I'm trying to find the normalisation constant $N$ for the following wavefunction:
$$
\psi\left(x\right) = \left\{
\begin{array}{lr}
N \left(x^2 – l^2\right)^2 &\: \left|x\right| \le l \\
0 &\: otherwise
\end{array}
\right.
$$
Using:
$$
\int_{-\infty}^{\infty} \left|\psi\left(x\right)\right|^2 \, dx = 1
$$
The answer should be:
$$
N = \sqrt{\frac{315}{256}} \frac{e^{i \phi}}{\sqrt{l}}
$$
However I get:
$$
\int_{-\infty}^{\infty} \left|\psi\left(x\right)\right|^2 \, dx \ = \int_{-l}^{l} N^2 \left(x^2 – l^2 \right)^4 \, dx \ = \ \frac{N^2}{10} \left[\frac{\left(x^2 – l^2\right)^5}{x}\right]_{\,-l}^{\,l} = 0
$$
Which is clearly wrong, and I do not understand where the phase could have come from. Am I approaching this completely the wrong way?
I have now corrected the integration, however I get (double checked with Mathematica):
$$
N = \sqrt{\frac{315}{256}} \frac{1}{l^\frac{9}{2}}
$$
Which is the wrong power of $l$ (it should be $\frac{1}{2}$). Substituting either the answer or my answer into the original integral does not yield $1$ either.
(I am working though these quantum physics notes: http://ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2013/lecture-notes/MIT8_04S13_Lec04.pdf)
Best Answer
For question v1:
Since you're integrating the non-negative function $(x^2-l^2)^4$, you shouldn't get zero. Your mistake must be your expression for the antiderivative. Expanding out the integrand is probably a safe way to start.
There's something funky about the supposed answer for $N$. Look at the units, along with the entire expression for $\psi$. The units of $\psi$ should be such that $\left | \psi \right |^2 dx$ is dimensionless. That tells you what the dimension of $N$ ought to be given that $\psi\sim l^4$. The correct solution you gave is off.