= Solution
The stationary equation is
$$
-\frac{\hbar^2}{2m}\psi''-U\delta(x)\psi=E\psi.
$$
The 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 $\psi''$, with no matching term in this potential, so $\psi$ is continuous. Integrating through the origin gives
$$
\psi'(0^+)-\psi'(0^-)=-\frac{2mU}{\hbar^2}\psi(0).
$$
For $E<0$ define $\kappa=\sqrt{-2mE}/\hbar$. Each side has general solution $c_+e^{\kappa x}+c_-e^{-\kappa x}$ with independent constants. Square integrability retains $Ce^{\kappa x}$ on the left and $Ce^{-\kappa x}$ on the right. The jump requires $\kappa=mU/\hbar^2$, and <wavefunction normalization> gives
$$
\boxed{E_b=-\frac{mU^2}{2\hbar^2},\qquad
\psi_b(x)=\sqrt{\frac{mU}{\hbar^2}}\,e^{-mU|x|/\hbar^2}}.
$$
There is one <bound state> for every $U>0$.
For $E>0$ write $k=\sqrt{2mE}/\hbar$. 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 $2ikr=-2(mU/\hbar^2)t$. With $g=mU/(\hbar^2k)$,
$$
r=\frac{ig}{1-ig},\qquad t=\frac1{1-ig},\qquad
\boxed{R=\frac{g^2}{1+g^2},\quad T=\frac1{1+g^2},\quad R+T=1}.
$$
The incident and transmitted velocities are equal, so squared amplitude ratios are current ratios here. At fixed positive <energy>, $U\to0$ 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>.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2013/paper-1-delta-scattering.png]
{title=Original reflection and transmission curves versus dimensionless attractive-well strength at fixed incident energy}
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