Demonstrate that the first derivative of the wave function is continuous at a point if the potential is finite at that point.
To do this, integrate the energy eigenvalue equation from \(-\epsilon\) to \(\epsilon\) and take the limit that \(\epsilon \rightarrow 0\).
Hint 1: You should apply the fundamental theorem of calculus, which says that \(\int_a^b \frac{df}{dx} \; dx = f(b) - f(a)\).
Hint 2: For a region with a very small width, you can consider the integration region to be a small retangle (similar to the idea behind a Reimann sum).
Show that for a delta function potential of the form:
\[V(x) = \beta \delta(x)\]
this boundary condition on the first derivative of the wavefunction is:
\[\lim_{\epsilon \rightarrow 0} \Bigg(\left. \frac{\partial \psi}{\partial x}\right|_{\epsilon} - \left. \frac{\partial \psi}{\partial x}\right|_{-\epsilon} \Bigg) = \frac{2m \beta}{\hbar^2} \psi(0) \]
that is, the discontinuity in the derivative is proportional to the value of the wavefunction at that point.
Consider a particle of mass \(m\) in a finite potential well: \begin{align*} V(x) = \begin{cases} V_0 & x < -a \\ 0 & -a < x < a \\ V_0 & a < x \\ \end{cases} \end{align*}
Consider the even solutions only.
Normalize the even solutions and use the boundary conditions to help you solve to the \(A\) and \(D\) parameters in terms of \(q\) and \(k\).