1. Consider the infinite square well on the interval \(-a\) to \(a\), which is defined by a potential $$\begin{equation} V(x) = \begin{cases} 0 & -a < x < a \\ \infty & \text{else} \end{cases}. \end{equation}$$
    1. Calculate the allowed energy levels and corresponding (normalized) wave functions.
    2. Show that the wave functions \(\psi_n(x)\) satisfy the relation $$\begin{equation} \int \psi_m^*(x) \psi_n(x) \d x = \delta_{mn} \end{equation}$$
    3. Suppose a particle has wave function $$\begin{equation} \psi(x) = \begin{cases} A(a^2-x^2) & -a < x < a \\ 0 & \text{else} \end{cases}. \end{equation}$$
      1. Normalize the wave function by calculating \(A\).
      2. If you measure the energy of the particle, what is the probability that you will find it in its \(n\)th energy eigenstate?
  2. Consider a particle in the \(n\)th energy eigenstate state of the infinite square well.
    1. Compute the momentum space eigenfunctions of $$\begin{equation} \psi_n(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right) \end{equation}$$
    2. Plot the probability distribution of a particle in the \(n\)th energy eigenstate in the infinite square well for \(n=1,2\) (using a computer).
    1. We simply compute the Fourier transform $$\begin{align} \psi_n(p) ={}& \int_{-\infty}^\infty \psi_n(x) e^{-ipx/\hbar} \frac{\d x}{\sqrt{2\pi\hbar}} \\ ={}& \sqrt{\frac{2}{a}}\int_0^a e^{-ipx/\hbar}\frac{e^{in\pi x/a} - e^{-in\pi x/a}}{2i} \frac{\d x}{\sqrt{2\pi\hbar}} \\ ={}& \frac{1}{2i\sqrt{\pi a\hbar}}\int_0^a\left(e^{ix(n\pi a/-p/\hbar)} - e^{-ix(n\pi/a+p/\hbar)}\right)\d x\\ ={}& \frac{1}{2i\sqrt{\pi a\hbar}}\left(\frac{e^{ix(n\pi/a-p/\hbar)}}{i(n\pi/a-p/\hbar)}\bigg|_0^a + \frac{e^{-ix(n\pi/a+p/\hbar)}}{i(n\pi/a+p/\hbar)}\bigg|_0^a\right) \\ ={}& \frac{1}{2i\sqrt{\pi a\hbar}}\left(\frac{e^{i(n\pi -pa/\hbar)}-1}{i(n\pi/a-p/\hbar)} + \frac{e^{-i(n\pi+pa/\hbar)}-1}{i(n\pi/a+p/\hbar)}\right) \\ ={}& \frac{1}{2\sqrt{\pi a\hbar}}\left(\frac{1-(-1)^ne^{-ipa/\hbar}}{n\pi/a-p/\hbar} + \frac{1-(-1)^ne^{-ipa/\hbar}}{n\pi/a+p/\hbar}\right) \\ ={}& \frac{1-(-1)^ne^{-ipa/\hbar}}{2\sqrt{\pi a\hbar}}\left(\frac{1}{n\pi/a-p/\hbar} + \frac{1}{n\pi/a+p/\hbar}\right) \\ ={}& \frac{1-(-1)^ne^{-ipa/\hbar}}{2\sqrt{\pi a\hbar}}\frac{2n\pi/a}{(n\pi/a)^2-(p/\hbar)^2} \\ ={}& \frac{1-(-1)^ne^{-ipa/\hbar}}{\sqrt{\pi a\hbar}}\frac{n\pi/a}{(n\pi/a)^2-(p/\hbar)^2} \\ ={}& \frac{1-(-1)^ne^{-ipa/\hbar}}{(n\pi)^2-(pa/\hbar)^2}n\sqrt{\frac{\pi a}{\hbar}} \end{align}$$ Then, we take the magnitude to find $$\begin{align} |\psi_n(p)|^2 ={}& \frac{|1-(-1)^ne^{-ipa/\hbar}|^2}{((n\pi)^2-(pa/\hbar)^2)^2}\frac{\pi a n^2}{\hbar} \\ ={}& \frac{2n^2\pi a/\hbar}{((n\pi)^2 - p^2a^2/\hbar^2)^2}(1-(-1)^n\cos(pa/\hbar)) \end{align}$$