Quantum physics: proving wave packet is normalized

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1. Jan 28, 2015

Jillds

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

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

3. 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.

2. Jan 28, 2015

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.

3. Jan 28, 2015

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.

4. Jan 28, 2015

BvU

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

5. Jan 29, 2015

Jillds

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

6. Jan 29, 2015

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

7. Jan 29, 2015

Jillds

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