Why Is My Normalization Constant Different from the Paper's Result?

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weezy
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Homework Statement



## \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
 
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weezy said:

Homework Statement



## \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##
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|
weezy said:
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
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.
 
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