Exploring Quantum Physics - Week 3 Extra Credits Question 5 - Part 1

Question 5 is a hard question, so we will deal with it in parts. Let's first compute the expectation of the position operator on the wavefunctions.

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

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 $dy = \frac{2 n \pi}{L} dx$.

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

Therefore the expectation of the position operator is $\frac{L}{2}$, not a surprising result.