Consider a particle with mass \(m\) in a one dimension potential: \begin{align*} V(x) &= -\gamma\delta(x) \end{align*}
where \(\gamma\) is positive.
For this potential, there is only one bound state. Solve for the energy eigenstate and value of the energy of the bound state.
Hint: There are actually two approaches you can use to solve this. The first approach is to work with the \(\delta\) function directly and use the appropriate boundary conditions for an infinite potential. The second approach is to start with the solution to a finite square well and then turn it into a delta function well by taking the limit that the width of the well goes to zero.
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*}
The even solutions are expressed as \begin{align*} \phi(x) = \begin{cases} Ae^{qx} & x < -a \\ D\cos kx & -a < x < a \\ Ae^{-qx} & a < x \\ \end{cases} \end{align*}
Normalize the even solutions and use the boundary conditions to help you solve to the \(A\) and \(D\) parameters in terms of \(q\) and \(k\).