# Normalizing wavefunction of (x^2)e^(-x^2)?

1. Sep 19, 2013

### lonewolf219

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

ψ(x,t)=Axe$^{-cx^2}$e$^{-iωt}$

2. Relevant equations

1=∫ψ*ψdx

3. The attempt at a solution

1=$\int$A$^{2}$x$^{2}$e$^{-2cx^2}$

I think I multiplied ψ*ψ correctly... I'm surprised a little that there is an x^2 with this problem. We have not discussed these integrals in class. I watched a video on how to integrate e$^{-x^2}$, but don't know what to do with the product x^2 and e^(-2cx^2)

2. Sep 19, 2013

### dextercioby

Did you take statistical mechanics ? You learn about integrating gaussians there. If not, then at least a course in mathematical methods of physics should have explained you this.

Anyways, think I(c) = int_R exp(-cx2) dx. Compute it then differentiate the result wrt c. (c is a parameter).

3. Sep 19, 2013

### kostas230

I'll give you a hint.
I presume you know that: $\int_{- \infty}^{+\infty} e^{-\lambda x^2}dx = \sqrt{\pi/ \lambda}$. Consider the function:
$$I(\lambda )=\int_{- \infty}^{+\infty}e^{-\lambda x^2}dx$$
Then:
$$\frac{d}{d\lambda }I(\lambda )=-\int_{- \infty}^{+\infty}x^2e^{-\lambda x^2}dx$$

4. Sep 19, 2013

### vela

Staff Emeritus
You could also integrate by parts.

5. Sep 19, 2013

### lonewolf219

Thanks for the help, guys! Didn't take statistical mechanics, unfortunately...

Does this mean that the derivative of $\sqrt{\frac{\pi}{\lambda}}$ is my answer?

6. Sep 19, 2013

### Jolb

vela is correct, but kostas230's trick is one of the most beautiful in mathematics and one of the most important in physics. The technique is referred to as using a "generating functional."

7. Sep 19, 2013

### kostas230

I don't know if you ever had to evaluate the Gaussian integral, but here's an elegant way to do it. Consider the integral:

$$I= \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} e^{-(x^2+y^2)}dxdy$$

By changing from cartesian coordinates to polar coordinates, the integral above becomes:
$$I= \int_{0}^{+\infty}\int_{0}^{2\pi}re^{-r^2}drd\theta$$

I trust that you are able to evaluate the integral. Can you find the gaussian integral $I'=\int_{-\infty}^{+\infty}e^{-x^2}dx$ using the above results? ;)

8. Sep 19, 2013

### lonewolf219

Thanks kostas230, appreciate your help... The result should be in the form of a Hermite polynomial, is that correct? I think the derivative of re$^{-r^2}$ is

e$^{-r^2}$(1-2r$^{2}$)

Not quite sure what to do next

9. Sep 20, 2013

### kostas230

No, I said: compute the integral, not find the derivative of the integrated function

10. Sep 20, 2013

### lonewolf219

If you are asking me to find the integral of re$^{-r^2}$drd$\theta$ from 0$\leq$r$\leq$$\infty$ and 0$\leq\theta$$\leq$2$\pi$ I believe it is $\pi$. And then the integral with a coefficient λ in the power would be $\frac{\pi}{\lambda}$

11. Sep 20, 2013

### Staff: Mentor

That is right, and it allows you to find the value of the original Gaussian integral, if you find the relation between that and the formula kostas230 used.

12. Sep 20, 2013

### lonewolf219

Thank you, mfb.

So is it that I$^{2}$= $\frac{\pi}{\lambda}$

So the derivative of $\frac{\pi}{\lambda}$ is -$\frac{\pi}{\lambda^2}$?

Since the derivative would be 2I for I$^{2}$, should I divide -$\frac{\pi}{\lambda^2}$ by 2, so that I=-$\frac{\pi}{2\lambda^2}$ ?

Although A is a positive value.. would the negative sign in front of kostas' integral cancel this one, so that I =$\frac{\pi}{2\lambda^2}$ ?

Last edited: Sep 20, 2013
13. Sep 20, 2013

### Staff: Mentor

I think you are mixing two different approaches here.

14. Sep 20, 2013

### kostas230

My first post was on an algorithmic process in finding integrals of the form: $I_n=\int_{-\infty}^{+\infty}x^n e^{-\lambda x^2}dx$ presuming that you know the integral $I_0=\int_{-\infty}^{+\infty}e^{-\lambda x^2}dx=\sqrt{\pi}$. My other posts where to show you a way to compute the integral $\int_{-\infty}^{+\infty}e^{-\lambda x^2}dx=\sqrt{\pi}$.