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I am having problems finding the normalisation constant $N$. I have tried this so far

use the substitution $x=a tan(u)$ so $dx=a sec^2(u)du$, so

$\displaystyle 1=\int_{-\infty}^{\infty}N^2 \frac{e^{i2px/\hbar}}{x^2+a^2}dx$

$= \displaystyle \int_{-\infty}^{\infty}N^2 \frac{e^{i2patan(u)/\hbar}}{a^2tan^2(u)+a^2}asec^2(u) du$

$= \displaystyle \int_{-\infty}^{\infty}N^2 \frac{e^{i2patan(u)/\hbar}}{a} du$

and from here I cannot see where to proceed

I thought it might be useful to use that sine is an odd function so we would have

$\displaystyle \int_{-\infty}^{\infty}N^2 \frac{cos(2px/\hbar)}{{x^2+a^2}}dx$

and using the same substitution doesnt seem to yield anything

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    $\begingroup$ It is really long since the last time I did quantum theory, but I am pretty sure you have to normalise |Ψ|², not Ψ $\endgroup$
    – Shady_arc
    May 18, 2014 at 12:18
  • $\begingroup$ you are right, still think i have a similiar problem $\endgroup$
    – Trajan
    May 18, 2014 at 12:39

2 Answers 2

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$|\phi|^2 = N^2e^{ipx/\hbar}e^{-ipx/\hbar}\frac1{\sqrt{x^2+a^2}^2} = N^2\frac1{x^2+a^2}$

Now we got to some familiar tabled integrals; $$\int_{-\infty}^{\infty}{\frac{N^2}{x^2+a^2}} = \left[N^2\frac1a\arctan \frac xa\right]_{-\infty}^{+\infty}$$

$arctan(x)$ changes from $-\frac{\pi}2$ to $\frac{\pi}2$, so the result is $N^2\frac\pi a$ if I have not got it wrong from the very beginning. Then you just equal that to $1$.

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As mentioned by user2388, the normalization condition reads $$ 1 = \int\limits_{-\infty}^{+\infty} |\psi(x)|^ 2 dx $$

From there the computation is standard:

$$ 1 = |N|^ 2 \int\limits_{-\infty}^{+\infty} \dfrac{1}{x^2+a^2} dx $$

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  • $\begingroup$ I need a break! $\endgroup$
    – Trajan
    May 18, 2014 at 12:57

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