Quantum physics: proving wave packet is normalized

[Jillds]

Homework Statement

Following gaussian wave packet: $\psi (x)= \frac{1}{\sqrt{\sqrt{\pi a^2}}} e^{-\frac{x^2}{2a^2}}$
Prove that this function is normalized.

Homework Equations

$\int_{- \infty}^{\infty} |\psi (x)|^2 dx = 1$

The Attempt at a Solution

Is $\frac{1}{\sqrt{\sqrt{\pi a^2}}} \int_{- \infty}^{\infty} e^{-\frac{x^2}{a^2}} dx$ equal to 1?

I have a solution given, but I don't really get how they even got that solution:

$\frac{1}{\sqrt{\pi a^2}} \sqrt{\pi a^2}$ = 1
But that solution totally ignores the double root in the first term, and I'm mond boggled how they managed to integrate the exponent to ending up as $\sqrt{\pi a^2}$? Actually I don't manage to clearly and unambiuously integrate that exponent myself.

 

[DEvens]

First, note that the equation in 2 has the square of the wave function. That is why the exponent is exp(-x^2 / a^2) instead of exp(-x^2 / [2*a^2] ). Right? So why didn't you square the multiplier out front as well?

As to how to do the integral: The problem is you have this integral of exp(-x^2/a^2) with respect to x. And that's hard. But note that it's an integral from -infinity to +infinity. And the hint is that the answer seems to have a square root in it. So what about doing the square of the integral? You can convert to an integral over r and theta in 2-D, and you should be able to do that integral. Then take the square root, and violins.

 

[Jillds]

You are correct. I forgot to square the constant before the exponent. My eye fell on the fact that it's a Gaussian wave packet, therefore it's a Gaussian integral, which is basically what you're epxlaining (I think). $\int_{-\infty}^{\infty} e^{-x^2}$ would be $\sqrt{\pi}$ and the $a^2$ would just be multiplied under the squareroot following the features of a Gaussion integral. So, yes the solution I had from the course is what I was so puzzled about.

 

[BvU]

Did you also learn how to actually do this integral $$
\int_{-\infty}^{\infty} e^{-x^2}\ \ {\rm ?} $$

 

[Jillds]

There is mention of it in later chapters, with regards to Dirac and Fourrier.

 

[BvU]

Nah, don't need all that complicated stuff for this one. it's a simple but very nifty trick, good to remember (I did:) ):

They change from the 1D integral to its square, a 2D integral. In polar coordinates the $dx\; dy$ transform to $r \;dr\; d\phi$ and then it's easy (using $r\;dr = {1\over 2}d(r^2)\$ ). See Wiki

 

[Jillds]

Thank you very much. Yes, with the polar coordinates it's not that hard. :)