Exploring Quantum Physics - Week 3 Extra Credit Question 5 - Part 2

Question 5 is a hard question, so we will deal with it in parts. In the second part we computes the expectation of position squared. This is almost identical to the previous post except a single difference in the power of $x$.

$\begin{eqnarray*} \langle x^2 \rangle &=& \int\limits_{0}^{L}{\psi(x, t)^* x^2 \psi(x, t) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}})^* x^2 (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{\frac{i E_n t}{\hbar}}) x^2 (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) x^2 (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& \frac{2}{L}\int\limits_{0}^{L}{x^2 \sin^2(\frac{n\pi x}{L}) dx} \\ &=& \frac{2}{L}\int\limits_{0}^{L}{x^2 \frac{1 - \cos(\frac{2n\pi x}{L})}{2} dx} \\ &=& \frac{1}{L}\int\limits_{0}^{L}{x^2 (1 - \cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{1}{L}\int\limits_{0}^{L}{x^2dx} - \frac{1}{L}\int\limits_{0}^{L}{x^2\cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{1}{L}(\frac{x^3}{3})|_{0}^{L} - \frac{1}{L}\int\limits_{0}^{L}{x^2\cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{L^2}{3} - \frac{1}{L}\int\limits_{0}^{L}{x^2\cos(\frac{2n\pi x}{L})) dx} \\ \end{eqnarray*}$

The remaining integral seems complicated, but we have done with the preparation in the previous post, to use that, we let $y = \frac{2 n \pi x}{L}$, when $x = 0$, we have $y = 0$, when $x = L$, we have $y = 2 n \pi$, lastly $dy = \frac{2 n \pi}{L} dx$.

$\begin{eqnarray*} \langle x^2 \rangle &=& \frac{L^2}{3} - \frac{1}{L}\int\limits_{0}^{L}{x^2\cos(\frac{2n\pi x}{L})) dx} \\ &=& \frac{L^2}{3} - \frac{1}{L}\int\limits_{0}^{2n\pi}{(\frac{L y}{2 n \pi})^2\cos(y) \frac{L}{2 n \pi} dy} \\ &=& \frac{L^2}{3} - \frac{1}{L}(\frac{L}{2 n \pi})^3\int\limits_{0}^{2n\pi}{y^2 \cos(y) dy} \\ &=& \frac{L^2}{3} - \frac{1}{L}(\frac{L}{2 n \pi})^3(t^2 \sin t + 2t\cos t - 2 \sin t)|_0^{2n\pi} \\ &=& \frac{L^2}{3} - \frac{1}{L}(\frac{L}{2 n \pi})^3(4n\pi) \\ &=& L^2(\frac{1}{3} - \frac{2}{(2 n \pi)^2}) \end{eqnarray*}$

Therefore the expectation of the position squared operator is $L^2(\frac{1}{3} - \frac{2}{(2 n \pi)^2})$