

Which, using your method, I get to be:

$$exp(-2\alpha x) \left(-\frac{x{^4}}{2\alpha} + \frac{x{^3}}{\alpha{^2}} - \frac{3x{^2}}{2\alpha{^3}} + \frac{3x}{2\alpha{^4}} - \frac{3}{4\alpha{^5}} \right)$$

Which I *think* leads to $$\frac{3A{^2}}{4\alpha{^5}} = 1$$ or $$A = \sqrt{4/3}\alpha{^\frac{5}{2}}$$

8. Dec 31, 2007

### Erythro73

Seems good, from my calculation, as I arrive, too, at
$1=\frac{3A^2}{4\alpha ^5}$

by two differents methods (that is, by the use of the gamma function and with the long method of integration by parts).

Last edited: Dec 31, 2007
9. Dec 31, 2007

### raintrek

Awesome, thanks for your help George Jones and Erythro73!

That technique is awesome - I'll definitely be using it again!!