I'm trying to find the normalisation constant $N$ for the following wavefunction:

$$ \psi\left(x\right) = \left\{ \begin{array}{lr} N \left(x^2 - l^2\right)^2 &\: \left|x\right| \le l \\ 0 &\: otherwise \end{array} \right. $$


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

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)

  • $\begingroup$ You did the integral wrong. The phase is there simply because you cannot determine the wavefunction up to an arbitrary phase, only the modulus squared. It's kind of peculiar they didn't include the exponential in the definition of $\psi(x)$ but rather in the definition of $N$. Usually it's the other way around. Try the integral again. $\endgroup$ – TeeJay Aug 11 '14 at 0:09

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.

  • $\begingroup$ Thank you, my integration was indeed wrong, however I now get l^9/2 instead of l^1/2. Why is this? (I have cross checked the integration with Mathematica). $\endgroup$ – alexdavey Aug 11 '14 at 0:55
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    $\begingroup$ There was a mistake in my Mathematica code. It appears that the given answer given is wrong, and it should be l^9/2, as substituting it back into the original expression gives one for only l^9/2. Thank you. $\endgroup$ – alexdavey Aug 11 '14 at 12:57

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