# Finding Normalization constant

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1. Jul 31, 2017

### weezy

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

$\psi(x) = N. (x^2 - l^2)^2$ for $|x| < l , 0$ otherwise

We have to find N such that this wavefunction is normalised.

2. The attempt at a solution

I tried expanding the $(x^2 - l^2)^2$ term inside the integral but this integral is extremely messy :

$\frac{1}{N^2} = \int_{- \infty}^{+ \infty} (x^2 - l^2)^2 dx$
from which I got:
$\frac{1}{N^2} = 2[ \frac{l^8}{5} - \frac{2l^2}{3} + 1]$

The paper which I'm following gives a completely different answer i.e.

$N = \sqrt{\frac{315}{216}} \frac{e^{i\psi}}{\sqrt{l}}$

And you can guess, I'm totally perplexed by this result. I don't know how the exponential enters the integral or how even that number is related to this integral. Would gladly appreciate some guidance!

This is the paper I'm following and the integral appears on page 2 eqn (0.5): https://ocw.mit.edu/courses/physics...pring-2013/lecture-notes/MIT8_04S13_Lec04.pdf

2. Jul 31, 2017

### ehild

You have to integrate $|\psi(x)|^2$ so you have $(x^2 - l^2)^4$ as integrand, and the integration limits are -l and +l, as the function is zero for x>|l|
The numerical result would be the same if you integrated the correct function correctly. The wavefunction can contain an arbitrary e multiplication factor, as it cancels when taking |ψ|2. By the way, the result in the book is not quite correct, as there should be √l9.