Problems with a delta-like potential may be solved by applying a Fourier transform.
A delta-like potential V(x) = \alpha \, \delta(x - x_0) has a Fourier transform:
<br />
\tilde{V}(k) = \alpha \, e^{-i k \, x_0}<br />
The Schroedinger equation in momentum space is:
<br />
\frac{\hbar^2 \, k^2}{2 \, m} a(k) + \int_{-\infty}^{\infty} \frac{dk'}{2 \pi} \, \tilde{V}(k - k') a(k') = E \, a(k)<br />
where a(k) = \int_{-\infty}^{\infty} \psi(x) e^{-i k x} \, dx is the wave function Fourier transform. The inverse Fourier transform is:
<br />
\psi(x) = \int_{-\infty}^{\infty} \frac{dk}{2 \pi} a(k) e^{i k \, x}<br />
Substituting the Fourier transform of the potential in our Schrodinger equation, and using the definition of an inverse Fourier transform of the wave function, we get:
<br />
\left(E - \frac{\hbar^2 \, k^2}{2 \, m} \right) \, a(k) = \alpha \, \psi(x_0) \, e^{-i k \, x_0}<br />
Substituting this into the definition for \psi(x_0), we get the self-consistency condition:
<br />
\psi(x_0) = \alpha \, \psi(x_0) \, \int_{-\infty}^{\infty} \frac{dk}{2\pi} \frac{1}{E - \frac{\hbar^2 k^2}{2 \, m}}<br />
If we assume that \psi(x_0) \neq 0 (otherwise a(k) \equiv 0, \forall k, which is a trivial solution), and we introduces the energy parameter \epsilon (with a dimension [k]2):
<br />
\epsilon \equiv \frac{2 m \, E}{\hbar^2}<br />
<br />
1 = -\frac{2 m \, \alpha}{\hbar^2} \, \int_{-\infty}^{\infty} \frac{dk}{2\pi} \frac{1}{k^2 - \epsilon}<br />
This is an implicit equation for \epsilon. Can you find its solutions?
