Consider a particle with mass \(m\) and energy \(E\) incident from the left on a square potential barrier with height \(V_0>0\): \begin{equation*} V(x) = \begin{cases} 0 & x<0 \\ V_0 & 0<x<a\\ 0 & a<x \end{cases} \end{equation*}
This is an example of an unbounded system, so there is no condition on the energy eigenvalue. There are two cases, \(E>V_0\) and \(E<V_0\). Consider only \(E>V_0\).
Set up a wave incident from the left, and one transmitted to the right, so the total wave function is:
\begin{equation*} \psi(x) = \begin{cases} e^{ik_1x} + Ae^{-ik_1x} & x<0 \\ Ce^{ik_3x} + De^{-ik_3x} & 0<x<a \\ Be^{ik_2x} & a<x\\ \end{cases} \end{equation*}
Use the energy eigenvalue equation to solve for the values \(k_1\), \(k_2\), \(k_3\) in terms of \(E\) and \(V_0\).
What are the boundary conditions that establish the relationship among the coefficients \(A\), \(B\), \(C\), and \(D\)?
The probability to observe the particle reflected is \begin{equation*} r \equiv \left|A\right|^2 \end{equation*} (remember we measure probabilities, and not amplitudes). Find \(r\). Also find the probability of transmission
\begin{equation*} t \equiv \left|B\right|^2 \end{equation*}
To make the algebra easy, you can assume that \(E=\frac{4}{3}, \,V_0=\frac{1}{2m}(\frac{2\pi\hbar}{a})^2\).
Show that \(r+t=1\).
(2 points each)
Consider a system of three unequal masses in a row between two fixed walls. The walls and masses are all connected by unequal springs.