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

[lonewolf219]

Homework Statement

ψ(x,t)=Axe$^{-cx^2}$e$^{-iωt}$

Homework Equations

1=∫ψ*ψdx

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)

 

[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).

 

[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:
$$\frac{d}{d\lambda }I(\lambda )=-\int_{- \infty}^{+\infty}x^2e^{-\lambda x^2}dx$$

 

[vela]

You could also integrate by parts.

 

[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?

 

[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."

 

[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? ;)

 

[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

 

[kostas230]

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

 

[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 $\frac{\pi}{\lambda}$

 

[mfb]

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.

 

[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 =$\frac{\pi}{2\lambda^2}$ ?

 

[mfb]

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

I think you are mixing two different approaches here.

 

[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 $\int_{-\infty}^{+\infty}e^{-\lambda x^2}dx=\sqrt{\pi}$.