How to Determine Normalization Factor for Wavefunction with Exponential Decay?

[raintrek]

Homework Statement

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...

 

[George Jones]

This looks OK.

Have you ever seen tabular integration by parts?

 

[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.

 

[raintrek]

This looks OK.

Have you ever seen tabular integration by parts?

Never...

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

 

[George Jones]

Never...

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

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

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

 

[George Jones]

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

Sorry, you need to see the http://groups.google.ca/group/sci.m...d3b7a8301/a443e74805de52a4?#a443e74805de52a4".

It looks a little complicated,but it's very easy and useful once you get the hang of it.

 

[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}}

Would someone be able to verify that? I'm paranoid about this being wrong now!

 

[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).

 

[raintrek]

Awesome, thanks for your help George Jones and Erythro73!

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