I'm stuck on this question for a physics assignment
- A particle in one dimension is prepared in a quantum state given by the wave function $$\psi(x) = \left(\frac{\beta}{\pi}\right)^{1/4}\exp\left(-\beta x^2/2\right),\quad -\infty <x<\infty,$$ where $\beta> 0 $.
[Given: $\int_{-\infty}^\infty dy\, e^{-y^2} = \sqrt{\pi}$ and $\int_{-\infty}^\infty dy\,y^2e^{-y^2} = \frac{\sqrt{\pi}}{2}$]
I'm looking for the expected value of the position squared for this waveform which is given by the $\int_{-\infty}^\infty x^2\psi^2 dx$ and I've rearranged it to get this $$\int_{-\infty}^\infty x^2\left(\frac{\beta}{\pi}\right)^{\frac{1}{2}} e^{-\beta x^2}$$ but can't see how to progress. Does anybody have any tips/solutions?
So we have $$ \sqrt{\frac{\beta}{\pi}}\int_{-\infty}^\infty x^2e^{-\beta x^2}\mathrm dx\\ \stackrel{\text{by parts and u substitution}}{=} \sqrt{\frac{\beta}{\pi}}xe^{-\beta x^2}\vert_{-\infty}^\infty+\frac{1}{2\beta}\int_{-\infty}^\infty e^{-\beta x^2}\mathrm dx\\ =\frac{1}{2\beta}\int_{-\infty}^\infty e^{-\beta x^2}\mathrm dx $$ can you finish from here using the fact about the gaussian integral? The substitution $u=\sqrt{\beta}x$ looks pretty appealing...