Introduction: Finding the One Bound State
We derive the bound state of the delta potential step by step, demonstrating the general technique for handling singular potentials.
Solution Method
For \(x \neq 0\): \(-\frac{\hbar^2}{2m}\psi'' = E\psi\) with \(E < 0\).
Let \(\kappa = \sqrt{-2mE}/\hbar > 0\). Then:
\[\psi(x) = \begin{cases} Ae^{\kappa x} & x < 0 \\ Be^{-\kappa x} & x > 0 \end{cases}\]Applying Boundary Conditions
Continuity at x = 0: \(A = B\)
Discontinuity of derivative:
\[-\kappa B - \kappa A = -\frac{2m\alpha}{\hbar^2}A\] \[-2\kappa A = -\frac{2m\alpha}{\hbar^2}A\] \[\kappa = \frac{m\alpha}{\hbar^2}\]Energy and Wavefunction
\[E = -\frac{\hbar^2\kappa^2}{2m} = -\frac{m\alpha^2}{2\hbar^2}\]Normalized wavefunction:
\[\psi(x) = \frac{\sqrt{m\alpha}}{\hbar}e^{-m\alpha|x|/\hbar^2}\]The Quantum Connection
The delta potential proves that any attractive potential in 1D has at least one bound state (in 3D this isn't true—attraction must be strong enough). The exponential decay \(e^{-\kappa|x|}\) is characteristic: the bound state is localized near the potential with decay length \(1/\kappa = \hbar^2/m\alpha\).