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

Question 5 is a hard question, so we will deal with it in parts. In the third part we computes the expectation of momentum, it is similar to the previous calculation by not quite.

$\begin{eqnarray*} \langle p \rangle &=& \int\limits_{0}^{L}{\psi(x, t)^* i\hbar\frac{\partial}{\partial x} \psi(x, t) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}})^* i\hbar\frac{\partial}{\partial x} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{\frac{i E_n t}{\hbar}}) i\hbar\frac{\partial}{\partial x} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})e^{-\frac{i E_n t}{\hbar}}) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) i\hbar\frac{\partial}{\partial x} (\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) dx} \\ &=& \int\limits_{0}^{L}{(\sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})) i\hbar\frac{n\pi}{L} (\sqrt{\frac{2}{L}}\cos(\frac{n\pi x}{L})) dx} \\ &=& \frac{2i\hbar n\pi}{L^2}\int\limits_{0}^{L}{\sin(\frac{n\pi x}{L})\cos(\frac{n\pi x}{L}) dx} \\ &=& \frac{2i\hbar n\pi}{L^2}\int\limits_{0}^{L}{\frac{1}{2}\sin(\frac{2n\pi x}{L}) dx} \\ &=& \frac{i\hbar n\pi}{L^2}\int\limits_{0}^{L}{\sin(\frac{2n\pi x}{L}) dx} \\ &=& \frac{i\hbar n\pi}{L^2}(\frac{L}{2n\pi})\cos(\frac{2n\pi x}{L})|_0^{L} \\ &=& 0 \end{eqnarray*}$

Therefore the expectation of the momentum operator is $0$, that of course doesn't mean the particle is not moving, it merely means the average velocity is 0, that is, it is not drifting left or right.