Exploring Quantum Physics - Week 3 Extra Credit Question 5 - Part 4

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

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

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

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

Therefore the expectation of the momentum squared operator is $\frac{n^2\hbar^2\pi^2}{L^2}$