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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
2 years, 6 months ago

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