Past exam of the physics A course of the University of Cambridge / 2013 / Paper 1 / Section B / B8

The stationary equation isThe second derivative represents kinetic energy; the delta term is an attractive point interaction, with $U$ having dimensions energy times length. A jump in $\psi$ would produce a delta derivative in ${\psi }^{\prime \prime }$, with no matching term in this potential, so $\psi$ is continuous. Integrating through the origin givesFor $E<0$ define $\kappa =\sqrt{-2mE}/\hslash$. Each side has general solution $c_{+}{e}^{\kappa x}+c_{-}{e}^{-\kappa x}$ with independent constants. Square integrability retains $C{e}^{\kappa x}$ on the left and $C{e}^{-\kappa x}$ on the right. The jump requires $\kappa =mU/{\hslash }^2$, and wavefunction normalization givesThere is one bound state for every $U>0$.

For $E>0$ write $k=\sqrt{2mE}/\hslash$. General solutions are independent combinations of $e^{ikx}$ and $e^{-ikx}$ on each side. For incidence from the left choose ${\psi }_L=e^{ikx}+r{e}^{-ikx}$ and ${\psi }_R=t{e}^{ikx}$. Continuity gives $t=1+r$ and the jump gives $2ikr=-2(mU/{\hslash }^2)t$. With $g=mU/({\hslash }^{2}k)$,The incident and transmitted velocities are equal, so squared amplitude ratios are current ratios here. At fixed positive energy, $U\to 0$ gives full transmission and increasing $U$ gives increasing reflection, approaching unity. Even an attractive point well reflects quantum waves; energy is conserved, and the static potential does not irreversibly capture the incident particle into its bound state.