# This made me happy :)

Given the wave function

$\Psi(x) = \frac{N}{x^2 + a^2}$

we are to find the values of $$n \in \left \{ 0 \right \} \bigcup \mathbb{N}$$ such that $$\left \langle x^n \right \rangle$$ has a meaningful physical value. And by $$\left \langle x^n \right \rangle$$ we mean the following integral

$\int_{-\infty}^{+\infty} \Psi^{*} x^{n} \Psi \; \mathrm{d}x$

First, normalization will result in

$\int_{-\infty}^{+\infty} \Psi^{*} \Psi \; \mathrm{d}x = N^{2} \lim _{t\rightarrow \infty} \left (\frac{\frac{ax}{a^2 + x^2} + \arctan \left ( \frac{x}{a} \right )}{2a^3} \right )_{-t}^{t} = \frac{N^{2}\pi}{2a^{3}} = 1 \;\; \therefore \;\; N = \sqrt{\frac{2a^{3}}{\pi}}$

We can find $$\left \langle x^n \right \rangle$$ taking this integral to the complex plane, and observing that if we allow the function $$\Psi$$ to take complex values and we define

$f(z) = \frac{z^{n}}{(z^{2} + a^{2})^{2}}$

Then the residue of $$f$$ at $$ia$$ will be

$\mathrm{Res}(f, ia) = \lim_{z\rightarrow ia} \frac{\mathrm{d} }{\mathrm{d} z} f(z) = (ia)^{n-1} \; \frac{1-n}{4a^{2}}$

And according to a suitable theorem, we are able to write

$\int_{-\infty}^{+\infty} \Psi^{*} x^{n} \Psi \; \mathrm{d}x = 2 \pi i \times N \times \mathrm{Res}(f, ia) = \frac{\sqrt{2\pi}}{2} i^{n} (1-n) a^{n-\frac{3}{2}}$

And then we are allowed to say that only when $$n$$ takes the values $$n = 2k$$ with $$k \in \left \{ 0 \right \} \bigcup \mathbb{N}$$ the integral has a physical meaning.

Note by Lucas Tell Marchi
3 years, 6 months ago

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